This question already has answers here:
Replace multiple strings with multiple other strings
(27 answers)
Closed 3 years ago.
I have a string prototype whose code is given below:
String.prototype.replaceAll = function(str1, str2, ignore) {
return this.replace(
new RegExp(
str1.replace(/([\/\,\!\\\^\$\{\}\[\]\(\)\.\*\+\?\|\<\>\-\&])/g,"\\$&"),(ignore?"gi":"g")),(typeof(str2)=="string")?str2.replace(/\$/g,"$$$$"
):str2
)};
Usage:
var a = "I am Javascript";
console.log(
a.replaceAll("am", "love")
); // => I love Javascript
But when it comes to multiple exchange of characters or words, I have to run the prototype multiple times to achieve it. But I have thought of something like this:
var a = "I am Java";
console.log(
a.replaceAll(["am" , "Java"], ["love", "Javascript"])
); // => I love Javascript
So can you help me to achieve it? Or there is any other alternative?
I'd prefer to store replacements as key-value pairs in an object or as an array of pairs. Regardless of the format, you can dynamically create a regex by joining the values you want to replace using | alternation. Then give replace a callback function and use its match parameter as a key to look up its corresponding pair in the swaps object.
const s = "I am Java";
const swaps = {am: "love", Java: "JS"};
const pattern = new RegExp(Object.keys(swaps).join("|"), "g");
console.log(s.replace(pattern, m => swaps[m]));
To handle case-insensitive replacements, ensure all keys in swaps are lowercase (either programmatically or manually, depending on usage) and lowercase the matches before keying in:
const s = "I am Java";
const swaps = {am: "love", java: "JS"};
const pattern = new RegExp(Object.keys(swaps).join("|"), "gi");
console.log(s.replace(pattern, m => swaps[m.toLowerCase()]));
This works.
String.prototype.replaceAll = function(str1, str2, ignore) {
let flags = 'g';
if (ignore) {
flags += 'i';
}
if (Array.isArray(str1) && Array.isArray(str2)) {
let newStr = this;
str1.map((element, index) => {
if (str2[index]) {
newStr = newStr.replace(new RegExp(element, flags), str2[index]);
}
return newStr;
});
return newStr;
}
else {
return this.replace(new RegExp(str1, flags), str2);
}
}
I want to match a number in a string:
'abc#2003, or something else #2017'
I want to get result [2003, 2007] with match function.
let strReg = 'abc#2003, or something else #2017';
let reg = new RegExp(/(?=(#\d+))\1/);
strReg.match(reg) //[ '#2003 ', '#2017 ' ]
let reg1 = new RegExp(/(?=#(\d+))\1/)
strReg.match(reg1) //null, but I expect ［2003, 2007］
the result mains '\1' match after '?=', ?=()\1 works, ?=#()\1 not.
javascript only supports backwards, how should I do to match '#' but ignore it?
I take it that you want an array of the results, so...
var s = "abc#2003, or something else #2017 not the 2001 though";
var re = /#(\d+)/g;
var result = [];
var match = re.exec(s);
while (match !== null) {
result.push(parseInt(match[1]));
match = re.exec(s);
}
console.log(result);
Outputs:
Array [ 2003, 2017 ]
match(0) is the entire match, match(1) is the captured group.
Also, see How do you access the matched groups in a JavaScript regular expression?
Inspired by javascript regex - look behind alternative?, if you want to do it as almost a one-liner:
var re = /(\d+)(?=#)/g; /* write the regex backwards */
var result = [];
s.split('').reverse().join('').match(re).forEach(function (el) { result.push(parseInt(el.split('').reverse().join(''))); });
console.log(result.reverse());
Caveat: Who wrote this programing saying? “Always code as if the guy who ends up maintaining your code will be a violent psychopath who knows where you live.”
Small change to your code does the job as follows:
/#(\d+)/g
number followed by # will be remembered as you required.
I want to create a regex with following logic:
1., If string contains T replace it with space
2., If string contains Z remove Z
I wrote two regex already, but I can't combine them:
string.replace(/\T/g,' ') && string.replace(/\Z/g,'');
EDIT: I want the regex code to be shorter
Doesn't seem this even needs regex. Just 2 chained replacements would do.
var str = '[T] and [Z] but not [T] and [Z]';
var result = str.replace('T',' ').replace('Z','');
console.log(result);
However, a simple replace only replaces the first occurence.
