Develop Reference

How to find an sentence inside of an array of words and join the words

Follow the example:
const arrayOfSentences = ['fallen little short', 'hold the nominal office of governess', 'shadow of authority', 'the real evils indeed']
const arrayOfwords = ['who', 'had','fallen', 'little','short','of','a','mother','in','affection','hold','the','nominal','office','of','governess','the','mildness','of','her','temper','had','hardly','allowed','her','to','impose','any','restraint;','and','the','shadow','of','authority','being','now','long','passed','away,','they','had']
the output should be the array of words but with the sentences merged, for example ("fallen", "little", "short" -> "fallen little short"):
const mergedArray = ['who', 'had','fallen little short','of','a','mother','in','affection','hold the nominal office of
governess','the','mildness','of','her','temper','had','hardly','allowed','her','to','impose','any','restraint;','and','the','shadow of authority','being','now','long','passed','away,','they','had']

Here's a sneaky way to do that:
const arrayOfSentences = ['fallen little short', 'hold the nominal office of governess', 'shadow of authority', 'the real evils indeed']
const arrayOfwords = ['who', 'had','fallen', 'little','short','of','a','mother','in','affection','hold','the','nominal','office','of','governess','the','mildness','of','her','temper','had','hardly','allowed','her','to','impose','any','restraint;','and','the','shadow','of','authority','being','now','long','passed','away,','they','had']
let sentence = arrayOfwords.join(' ');
arrayOfSentences.forEach(s => {
sentence = sentence.replace(s, s.replace(/\s/g, '|'));
});
const mergedArray = sentence.split(' ').map(s => s.replace(/\|/g, ' '));
console.log(mergedArray);
Basically what it does is to make the whole arrayOfwords a sentence.
Then loop through the arrayOfSentences to find the appearances in that sentence and replace them to some placeholders.
Then re-split the sentence to an array, and replace the placeholders as what it was in arrayOfSentences.

Replace regular expression matches with array of values

I have a regular expression to find text with ??? inside a string.
const paragraph = 'This ??? is ??? and ???. Have you seen the ????';
const regex = /(\?\?\?)/g;
const found = paragraph.match(regex);
console.log(found);
Is there a way to replace every match with a value from an array in order?
E.g. with an array of ['cat', 'cool', 'happy', 'dog'], I want the result to be 'This cat is cool and happy. Have you seen the dog?'.
I saw String.prototype.replace() but that will replace every value.

Use a replacer function that shifts from the array of replacement strings (shift removes and returns the item at the 0th index):
const paragraph = 'This ??? is ??? and ???. Have you seen the ????';
const regex = /(\?\?\?)/g;
const replacements = ['cat', 'cool', 'happy', 'dog'];
const found = paragraph.replace(regex, () => replacements.shift());
console.log(found);
(if there are not enough items in the array to replace all, the rest of the ???s will be replaced by undefined)

How can I split commas and periods from words inside of string using split?

I am trying to change specific word in a string with something else. For example, I want to change 'John' in let name = 'Hi, my name is John.'; to 'Jack'.
I know how to split a string by words or characters. I also know how to remove commas, periods, and other symbols in a string. However, if I split the given string with a separator (" "), I will have 'John.' which I do not want. (I know I can switch 'John.' with 'Jack.' but assume that I have an key and value pairs in an object and I am using the values which are names {Father: Jack, Mother: Susan, ...}
I don't know how to separate a string word by word including commas and periods.
For example, if I was given an input which is a string:
'Hi, my name is John.'
I want to split the input as below:
['Hi', ',', 'my', 'name', 'is', 'John', '.']
Does anyone know how to do it?
Below is the challenge I am working on.
Create a function censor that accepts no arguments. censor will return a function that will accept either two strings, or one string. When two strings are given, the returned function will hold onto the two strings as a pair, for future use. When one string is given, the returned function will return the same string, except all instances of a first string (of a saved pair) will be replaced with the second string (of a saved pair).
//Your code here
const changeScene = censor();
changeScene('dogs', 'cats');
changeScene('quick', 'slow');
console.log(changeScene('The quick, brown fox jumps over the lazy dogs.')); // should log: 'The slow, brown fox jumps over the lazy cats.'

