Develop Reference

JavaScript Split with RegEx without Global Match

I have an expression.
var expression = "Q101='You will have an answer here like a string for instance.'"
I have a regular expression that searches the expression.
var regEx = new regExp(/=|<>|like/)
I want to split the expression using the regular expression.
var result = expression.split(regExp)
This will return the following:
["Q101", "'You will have an answer here ", " a string for instance'"]
This is not what I want.
I should have:
["Q101", "'You will have an answer here like a string for instance'"]
How do I use the regular expression above to split only on the first match?

Since you only want to grab the two parts either side of the first delimiter it might be easier to use String.match and discard the whole match:
var expression = "Q101='You will have an answer here like a string for instance.'";
var parts = expression.match(/^(.*?)(?:=|<>|like)(.*)$/);
parts.shift();
console.log(parts);
expression = "Q101like'This answer uses like twice'";
parts = expression.match(/^(.*?)(?:=|<>|like)(.*)$/);
parts.shift();
console.log(parts);

JavaScript's split method won't quite do what you want, because it will either split on all matches, or stop after N matches. You need an extra step to find the first match, then split once by the first match using a custom function:
function splitMatch(string, match) {
var splitString = match[0];
var result = [
expression.slice(0, match.index),
expression.slice(match.index + splitString.length)
];
return result;
}
var expression = "Q101='You will have an answer here like a string for instance.'"
var regEx = new RegExp(/=|<>|like/)
var match = regEx.exec(expression)
if (match) {
var result = splitMatch(expression, match);
console.log(result);
}

While JavaScript's split method does have an optional limit parameter, it simply discards the parts of the result that make it too long (unlike, e.g. Python's split). To do this in JS, you'll need to split it manually, considering the length of the match —
const exp = "Q101='You will have an answer here like a string for instance.'"
const splitRxp = /=|<>|like/
const splitPos = exp.search(splitRxp)
const splitStr = exp.match(splitRxp)[0]
const result = splitPos != -1 ? (
[
exp.substring(0, splitPos),
exp.substring(splitPos + splitStr.length),
]
) : (
null
);
console.log(result)

Javascript regex capture giving unexpected results

I am trying to capture all data before the first _. What I have so far is
const regex = /(.*)(?=_)/g;
var s = "Mike_Jones_Jr";
console.log(s.match(regex));
The output is an array Array ["Mike_Jones","" ]
What I was expecting was Mike

Use /^[^_]*/
^ looks from the beginning of the string
[^_] negates the _
* gives any number of characters
const regex = /^[^_]*/;
var s = "Mike_Jones_Jr";
console.log(s.match(regex));

var s = "Mike_Jones_Jr";
console.log(s.split('_')[0]);

Create a capture group ((something between parentheses)) that starts at the beginning of the line (^) and is lazy (.*?), then grab the second item in the matching array.
const regex = /(^.*?)_/s
console.log('Mike_Jones_Jr'.match(regex)[1] || '')
console.log(`Mike
_Jones_Jr`.match(regex)[1] || '')

You can simply use split,
Note:- Second parameter is to limit the number of elements in final outptut
var s = "Mike_Jones_Jr";
console.log( s.split('_', 1) );
If you want to do using regex, you can drop the g flag
const regex = /^[^_]*(?=_)/;
var s = "Mike_Jones_Jr";
console.log(s.match(regex));
console.log("_ melpomene is awesome".match(regex));

js regex discrepancy

I have a JS regex match that seems to include the brackets incorrectly. I tested it out at Regex101 and it seems to work appropriately there but when I run it I get this alert response:
[#],[Type,' '],[Problem w/ICD],['- ',Assessment],[' : ',Comment],[LF],[LF]
var temp = "[#]. [Type,' '][Problem w/ICD]['- ',Assessment][' : ',Comment][LF][LF]";
var rep = temp.match(/\[(.*?)\]/g);
alert(rep);
Why are the brackets included when they are outside the capture group?