To replace all, regex still comes in handy. By making use of the global g flag.
Note that the characters aren't escaped with \. There's no need.
var str = '[T] and [Z] and another [T] and [Z]';
var result = str.replace(/T/g,' ').replace(/Z/g,'');
console.log(result);
// By using regex we could also ignore lower/upper-case. (the i flag)
// Also, if more than 1 letter needs replacement, a character class [] makes it simple.
var str2 = '(t) or (Ⓣ) and (z) or (Ⓩ). But also uppercase (T) or (Z)';
var result2 = str2.replace(/[tⓉ]/gi,' ').replace(/[zⓏ]/gi,'');
console.log(result2);
But if the intention is to process really big strings, and performance matters?
Then I found out in another challenge that using an unnamed callback function inside 1 regex replace can prove to be faster. When compared to using 2 regex replaces.
Probably because if it's only 1 regex then it only has to process the huge string once.
Example snippet:
console.time('creating big string');
var bigstring = 'TZ-'.repeat(2000000);
console.timeEnd('creating big string');
console.log('bigstring length: '+bigstring.length);
console.time('double replace big string');
var result1 = bigstring.replace(/[t]/gi,'X').replace(/[z]/gi,'Y');
console.timeEnd('double replace big string');
console.time('single replace big string');
var result2 = bigstring.replace(/([t])|([z])/gi, function(m, c1, c2){
if(c1) return 'X'; // if capture group 1 has something
return 'Y';
});
console.timeEnd('single replace big string');
var smallstring = 'TZ-'.repeat(5000);
console.log('smallstring length: '+smallstring.length);
console.time('double replace small string');
var result3 = smallstring.replace(/T/g,'X').replace(/Z/g,'Y');
console.timeEnd('double replace small string');
console.time('single replace small string');
var result4 = smallstring.replace(/(T)|(Z)/g, function(m, c1, c2){
if(c1) return 'X';
return 'Y';
});
console.timeEnd('single replace small string');
Do you look for something like this?
ES6
var key = {
'T': ' ',
'Z': ''
}
"ATAZATA".replace(/[TZ]/g, (char) => key[char] || '');
Vanilla
"ATAZATA".replace(/[TZ]/g,function (char) {return key[char] || ''});
or
"ATAZATA".replace(/[TZ]/g,function (char) {return char==='T'?' ':''});
you can capture both and then decide what to do in the callback:
string.replace(/[TZ]/g,(m => m === 'T' ? '' : ' '));
var string = 'AZorro Tab'
var res = string.replace(/[TZ]/g,(m => m === 'T' ? '' : ' '));
console.log(res)
-- edit --
Using a dict substitution you can also do:
var string = 'AZorro Tab'
var dict = { T : '', Z : ' '}
var re = new RegExp(`[${ Object.keys(dict).join('') }]`,'g')
var res = string.replace(re,(m => dict[m] ) )
console.log(res)
Second Update
I have developed the following function to use in production, perhaps it can help someone else. It's basically a loop of the native's replaceAll Javascript function, it does not make use of regex:
function replaceMultiple(text, characters){
for (const [i, each] of characters.entries()) {
const previousChar = Object.keys(each);
const newChar = Object.values(each);
text = text.replaceAll(previousChar, newChar);
}
return text
}
Usage is very simple:
const text = '#Please send_an_information_pack_to_the_following_address:';
const characters = [
{
"#":""
},
{
"_":" "
},
]
const result = replaceMultiple(text, characters);
console.log(result); //'Please send an information pack to the following address:'
Update
You can now use replaceAll natively.
Outdated Answer
Here is another version using String Prototype. Enjoy!
String.prototype.replaceAll = function(obj) {
let finalString = '';
let word = this;
for (let each of word){
for (const o in obj){
const value = obj[o];
if (each == o){
each = value;
}
}
finalString += each;
}
return finalString;
};
'abc'.replaceAll({'a':'x', 'b':'y'}); //"xyc"
This question already has answers here:
Split string once in javascript?
(17 answers)
Closed 8 years ago.
I know there are several ways to split an array in jQuery but I have a special case:
If I have for example this two strings:
"G09.4 What"
"A04.3 A new Code"
When I split the first by ' ' I can simply choose the code in front with [0] what would be G09.4. And when I call [1] I get the text: What
But when I do the same with the second string I get for [1] A but I want to retrieve A new Code.