I think your real question is "How do I replace a substring with another string?"
Checkout the replace method:
let inputString = "Hi, my name is John.";
let switch1 = ["John", "Jack"];
let switched = inputString.replace(switch1[0], switch1[1]);
console.log(switched); // Hi, my name is Jack.
UPDATE: If you want to get ALL occurrences (g), be case insensitive (i), and use boundaries so that it isn't a word within another word (\\b), you can use RegExp:
let inputString = "I'm John, or johnny, but I prefer john.";
let switch1 = ["John", "Jack"];
let re = new RegExp(`\\b${switch1[0]}\\b`, 'gi');
console.log(inputString.replace(re, switch1[1])); // I'm Jack, or johnny, but I prefer Jack.

You can Try This ...
var string = 'Hi, my name is John.';
//var arr = string.split(/,|\.| /);
var arr = string.split(/([,.\s])/);
console.log(arr);

Using 'Hi, my name is John.'.split(/[,. ]/); will do the job. It will split commas and periods and spaces.
Edit: For those who want to keep the comma and period, here is my wildly inefficient method.
var str = 'Hi, my name is John.'
str = str.replace('.', 'period');
str = str.replace(',', 'comma');
str = str.split(/[,. ]/);
for (var i = 0; i < str.length; i++) {
if (str[i].indexOf('period') > -1) {
str[i] = str[i].replace('period', '');
str.splice(i+1, 0, ".");
} else if (str[i].indexOf('comma') > -1) {
str[i] = str[i].replace('comma', '');
str.splice(i+1, 0, ",");
}
}
console.log(str);

Why is this regex matching also words within a non-capturing group?

I have this string (notice the multi-line syntax):
var str = ` Number One: Get this
Number Two: And this`;
And I want a regex that returns (with match):
[str, 'Get this', 'And this']
So I tried str.match(/Number (?:One|Two): (.*)/g);, but that's returning:
["Number One: Get this", "Number Two: And this"]
There can be any whitespace/line-breaks before any "Number" word.
Why doesn't it return only what is inside of the capturing group? Am I misundersating something? And how can I achieve the desired result?

Per the MDN documentation for String.match:
If the regular expression includes the g flag, the method returns an Array containing all matched substrings rather than match objects. Captured groups are not returned. If there were no matches, the method returns null.
(emphasis mine).
So, what you want is not possible.
The same page adds:
if you want to obtain capture groups and the global flag is set, you need to use RegExp.exec() instead.
so if you're willing to give on using match, you can write your own function that repeatedly applies the regex, gets the captured substrings, and builds an array.
Or, for your specific case, you could write something like this:
var these = str.split(/(?:^|\n)\s*Number (?:One|Two): /);
these[0] = str;

Replace and store the result in a new string, like this:
var str = ` Number One: Get this
Number Two: And this`;
var output = str.replace(/Number (?:One|Two): (.*)/g, "$1");
console.log(output);
which outputs:
Get this
And this
If you want the match array like you requested, you can try this:
var getMatch = function(string, split, regex) {
var match = string.replace(regex, "$1" + split);
match = match.split(split);
match = match.reverse();
match.push(string);
match = match.reverse();
match.pop();
return match;
}
var str = ` Number One: Get this
Number Two: And this`;
var regex = /Number (?:One|Two): (.*)/g;
var match = getMatch(str, "#!SPLIT!#", regex);
console.log(match);
which displays the array as desired:
[ ' Number One: Get this\n Number Two: And this',
' Get this',
'\n And this' ]
Where split (here #!SPLIT!#) should be a unique string to split the matches. Note that this only works for single groups. For multi groups add a variable indicating the number of groups and add a for loop constructing "$1 $2 $3 $4 ..." + split.

