Develop Reference

what is best way to match words from list to words from sentence in javascript?

i have two sentences and i would like to find all the words they share regardless of capitalization or punctuation.
currently this is what I am doing:
searchWords = sentence1.split(" ");
var wordList = sentence2.split(" ");
const matchList = wordList.filter(value => -1 !== searchWords.indexOf(value));
it works ok but obviously capitalization and punctuation cause issues.
i know i need to incorporate something like .match() in there but i don't know how to work with it. I am sure this is something someone has done before just havent found the code yet, any refrences are also appreciated.
Thank you,
Best
This dude.

If you're looking for any words that match you can use RegExp with String.prototype.replace and verify a match using String.prototype.search with a created RegExp and an i flag to allow case insensitivity.
function compare(str1, str2, matches = []) {
str1.replace(/(\w+)/g, m => str2.search(new RegExp(m, "i")) >= 0 && matches.push(m));
return matches;
}
console.log( compare("Hello there this is a test", "Hello Test this is a world") );
If you're looking for specific words that match you can use functional composition to split each string into an Array, filter each by possible matches, and then filter one against the other.
function compare(str1, str2, matchables) {
let containFilter = (a) => (i) => a.includes(i),
matchFilter = s => s.toLowerCase().split(" ").filter(containFilter(matchables));
return matchFilter(str1).filter(containFilter( matchFilter(str2) ));
}
let matchables = ["hello", "test", "world"];
console.log( compare("Hello there this is a test", "Hi Test this is a world", matchables) );

I think you may be over-thinking this. Would just converting both sentences to an array and using a for loop to cycle through the words work? For example:
var searchWords = sentence1.split(" ");
var wordList = sentence2.toLowerCase().split(" ");
var commonWords = [];
for(var i = 0; i < searchWords.length; i++){
if(wordList.includes(searchWords[i].toLowerCase())){
commonWords.push(searchWords[i])
}
}
console.log(commonWords);
Or some variation of that.
As for the punctuation, you could probably add .replace(/[^A-Za-z0-9\s]/g,"") to the end of searchWords[i].toLowerCase() as mentioned in the following answer: https://stackoverflow.com/a/33408855/10601203

Get array of elements from a string and look up values

I have a calculation string from a database like:
var calc = "{a}+{b}==2"
and I want to pull all the elements with "{}" so that I can look up their values from a database. What's the fastest way of doing this, so I end up with an ordered array that I can look up, and replace the values back in the string.
I've considered:
- For loop, looking for { then finding the next }
- Split with a map
- IndexOf

Using regex
var exp = /{([^}]+)}/g ,index;
while(index = exp.exec("{a}+{b}==2")) {
console.log(index[1]);
}
.
Demo

Not sure if it's the "fastest" way, but you should consider using a regex.
Something like:
var calc = "{a}+{b}==2";
var re = /{([^}]+)}/g;
var res;
var result = [];
while (res = re.exec(calc))
{
result.push(res[1]);
}
console.log(result);
Your regex may need to be refined based on the actual definition of the {} expressions (based on allowed characters, quoting, etc.).
Once you have received the values back, you can then use replace to replace the values.
var values = {a: 1, b: 3};
var replaced = calc.replace(re,function(match,name) { return values[name]; });
console.log(replaced);
NB: be very careful if you plan to then send this to eval or the like.

Regex comes to the mind first but one other way of implementing this job in O(n) time could be;
function getDatas(s){
var dataOn = false;
return Array.prototype.reduce.call(s,(d,c) => dataOn ? c !== "}" ? (d[d.length-1] += c,d)
: (dataOn = false, d)
: c === "{" ? (dataOn = true, d.push(""),d)
: d, []);
}
var calc = "{a}+{b}+{colorSpace}==2",
result = getDatas(calc);
console.log(result);
So out of curiosity i have done some tests on JSBen and it seems that #jcaron's regex is indeed much efficient. You may extend those tests with any of your other ideas like indexOf or for loop.

regex to find specific strings in javascript

disclaimer - absolutely new to regexes....
I have a string like this:
subject=something||x-access-token=something
For this I need to extract two values. Subject and x-access-token.
As a starting point, I wanted to collect two strings: subject= and x-access-token=. For this here is what I did:
/[a-z,-]+=/g.exec(mystring)
It returns only one element subject=. I expected both of them. Where i am doing wrong?

The g modifier does not affect exec, because exec only returns the first match by specification. What you want is the match method:
mystring.match(/[a-z,-]+=/g)

No regex necessary. Write a tiny parser, it's easy.
function parseValues(str) {
var result = {};
str.split("||").forEach(function (item) {
var parts = item.split("=");
result[ parts[0] /* key */ ] = parts[1]; /* value */
});
return result;
}
usage
var obj = parseValues("subject=something||x-access-token=something-else");
// -> {subject: "something", x-access-token: "something-else"}
var subj = obj.subject;
// -> "something"
var token = obj["x-access-token"];
// -> "something-else"
Additional complications my arise when there is an escaping schema involved that allows you to have || inside a value, or when a value can contain an =.
You will hit these complications with regex approach as well, but with a parser-based approach they will be much easier to solve.