So how can I retrieve for each string the code and the separate text?
Use
var someString = "A04.3 A new Code";
var index = someString.indexOf(" "); // Gets the first index where a space occours
var id = someString.substr(0, index); // Gets the first part
var text = someString.substr(index + 1); // Gets the text part
You can split the string and shift off the first entry in the returned array. Then join the leftovers e.g.
var chunks = "A04.3 A new Code".split(/\s+/);
var arr = [chunks.shift(), chunks.join(' ')];
// arr[0] = "A04.3"
// arr[1] = "A new Code"
Instead of splitting the string on the space, use a combination of indexOf and slice:
var s = "A04.3 A new Code";
var i = s.indexOf(' ');
var partOne = s.slice(0, i).trim();
var partTwo = s.slice(i + 1, s.length).trim();
You can use match() and capture what you need via a regular expression:
"G09.4 What".match(/^(\S+)\s+(.+)/)
// => ["G09.4 What", "G09.4", "What"]
"A04.3 A new Code".match(/^(\S+)\s+(.+)/)
// => ["A04.3 A new Code", "A04.3", "A new Code"]
As you can see the two items you want are in [1] and [2] of the returned arrays.
What about this one:
function split2(str, delim) {
var parts=str.split(delim);
return [parts[0], parts.splice(1,parts.length).join(delim)];
}
FIDDLE
Or for more performance, try this:
function split2s(str, delim) {
var p=str.indexOf(delim);
if (p !== -1) {
return [str.substring(0,p), str.substring(p+1)];
} else {
return [str];
}
}
You can get the code and then remove it from the original string leaving you with both the code and the string without the code.
var originalString = "A04.3 A new Code",
stringArray = originalString.split(' '),
code,
newString;
code = stringArray[0];
newString = originalString.replace(code, '');
I'm trying to parse the following kind of string:
[key:"val" key2:"val2"]
where there are arbitrary key:"val" pairs inside. I want to grab the key name and the value.
For those curious I'm trying to parse the database format of task warrior.
Here is my test string:
[description:"aoeu" uuid:"123sth"]
which is meant to highlight that anything can be in a key or value aside from space, no spaces around the colons, and values are always in double quotes.
In node, this is my output:
[deuteronomy][gatlin][~]$ node
> var re = /^\[(?:(.+?):"(.+?)"\s*)+\]$/g
> re.exec('[description:"aoeu" uuid:"123sth"]');
[ '[description:"aoeu" uuid:"123sth"]',
'uuid',
'123sth',
index: 0,
input: '[description:"aoeu" uuid:"123sth"]' ]
But description:"aoeu" also matches this pattern. How can I get all matches back?
Continue calling re.exec(s) in a loop to obtain all the matches:
var re = /\s*([^[:]+):\"([^"]+)"/g;
var s = '[description:"aoeu" uuid:"123sth"]';
var m;
do {
m = re.exec(s);
if (m) {
console.log(m[1], m[2]);
}
} while (m);
Try it with this JSFiddle: https://jsfiddle.net/7yS2V/
str.match(pattern), if pattern has the global flag g, will return all the matches as an array.
For example:
const str = 'All of us except #Emran, #Raju and #Noman were there';
console.log(
str.match(/#\w*/g)
);
// Will log ["#Emran", "#Raju", "#Noman"]
To loop through all matches, you can use the replace function:
var re = /\s*([^[:]+):\"([^"]+)"/g;
var s = '[description:"aoeu" uuid:"123sth"]';
s.replace(re, function(match, g1, g2) { console.log(g1, g2); });
This is a solution
var s = '[description:"aoeu" uuid:"123sth"]';
var re = /\s*([^[:]+):\"([^"]+)"/g;
var m;
while (m = re.exec(s)) {
console.log(m[1], m[2]);
}
This is based on lawnsea's answer, but shorter.
Notice that the `g' flag must be set to move the internal pointer forward across invocations.
str.match(/regex/g)
returns all matches as an array.
If, for some mysterious reason, you need the additional information comes with exec, as an alternative to previous answers, you could do it with a recursive function instead of a loop as follows (which also looks cooler :).
function findMatches(regex, str, matches = []) {
const res = regex.exec(str)
res && matches.push(res) && findMatches(regex, str, matches)
return matches
}
// Usage
const matches = findMatches(/regex/g, str)
as stated in the comments before, it's important to have g at the end of regex definition to move the pointer forward in each execution.