Try
var str = " Number One: Get this\
Number Two: And this";
// `/\w+\s+\w+(?=\s|$)/g` match one or more alphanumeric characters ,
// followed by one or more space characters ,
// followed by one or more alphanumeric characters ,
// if following space or end of input , set `g` flag
// return `res` array `["Get this", "And this"]`
var res = str.match(/\w+\s+\w+(?=\s|$)/g);
document.write(JSON.stringify(res));

Replace multiple strings with multiple other strings

I'm trying to replace multiple words in a string with multiple other words. The string is "I have a cat, a dog, and a goat."
However, this does not produce "I have a dog, a goat, and a cat", but instead it produces "I have a cat, a cat, and a cat". Is it possible to replace multiple strings with multiple other strings at the same time in JavaScript, so that the correct result will be produced?
var str = "I have a cat, a dog, and a goat.";
str = str.replace(/cat/gi, "dog");
str = str.replace(/dog/gi, "goat");
str = str.replace(/goat/gi, "cat");
//this produces "I have a cat, a cat, and a cat"
//but I wanted to produce the string "I have a dog, a goat, and a cat".

Specific Solution
You can use a function to replace each one.
var str = "I have a cat, a dog, and a goat.";
var mapObj = {
cat:"dog",
dog:"goat",
goat:"cat"
};
str = str.replace(/cat|dog|goat/gi, function(matched){
return mapObj[matched];
});
jsfiddle example
Generalizing it
If you want to dynamically maintain the regex and just add future exchanges to the map, you can do this
new RegExp(Object.keys(mapObj).join("|"),"gi");
to generate the regex. So then it would look like this
var mapObj = {cat:"dog",dog:"goat",goat:"cat"};
var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
str = str.replace(re, function(matched){
return mapObj[matched];
});
And to add or change any more replacements you could just edit the map. 
fiddle with dynamic regex
Making it Reusable
If you want this to be a general pattern you could pull this out to a function like this
function replaceAll(str,mapObj){
var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
return str.replace(re, function(matched){
return mapObj[matched.toLowerCase()];
});
}
So then you could just pass the str and a map of the replacements you want to the function and it would return the transformed string.
fiddle with function
To ensure Object.keys works in older browsers, add a polyfill eg from MDN or Es5.

As an answer to:
looking for an up-to-date answer
If you are using "words" as in your current example, you might extend the answer of Ben McCormick using a non capture group and add word boundaries \b at the left and at the right to prevent partial matches.
\b(?:cathy|cat|catch)\b
\b A word boundary to prevent a partial match
(?: Non capture group
cathy|cat|catch match one of the alternatives
) Close non capture group
\b A word boundary to prevent a partial match
Example for the original question:
let str = "I have a cat, a dog, and a goat.";
const mapObj = {
cat: "dog",
dog: "goat",
goat: "cat"
};
str = str.replace(/\b(?:cat|dog|goat)\b/gi, matched => mapObj[matched]);
console.log(str);
Example for the example in the comments that not seems to be working well:
let str = "I have a cat, a catch, and a cathy.";
const mapObj = {
cathy: "cat",
cat: "catch",
catch: "cathy"
};
str = str.replace(/\b(?:cathy|cat|catch)\b/gi, matched => mapObj[matched]);
console.log(str);

Use numbered items to prevent replacing again.
eg
let str = "I have a %1, a %2, and a %3";
let pets = ["dog","cat", "goat"];
then
str.replace(/%(\d+)/g, (_, n) => pets[+n-1])
How it works:-
%\d+ finds the numbers which come after a %. The brackets capture the number.
This number (as a string) is the 2nd parameter, n, to the lambda function.
The +n-1 converts the string to the number then 1 is subtracted to index the pets array.
The %number is then replaced with the string at the array index.
The /g causes the lambda function to be called repeatedly with each number which is then replaced with a string from the array.
In modern JavaScript:-
replace_n=(str,...ns)=>str.replace(/%(\d+)/g,(_,n)=>ns[n-1])