You have to execute exec twice to get 2 extracted strings.
According to MDN: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/exec
If your regular expression uses the "g" flag, you can use the exec() method multiple times to find successive matches in the same string.
Usually, people extract all strings matching the pattern one by one with a while loop. Please execute following code in browser console to see how it works.
var regex = /[a-z,-]+=/g;
var string = "subject=something||x-access-token=something";
while(matched = regex.exec(string)) console.log(matched);

You can convert the string into a valid JSON string, then parse it to retrieve an object containing the expected data.
var str = 'subject=something||x-access-token=something';
var obj = JSON.parse('{"' + str.replace(/=/g, '":"').replace(/\|\|/g, '","') + '"}');
console.log(obj);

I don't think you need regexp here, just use the javascript builtin function "split".
var s = "subject=something1||x-access-token=something2";
var r = s.split('||'); // r now is an array: ["subject=something1", "x-access-token=something2"]
var i;
for(i=0; i<r.length; i++){
// for each array's item, split again
r[i] = r[i].split('=');
}
At the end you have a matrix like the following:
y x 0 1
0 subject something1
1 x-access-token something2
And you can access the elements using x and y:
"subject" == r[0][0]
"x-access-token" == r[1][0]
"something2" == r[1][1]

If you really want to do it with a pure regexp:
var input = 'subject=something1||x-access-token=something2'
var m = /subject=(.*)\|\|x-access-token=(.*)/.exec(input)
var subject = m[1]
var xAccessToken = m[2]
console.log(subject);
console.log(xAccessToken);
However, it would probably be cleaner to split it instead:
console.log('subject=something||x-access-token=something'
.split(/\|\|/)
.map(function(a) {
a = a.split(/=/);
return { key: a[0], val: a[1] }
}));

Javascript regex to replace "split"

I would like to use Javascript Regex instead of split.
Here is the example string:
var str = "123:foo";
The current method calls:
str.split(":")[1]
This will return "foo", but it raises an Error when given a bad string that doesn't have a :.
So this would raise an error:
var str = "fooblah";
In the case of "fooblah" I'd like to just return an empty string.
This should be pretty simple, but went looking for it, and couldn't figure it out. Thank you in advance.

Remove the part up to and including the colon (or the end of the string, if there's no colon):
"123:foo".replace(/.*?(:|$)/, '') // "foo"
"foobar" .replace(/.*?(:|$)/, '') // ""
How this regexp works:
.* Grab everything
? non-greedily
( until we come to
: a colon
| or
$ the end of the string
)

A regex won't help you. Your error likely arises from trying to use undefined later. Instead, check the length of the split first.
var arr = str.split(':');
if (arr.length < 2) {
// Do something to handle a bad string
} else {
var match = arr[1];
...
}

Here's what I've always used, with different variations; this is just a simple version of it:
function split(str, d) {
var op = "";
if(str.indexOf(d) > 0) {
op = str.split(d);
}
return(op);
}
Fairly simple, either returns an array or an empty string.

var str1 = "123:foo", str2 = "fooblah";
var res = function (s) {
return /:/.test(s) && s.replace(/.*(?=:):/, "") || ""
};
console.log(res(str1), res(str2))

Here is a solution using a single regex, with the part you want in the capturing group:
^[^:]*:([^:]+)

javascript get string before a character

I have a string that and I am trying to extract the characters before the quote.
Example is extract the 14 from 14' - €14.99
I am using the follwing code to acheive this.
$menuItem.text().match(/[^']*/)[0]
My problem is that if the string is something like €0.88 I wish to get an empty string returned. However I get back the full string of €0.88.
What I am I doing wrong with the match?

This is the what you should use to split:
string.slice(0, string.indexOf("'"));
And then to handle your non existant value edge case:
function split(str) {
var i = str.indexOf("'");
if(i > 0)
return str.slice(0, i);
else
return "";
}
Demo on JsFiddle

Nobody seems to have presented what seems to me as the safest and most obvious option that covers each of the cases the OP asked about so I thought I'd offer this:
function getCharsBefore(str, chr) {
var index = str.indexOf(chr);
if (index != -1) {
return(str.substring(0, index));
}
return("");
}

try this
str.substring(0,str.indexOf("'"));

Here is an underscore mixin in coffescript
_.mixin
substrBefore : ->
[char, str] = arguments
return "" unless char?
fn = (s)-> s.substr(0,s.indexOf(char)+1)
return fn(str) if str?
fn
or if you prefer raw javascript : http://jsfiddle.net/snrobot/XsuQd/
You can use this to build a partial like:
var beforeQuote = _.substrBefore("'");
var hasQuote = beforeQuote("14' - €0.88"); // hasQuote = "14'"
var noQoute = beforeQuote("14 €0.88"); // noQuote = ""
Or just call it directly with your string
var beforeQuote = _.substrBefore("'", "14' - €0.88"); // beforeQuote = "14'"
I purposely chose to leave the search character in the results to match its complement mixin substrAfter (here is a demo: http://jsfiddle.net/snrobot/SEAZr/ ). The later mixin was written as a utility to parse url queries. In some cases I am just using location.search which returns a string with the leading ?.

I use "split":
let string = "one-two-three";
let um = string.split('-')[0];
let dois = string.split('-')[1];
let tres = string.split('-')[2];
document.write(tres) //three