We are finally beginning to see a built-in matchAll function, see here for the description and compatibility table. It looks like as of May 2020, Chrome, Edge, Firefox, and Node.js (12+) are supported but not IE, Safari, and Opera. Seems like it was drafted in December 2018 so give it some time to reach all browsers, but I trust it will get there.
The built-in matchAll function is nice because it returns an iterable. It also returns capturing groups for every match! So you can do things like
// get the letters before and after "o"
let matches = "stackoverflow".matchAll(/(\w)o(\w)/g);
for (match of matches) {
console.log("letter before:" + match[1]);
console.log("letter after:" + match[2]);
}
arrayOfAllMatches = [...matches]; // you can also turn the iterable into an array
It also seem like every match object uses the same format as match(). So each object is an array of the match and capturing groups, along with three additional properties index, input, and groups. So it looks like:
[<match>, <group1>, <group2>, ..., index: <match offset>, input: <original string>, groups: <named capture groups>]
For more information about matchAll there is also a Google developers page. There are also polyfills/shims available.
If you have ES9
(Meaning if your system: Chrome, Node.js, Firefox, etc supports Ecmascript 2019 or later)
Use the new yourString.matchAll( /your-regex/g ).
If you don't have ES9
If you have an older system, here's a function for easy copy and pasting
function findAll(regexPattern, sourceString) {
let output = []
let match
// auto-add global flag while keeping others as-is
let regexPatternWithGlobal = RegExp(regexPattern,[...new Set("g"+regexPattern.flags)].join(""))
while (match = regexPatternWithGlobal.exec(sourceString)) {
// get rid of the string copy
delete match.input
// store the match data
output.push(match)
}
return output
}
example usage:
console.log( findAll(/blah/g,'blah1 blah2') )
outputs:
[ [ 'blah', index: 0 ], [ 'blah', index: 6 ] ]
Based on Agus's function, but I prefer return just the match values:
var bob = "> bob <";
function matchAll(str, regex) {
var res = [];
var m;
if (regex.global) {
while (m = regex.exec(str)) {
res.push(m[1]);
}
} else {
if (m = regex.exec(str)) {
res.push(m[1]);
}
}
return res;
}
var Amatch = matchAll(bob, /(&.*?;)/g);
console.log(Amatch); // yeilds: [>, <]
Iterables are nicer:
const matches = (text, pattern) => ({
[Symbol.iterator]: function * () {
const clone = new RegExp(pattern.source, pattern.flags);
let match = null;
do {
match = clone.exec(text);
if (match) {
yield match;
}
} while (match);
}
});
Usage in a loop:
for (const match of matches('abcdefabcdef', /ab/g)) {
console.log(match);
}
Or if you want an array:
[ ...matches('abcdefabcdef', /ab/g) ]
Here is my function to get the matches :
function getAllMatches(regex, text) {
if (regex.constructor !== RegExp) {
throw new Error('not RegExp');
}
var res = [];
var match = null;
if (regex.global) {
while (match = regex.exec(text)) {
res.push(match);
}
}
else {
if (match = regex.exec(text)) {
res.push(match);
}
}
return res;
}
// Example:
var regex = /abc|def|ghi/g;
var res = getAllMatches(regex, 'abcdefghi');
res.forEach(function (item) {
console.log(item[0]);
});
If you're able to use matchAll here's a trick:
Array.From has a 'selector' parameter so instead of ending up with an array of awkward 'match' results you can project it to what you really need:
Array.from(str.matchAll(regexp), m => m[0]);
If you have named groups eg. (/(?<firstname>[a-z][A-Z]+)/g) you could do this:
Array.from(str.matchAll(regexp), m => m.groups.firstName);
Since ES9, there's now a simpler, better way of getting all the matches, together with information about the capture groups, and their index:
const string = 'Mice like to dice rice';
const regex = /.ice/gu;
for(const match of string.matchAll(regex)) {
console.log(match);
}
// ["mice", index: 0, input: "mice like to dice rice", groups:
undefined]
// ["dice", index: 13, input: "mice like to dice rice",
groups: undefined]
// ["rice", index: 18, input: "mice like to dice
rice", groups: undefined]
It is currently supported in Chrome, Firefox, Opera. Depending on when you read this, check this link to see its current support.