This may not meet your exact need in this instance, but I've found this a useful way to replace multiple parameters in strings, as a general solution. It will replace all instances of the parameters, no matter how many times they are referenced:
String.prototype.fmt = function (hash) {
var string = this, key; for (key in hash) string = string.replace(new RegExp('\\{' + key + '\\}', 'gm'), hash[key]); return string
}
You would invoke it as follows:
var person = '{title} {first} {last}'.fmt({ title: 'Agent', first: 'Jack', last: 'Bauer' });
// person = 'Agent Jack Bauer'

using Array.prototype.reduce():
UPDATED (much better) answer (using object):
This function will replace all occurrences and is case insensitive
/**
* Replaces all occurrences of words in a sentence with new words.
* #function
* #param {string} sentence - The sentence to modify.
* #param {Object} wordsToReplace - An object containing words to be replaced as the keys and their replacements as the values.
* #returns {string} - The modified sentence.
*/
function replaceAll(sentence, wordsToReplace) {
return Object.keys(wordsToReplace).reduce(
(f, s, i) =>
`${f}`.replace(new RegExp(s, 'ig'), wordsToReplace[s]),
sentence
)
}
const americanEnglish = 'I popped the trunk of the car in a hurry and in a hurry I popped the trunk of the car'
const wordsToReplace = {
'popped': 'opened',
'trunk': 'boot',
'car': 'vehicle',
'hurry': 'rush'
}
const britishEnglish = replaceAll(americanEnglish, wordsToReplace)
console.log(britishEnglish)
// I opened the boot of the vehicle in a rush and in a rush I opened the boot of the vehicle
ORIGINAL answer (using array of objects):
const arrayOfObjects = [
{ plants: 'men' },
{ smart:'dumb' },
{ peace: 'war' }
]
const sentence = 'plants are smart'
arrayOfObjects.reduce(
(f, s) => `${f}`.replace(Object.keys(s)[0], s[Object.keys(s)[0]]), sentence
)
// as a reusable function
const replaceManyStr = (obj, sentence) => obj.reduce((f, s) => `${f}`.replace(Object.keys(s)[0], s[Object.keys(s)[0]]), sentence)
const result = replaceManyStr(arrayOfObjects , sentence1)
Example
// ///////////// 1. replacing using reduce and objects
// arrayOfObjects.reduce((f, s) => `${f}`.replace(Object.keys(s)[0], s[Object.keys(s)[0]]), sentence)
// replaces the key in object with its value if found in the sentence
// doesn't break if words aren't found
// Example
const arrayOfObjects = [
{ plants: 'men' },
{ smart:'dumb' },
{ peace: 'war' }
]
const sentence1 = 'plants are smart'
const result1 = arrayOfObjects.reduce((f, s) => `${f}`.replace(Object.keys(s)[0], s[Object.keys(s)[0]]), sentence1)
console.log(result1)
// result1:
// men are dumb
// Extra: string insertion python style with an array of words and indexes
// usage
// arrayOfWords.reduce((f, s, i) => `${f}`.replace(`{${i}}`, s), sentence)
// where arrayOfWords has words you want to insert in sentence
// Example
// replaces as many words in the sentence as are defined in the arrayOfWords
// use python type {0}, {1} etc notation
// five to replace
const sentence2 = '{0} is {1} and {2} are {3} every {5}'
// but four in array? doesn't break
const words2 = ['man','dumb','plants','smart']
// what happens ?
const result2 = words2.reduce((f, s, i) => `${f}`.replace(`{${i}}`, s), sentence2)
console.log(result2)
// result2:
// man is dumb and plants are smart every {5}
// replaces as many words as are defined in the array
// three to replace
const sentence3 = '{0} is {1} and {2}'
// but five in array
const words3 = ['man','dumb','plant','smart']
// what happens ? doesn't break
const result3 = words3.reduce((f, s, i) => `${f}`.replace(`{${i}}`, s), sentence3)
console.log(result3)
// result3:
// man is dumb and plants

This worked for me:
String.prototype.replaceAll = function(search, replacement) {
var target = this;
return target.replace(new RegExp(search, 'g'), replacement);
};
function replaceAll(str, map){
for(key in map){
str = str.replaceAll(key, map[key]);
}
return str;
}
//testing...
var str = "bat, ball, cat";
var map = {
'bat' : 'foo',
'ball' : 'boo',
'cat' : 'bar'
};
var new = replaceAll(str, map);
//result: "foo, boo, bar"

With my replace-once package, you could do the following:
const replaceOnce = require('replace-once')
var str = 'I have a cat, a dog, and a goat.'
var find = ['cat', 'dog', 'goat']
var replace = ['dog', 'goat', 'cat']
replaceOnce(str, find, replace, 'gi')
//=> 'I have a dog, a goat, and a cat.'