The brackets are included because when using string#match and a regex with /g modifier, you will lose capturing groups.
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.
You need to use a RegExp#exec() in a loop, and access the first capturing group via index 1.
var re = /\[(.*?)\]/g;
var str = '[#]. [Type,\' \'][Problem w/ICD][\'- \',Assessment][\' : \',Comment][LF][LF]';
var m;
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
console.log(res);
Result:
["#", "Type,' '", "Problem w/ICD", "'- ',Assessment", "' : ',Comment", "LF", "LF"]

Selecting numbers only in a JavaScript string

I want to select all the digits from a given string. I tried with the code below, but it does not return all the numbers in the string:
var match = /\d+/.exec("+12 (345)-678.90[]");
console.log(match.toString());
It only returns 12, while I expect it to return 1234567890.

simple implementation will be
var value='+12 (345)-678.90[]'.replace(/\D+/g, '');
console.log(value);

You need to use global flag, it will return you an array of matched data the you can use join() it.
"+12 (345)-678.90[]".match(/\d+/g).join('');
alert("+12 (345)-678.90[]".match(/\d+/g).join(''))

Use the global flag:
"+12 (345)-678.90[]".match(/\d+/g)

The \d+ pattern will return consecutive digits only, and since you running exec once without g option, it will only give you the first occurrence of consecutive digits.
Use this:
var re = /\d+/g;
var str = '+12 (345)-678.90[]';
var res = "";
while ((m = re.exec(str)) !== null) {
res += m[0];
}
alert(res);
Output is 1234567890, as we append found digit sequences to the res variable.

How can I remove all characters up to and including the 3rd slash in a string?

I'm having trouble with removing all characters up to and including the 3 third slash in JavaScript. This is my string:
http://blablab/test
The result should be:
test
Does anybody know the correct solution?

To get the last item in a path, you can split the string on / and then pop():
var url = "http://blablab/test";
alert(url.split("/").pop());
//-> "test"
To specify an individual part of a path, split on / and use bracket notation to access the item:
var url = "http://blablab/test/page.php";
alert(url.split("/")[3]);
//-> "test"
Or, if you want everything after the third slash, split(), slice() and join():
var url = "http://blablab/test/page.php";
alert(url.split("/").slice(3).join("/"));
//-> "test/page.php"

var string = 'http://blablab/test'
string = string.replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'')
alert(string)
This is a regular expression. I will explain below
The regex is /[\s\S]*\//
/ is the start of the regex
Where [\s\S] means whitespace or non whitespace (anything), not to be confused with . which does not match line breaks (. is the same as [^\r\n]).
* means that we match anywhere from zero to unlimited number of [\s\S]
\/ Means match a slash character
The last / is the end of the regex

var str = "http://blablab/test";
var index = 0;
for(var i = 0; i < 3; i++){
index = str.indexOf("/",index)+1;
}
str = str.substr(index);
To make it a one liner you could make the following:
str = str.substr(str.indexOf("/",str.indexOf("/",str.indexOf("/")+1)+1)+1);

You can use split to split the string in parts and use slice to return all parts after the third slice.
var str = "http://blablab/test",
arr = str.split("/");
arr = arr.slice(3);
console.log(arr.join("/")); // "test"
// A longer string:
var str = "http://blablab/test/test"; // "test/test";

You could use a regular expression like this one:
'http://blablab/test'.match(/^(?:[^/]*\/){3}(.*)$/);
// -> ['http://blablab/test', 'test]
A string’s match method gives you either an array (of the whole match, in this case the whole input, and of any capture groups (and we want the first capture group)), or null. So, for general use you need to pull out the 1th element of the array, or null if a match wasn’t found:
var input = 'http://blablab/test',
re = /^(?:[^/]*\/){3}(.*)$/,
match = input.match(re),
result = match && match[1]; // With this input, result contains "test"

let str = "http://blablab/test";
let data = new URL(str).pathname.split("/").pop();
console.log(data);