Use this...
var all_matches = your_string.match(re);
console.log(all_matches)
It will return an array of all matches...That would work just fine....
But remember it won't take groups in account..It will just return the full matches...
I would definatly recommend using the String.match() function, and creating a relevant RegEx for it. My example is with a list of strings, which is often necessary when scanning user inputs for keywords and phrases.
// 1) Define keywords
var keywords = ['apple', 'orange', 'banana'];
// 2) Create regex, pass "i" for case-insensitive and "g" for global search
regex = new RegExp("(" + keywords.join('|') + ")", "ig");
=> /(apple|orange|banana)/gi
// 3) Match it against any string to get all matches
"Test string for ORANGE's or apples were mentioned".match(regex);
=> ["ORANGE", "apple"]
Hope this helps!
This isn't really going to help with your more complex issue but I'm posting this anyway because it is a simple solution for people that aren't doing a global search like you are.
I've simplified the regex in the answer to be clearer (this is not a solution to your exact problem).
var re = /^(.+?):"(.+)"$/
var regExResult = re.exec('description:"aoeu"');
var purifiedResult = purify_regex(regExResult);
// We only want the group matches in the array
function purify_regex(reResult){
// Removes the Regex specific values and clones the array to prevent mutation
let purifiedArray = [...reResult];
// Removes the full match value at position 0
purifiedArray.shift();
// Returns a pure array without mutating the original regex result
return purifiedArray;
}
// purifiedResult= ["description", "aoeu"]
That looks more verbose than it is because of the comments, this is what it looks like without comments
var re = /^(.+?):"(.+)"$/
var regExResult = re.exec('description:"aoeu"');
var purifiedResult = purify_regex(regExResult);
function purify_regex(reResult){
let purifiedArray = [...reResult];
purifiedArray.shift();
return purifiedArray;
}
Note that any groups that do not match will be listed in the array as undefined values.
This solution uses the ES6 spread operator to purify the array of regex specific values. You will need to run your code through Babel if you want IE11 support.
Here's a one line solution without a while loop.
The order is preserved in the resulting list.
The potential downsides are
It clones the regex for every match.
The result is in a different form than expected solutions. You'll need to process them one more time.
let re = /\s*([^[:]+):\"([^"]+)"/g
let str = '[description:"aoeu" uuid:"123sth"]'
(str.match(re) || []).map(e => RegExp(re.source, re.flags).exec(e))
[ [ 'description:"aoeu"',
'description',
'aoeu',
index: 0,
input: 'description:"aoeu"',
groups: undefined ],
[ ' uuid:"123sth"',
'uuid',
'123sth',
index: 0,
input: ' uuid:"123sth"',
groups: undefined ] ]
My guess is that if there would be edge cases such as extra or missing spaces, this expression with less boundaries might also be an option:
^\s*\[\s*([^\s\r\n:]+)\s*:\s*"([^"]*)"\s*([^\s\r\n:]+)\s*:\s*"([^"]*)"\s*\]\s*$
If you wish to explore/simplify/modify the expression, it's been
explained on the top right panel of
regex101.com. If you'd like, you
can also watch in this
link, how it would match
against some sample inputs.
Test
const regex = /^\s*\[\s*([^\s\r\n:]+)\s*:\s*"([^"]*)"\s*([^\s\r\n:]+)\s*:\s*"([^"]*)"\s*\]\s*$/gm;
const str = `[description:"aoeu" uuid:"123sth"]
[description : "aoeu" uuid: "123sth"]
[ description : "aoeu" uuid: "123sth" ]
[ description : "aoeu" uuid : "123sth" ]
[ description : "aoeu"uuid : "123sth" ] `;
let m;
while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}
RegEx Circuit
jex.im visualizes regular expressions:
const re = /^\[(?:(.+?):"(.+?)"\s*)+\]$/g
const matches = [...re.exec('[description:"aoeu" uuid:"123sth"]').entries()]
console.log(matches)
Basically, this is ES6 way to convert Iterator returned by exec to a regular Array
Here is my answer:
var str = '[me nombre es] : My name is. [Yo puedo] is the right word';
var reg = /\[(.*?)\]/g;
var a = str.match(reg);
a = a.toString().replace(/[\[\]]/g, "").split(','));