NOTE!
If you are using dynamically provided mapping, NONE of the solutions here are sufficient enough!
In this case, there are two ways to solve this problem, (1) using split-join technique, (2) using Regex with special character escaping technique.
This one is a split-join technique, which is much more faster than the other one (at least 50% faster):
var str = "I have {abc} a c|at, a d(og, and a g[oat] {1} {7} {11."
var mapObj = {
'c|at': "d(og",
'd(og': "g[oat",
'g[oat]': "c|at",
};
var entries = Object.entries(mapObj);
console.log(
entries
.reduce(
// Replace all the occurrences of the keys in the text into an index placholder using split-join
(_str, [key], i) => _str.split(key).join(`{${i}}`),
// Manipulate all exisitng index placeholder -like formats, in order to prevent confusion
str.replace(/\{(?=\d+\})/g, '{-')
)
// Replace all index placeholders to the desired replacement values
.replace(/\{(\d+)\}/g, (_,i) => entries[i][1])
// Undo the manipulation of index placeholder -like formats
.replace(/\{-(?=\d+\})/g, '{')
);
This one, is the Regex special character escaping technique, which also works, but much slower:
var str = "I have a c|at, a d(og, and a g[oat]."
var mapObj = {
'c|at': "d(og",
'd(og': "g[oat",
'g[oat]': "c|at",
};
console.log(
str.replace(
new RegExp(
// Convert the object to array of keys
Object.keys(mapObj)
// Escape any special characters in the search key
.map(key => key.replace(/[-[\]{}()*+?.,\\^$|#\s]/g, '\\$&'))
// Create the Regex pattern
.join('|'),
// Additional flags can be used. Like `i` - case-insensitive search
'g'
),
// For each key found, replace with the appropriate value
match => mapObj[match]
)
);
The advantage of the latter, is that it can also work with case-insensitive search.

Just in case someone is wondering why the original poster's solution is not working:
var str = "I have a cat, a dog, and a goat.";
str = str.replace(/cat/gi, "dog");
// now str = "I have a dog, a dog, and a goat."
str = str.replace(/dog/gi, "goat");
// now str = "I have a goat, a goat, and a goat."
str = str.replace(/goat/gi, "cat");
// now str = "I have a cat, a cat, and a cat."

one possible solution could be by using the mapper expression function.
const regex = /(?:cat|dog|goat)/gmi;
const str = `I have a cat, a dog, and a goat.`;
let mapper = (key) => {
switch (key) {
case "cat":
return "dog"
case "dog":
return "goat";
case "goat":
return "cat"
}
}
let result = str.replace(regex, mapper);
console.log('Substitution result: ', result);
//Substitution result1: I have a dog, a goat, and a cat.

You can find and replace string using delimiters.
var obj = {
'firstname': 'John',
'lastname': 'Doe'
}
var text = "Hello {firstname}, Your firstname is {firstname} and lastname is {lastname}"
console.log(mutliStringReplace(obj,text))
function mutliStringReplace(object, string) {
var val = string
var entries = Object.entries(object);
entries.forEach((para)=> {
var find = '{' + para[0] + '}'
var regExp = new RegExp(find,'g')
val = val.replace(regExp, para[1])
})
return val;
}

This solution can be adapted to only replace whole words - so for example, "catch", "ducat" or "locator" wouldn't be found when searching for "cat". This can be done by using negative lookbehind (?<!\w) and negative lookahead (?!\w) on word characters before and after each word in the regular expression:
(?<!\w)(cathy|cat|ducat|locator|catch)(?!\w)
JSFiddle demo: http://jsfiddle.net/mfkv9r8g/1/

Try my solution. feel free to improve
function multiReplace(strings, regex, replaces) {
return str.replace(regex, function(x) {
// check with replaces key to prevent error, if false it will return original value
return Object.keys(replaces).includes(x) ? replaces[x] : x;
});
}
var str = "I have a Cat, a dog, and a goat.";
//(json) use value to replace the key
var replaces = {
'Cat': 'dog',
'dog': 'goat',
'goat': 'cat',
}
console.log(multiReplace(str, /Cat|dog|goat/g, replaces))

user regular function to define the pattern to replace and then use replace function to work on input string,
var i = new RegExp('"{','g'),
j = new RegExp('}"','g'),
k = data.replace(i,'{').replace(j,'}');

I have modified Ben McCormick's answer to work with your new test case.
I simply added word boundaries to the regular expression:
/\b(cathy|cat|catch)\b/gi
"Run code snippet" to see the results below:
var str = "I have a cat, a catch, and a cathy.";
var mapObj = {
cathy:"cat",
cat:"catch",
catch:"cathy"
};
str = str.replace(/\b(cathy|cat|catch)\b/gi, function(matched){
return mapObj[matched];
});
console.log(str);

const str = 'Thanks for contributing an answer to Stack Overflow!'
const substr = ['for', 'to']
function boldString(str, substr) {
let boldStr
boldStr = str
substr.map(e => {
const strRegExp = new RegExp(e, 'g');
boldStr= boldStr.replace(strRegExp, `<strong>${e}</strong>`);
}
)
return boldStr
}

String.prototype.replaceSome = function() {
var replaceWith = Array.prototype.pop.apply(arguments),
i = 0,
r = this,
l = arguments.length;
for (;i<l;i++) {
r = r.replace(arguments[i],replaceWith);
}
return r;
}
/*
replaceSome method for strings
it takes as ,much arguments as we want and replaces all
of them with the last argument we specified
2013 CopyRights saved for: Max Ahmed
this is an example:
var string = "[hello i want to 'replace x' with eat]";
var replaced = string.replaceSome("]","[","'replace x' with","");
document.write(string + "<br>" + replaced); // returns hello i want to eat (without brackets)
*/
jsFiddle: http://jsfiddle.net/CPj89/

<!DOCTYPE html>
<html>
<body>
<p id="demo">Mr Blue
has a blue house and a blue car.</p>
<button onclick="myFunction()">Try it</button>
<script>
function myFunction() {
var str = document.getElementById("demo").innerHTML;
var res = str.replace(/\n| |car/gi, function myFunction(x){
if(x=='\n'){return x='<br>';}
if(x==' '){return x='&nbsp';}
if(x=='car'){return x='BMW'}
else{return x;}//must need
});
document.getElementById("demo").innerHTML = res;
}
</script>
</body>
</html>

I wrote this npm package stringinject https://www.npmjs.com/package/stringinject which allows you to do the following
var string = stringInject("this is a {0} string for {1}", ["test", "stringInject"]);
which will replace the {0} and {1} with the array items and return the following string
"this is a test string for stringInject"
or you could replace placeholders with object keys and values like so:
var str = stringInject("My username is {username} on {platform}", { username: "tjcafferkey", platform: "GitHub" });
"My username is tjcafferkey on Github"

var str = "I have a cat, a dog, and a goat.";
str = str.replace(/goat/i, "cat");
// now str = "I have a cat, a dog, and a cat."
str = str.replace(/dog/i, "goat");
// now str = "I have a cat, a goat, and a cat."
str = str.replace(/cat/i, "dog");
// now str = "I have a dog, a goat, and a cat."

You can use https://www.npmjs.com/package/union-replacer for this purpose. It is basically a string.replace(regexp, ...) counterpart, which allows multiple replaces to happen in one pass while preserving full power of string.replace(...).
Disclosure: I am the author. The library was developed to support more complex user-configurable replacements and it addresses all the problematic things like capture groups, backreferences and callback function replacements.
The solutions above are good enough for exact string replacements though.

by using prototype function we can replace easily by passing object with keys and values and replacable text
String.prototype.replaceAll=function(obj,keydata='key'){
const keys=keydata.split('key');
return Object.entries(obj).reduce((a,[key,val])=> a.replace(`${keys[0]}${key}${keys[1]}`,val),this)
}
const data='hids dv sdc sd ${yathin} ${ok}'
console.log(data.replaceAll({yathin:12,ok:'hi'},'${key}'))

All solutions work great, except when applied in programming languages that closures (e.g. Coda, Excel, Spreadsheet's REGEXREPLACE).
Two original solutions of mine below use only 1 concatenation and 1 regex.
Method #1: Lookup for replacement values
The idea is to append replacement values if they are not already in the string. Then, using a single regex, we perform all needed replacements:
var str = "I have a cat, a dog, and a goat.";
str = (str+"||||cat,dog,goat").replace(
/cat(?=[\s\S]*(dog))|dog(?=[\s\S]*(goat))|goat(?=[\s\S]*(cat))|\|\|\|\|.*$/gi, "$1$2$3");
document.body.innerHTML = str;
Explanations:
cat(?=[\s\S]*(dog)) means that we look for "cat". If it matches, then a forward lookup will capture "dog" as group 1, and "" otherwise.
Same for "dog" that would capture "goat" as group 2, and "goat" that would capture "cat" as group 3.
We replace with "$1$2$3" (the concatenation of all three groups), which will always be either "dog", "cat" or "goat" for one of the above cases
If we manually appended replacements to the string like str+"||||cat,dog,goat", we remove them by also matching \|\|\|\|.*$, in which case the replacement "$1$2$3" will evaluate to "", the empty string.
Method #2: Lookup for replacement pairs
One problem with Method #1 is that it cannot exceed 9 replacements at a time, which is the maximum number of back-propagation groups.
Method #2 states not to append just replacement values, but replacements directly:
var str = "I have a cat, a dog, and a goat.";
str = (str+"||||,cat=>dog,dog=>goat,goat=>cat").replace(
/(\b\w+\b)(?=[\s\S]*,\1=>([^,]*))|\|\|\|\|.*$/gi, "$2");
document.body.innerHTML = str;
Explanations:
(str+"||||,cat=>dog,dog=>goat,goat=>cat") is how we append a replacement map to the end of the string.
(\b\w+\b) states to "capture any word", that could be replaced by "(cat|dog|goat) or anything else.
(?=[\s\S]*...) is a forward lookup that will typically go to the end of the document until after the replacement map.
,\1=> means "you should find the matched word between a comma and a right arrow"
([^,]*) means "match anything after this arrow until the next comma or the end of the doc"
|\|\|\|\|.*$ is how we remove the replacement map.

We can also use split() and join() methods:
var str = "I have a cat, a dog, and a goat.";
str=str.split("cat").map(x => {return x.split("dog").map(y => {return y.split("goat").join("cat");}).join("goat");}).join("dog");
console.log(str);

you can try this. buy not smart.
var str = "I have a cat, a dog, and a goat.";
console.log(str);
str = str.replace(/cat/gi, "XXX");
console.log(str);
str = str.replace(/goat/gi, "cat");
console.log(str);
str = str.replace(/dog/gi, "goat");
console.log(str);
str = str.replace(/XXX/gi, "dog");
console.log(str);
Out put:
I have a dog, a goat, and a cat.

I expanded on #BenMcCormicks a bit. His worked for regular strings but not if I had escaped characters or wildcards. Here's what I did
str = "[curl] 6: blah blah 234433 blah blah";
mapObj = {'\\[curl] *': '', '\\d: *': ''};
function replaceAll (str, mapObj) {
var arr = Object.keys(mapObj),
re;
$.each(arr, function (key, value) {
re = new RegExp(value, "g");
str = str.replace(re, function (matched) {
return mapObj[value];
});
});
return str;
}
replaceAll(str, mapObj)
returns "blah blah 234433 blah blah"
This way it will match the key in the mapObj and not the matched word'

Solution with Jquery (first include this file): Replace multiple strings with multiple other strings:
var replacetext = {
"abc": "123",
"def": "456"
"ghi": "789"
};
$.each(replacetext, function(txtorig, txtnew) {
$(".eng-to-urd").each(function() {
$(this).text($(this).text().replace(txtorig, txtnew));
});
});