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Regex to match character unless it is preceded by an odd number of another specific character

Another way to state my problem is to match a character always when it is preceded by an even number (0, 2, 4, ...) of another specific character.
In my case I want to match all ' characters in string unless it is preceded by an odd number (1, 3, 5 ...) of ?
example:
- ?' => shouldn't match (preceded by one ?)
- ??' => Should match (preceded by 2 ?)
- ?????' => Shouldn't match (preceded by 5 ?)
Lets consider this scenario:
We have this string : ' ??' ????' ?' ??????' then the regex should match all ' characters in this case except for the 4th one, so for example if I want to use String.split(regex) the result would be ['', '??', '????', ?' '??????']
Currently I was using this regex: (?<!\?)', but the problem is that it matches only if there is no ? before '

You can use
/(?<=(?<!\?)(?:\?\?)*)'/g
See the regex demo. Details:
(?<=(?<!\?)(?:\?\?)*) - a positive lookbehind that matches a location that is preceded with any zero or more occurrences of double ? not immediately preceded with another ?
' - a ' char.
Sample code:
const texts = ["The ?' should not match","The ??' should match","?????' => The ?????' should not match"];
const rx = /(?<=(?<!\?)(?:\?\?)*)'/g
for (var text of texts) {
console.log(text, '=>', rx.test(text));
}
If you need replacing, it is possible with
const texts = ["The ?' should not match","The ??' should match","?????' => The ?????' should not match"];
const rx = /(?<=(?<!\?)(?:\?\?)*)'/g
for (var text of texts) {
console.log(text, '=>', text.replace(rx, '<MATCH>$&</MATCH>'));
}

You may use this regex with a lookbehind condition:
(?<=([^?]|^)(?:\?\?)*)'
RegEx Demo
RegEx Explanation
(?<=: Start lookbehind condition
([^?]|^): Match a non-? character or start
(?:\?\?)*: Match 0 or more pairs of ?
): End lookbehind condition
': Match a '

Check if first and last character contains given special char

I have input string
..-----''''''.......VAibhavs.sharma'..'-.'-.''-....''
I want to check if the first and last char place contains - or ' or ..
If yes then trim until we get name.
Expected output : VAibhavs.sharma
I am using like this.
while (
myString.charAt(0) == "." ||
myString.charAt(0) == "'" ||
myString.charAt(0) == "-" ||
myString.charAt(myString.length - 1) == "." ||
myString.charAt(myString.length - 1) == "'" ||
myString.charAt(myString.length - 1) == "-"
)
I know this is not correct way. How can I use regex?
I tried /^\'$. But this only checks or first char for a single special char.

You can use regular expression:
input = "..-----''''''.......VAibhavs.sharma'..'-.'-.''-....''"
output = input.replace(/^[-'\.]+/,"").replace(/[-'\.]+$/,"")
console.log(output)
[-'\.] ... -, ' or . character
+ ... one or more times
^ ... beginning of the string
$ ... end of the string
EDIT:
using match:
input = "..-----''''''.......VAibhavs.sharma'..'-.'-.''-....''"
output = input.match(/^[-'\.]+(.*?)[-'\.]+$/)[1]
console.log(output)
(...) ... (1st) group
.*? ... any chacter, zero or more times, ? means non-greedy
.match(...)[1] ... 1 means 1st group

There is already one accepted answer but still, this is how I would do.
var pattern = /\b[A-Za-z.]+\b/gm;
var str = "..-----''''''.......VAibhavs.sharma'..'-.'-.''-....''";
console.log(str.match(pattern));
// Output
// ["VAibhavs.sharma"]
\b is a zero-width word boundary. It matches positions where one side is a word character (usually a letter, digit or underscore) and the other side is not a word character (for instance, it may be the beginning of the string or a space character).

JS regex return string from url

I have the following URL structure:
https://api.bestschool.com/student/1102003120009/tests/json
I want to cut the student ID from the URL. So far I've came up with this:
/(student\/.*[^\/]*)/
which returns
student/1102003120009/tests/json
I only want the ID.

Your regex (student\/.*[^\/]*) matches and captures into Group 1 a literal sequence student/, then matches any characters other than a newline, 0 or more occurrences (.*) - that can match the whole line at once! - and then 0 or more characters other than /. It does not work because of .*. Also, a capturing group should be moved to the [^\/]* pattern.
You can use the following regex and grab Group 1 value:
student\/([^\/]*)
See regex demo
The regex matches student/ literally, and then matches and captures into Group 1 zero or more symbols other than /.
Alternatively, if you want to avoid using capturing, and assuming that the ID is always numeric and is followed by /tests/, you can use the following regex:
\d+(?=\/tests\/)
The \d+ matches 1 or more digits, and (?=\/tests\/) checks if right after the digits there is a /tests/ character sequence.
var re = /student\/([^\/]*)/;
var str = 'https://api.bestschool.com/student/1102003120009/tests/json';
var m = str.match(re);
if (m !== null) {
document.getElementById("r").innerHTML = "First method : " + m[1] + "<br/>";
}
var m2 = str.match(/\d+(?=\/tests\/)/);
if (m2 !== null) {
document.getElementById("r").innerHTML += "Second method: " + m2;
}
<div id="r"/>

JS string replace only replacing every other occurence

I have the following JS:
"a a a a".replace(/(^|\s)a(\s|$)/g, '$1')
I expect the result to be '', but am instead getting 'a a'. Can anyone explain to me what I am doing wrong?
Clarification: What I am trying to do is remove all occurrences of 'a' that are surronded by whitespace (i.e. a whole token)

It's because this regex /(^|\s)a(\s|$)/g match the previous char and the next char to each a
in string "a a a a" the regex matches :
"a " , then the string to check become "a a a"$ (but now the start of the string is not the beginning and there is not space before)
" a " (the third a) , then become "a"$ (that not match because no space before)
Edit:
Little bit tricky but working (without regex):
var a = "a a a a";
// Handle beginning case 'a '
var startI = a.indexOf("a ");
if (startI === 0){
var off = a.charAt(startI + 2) !== "a" ? 2 : 1; // test if "a" come next to keep the space before
a = a.slice(startI + off);
}
// Handle middle case ' a '
var iOf = -1;
while ((iOf = a.indexOf(" a ")) > -1){
var off = a.charAt(iOf + 3) !== "a" ? 3 : 2; // same here
a = a.slice(0, iOf) + a.slice(iOf+off, a.length);
}
// Handle end case ' a'
var endI = a.indexOf(" a");
if (endI === a.length - 2){
a = a.slice(0, endI);
}
a; // ""

First "a " matches.
Then it will try to match against "a a a", which will skip first a, and then match "a ".
Then it will try to match against "a", which will not match.
First match will be replaced to beginning of line. => "^"
Then we have "a" that didn't match => "a"
Second match will be replaced to " " => " "
Then we have "a" that didn't match => "a"
The result will be "a a".
To get your desired result you can do this:
"a a a a".replace(/(?:\s+a(?=\s))+\s+|^a\s+(?=[^a]|$|a\S)|^a|\s*a$/g, '')

As others have tried to point out, the issue is that the regex consumes the surrounding spaces as part of the match. Here's a [hopefully] more straight forward explanation of why that regex doesn't work as you expect:
First let's breakdown the regex, it says match the a space or start of string, followed by an 'a' followed by a space or the end of the string.
Now let's apply it to the string. I've added character indexes beneath the string to make things easier to talk about:
a a a a
0123456
The regex looks at the 0 index char, and finds an 'a' at that location, followed by a space at index 2. This is a match because it is the start of the string, followed by an a followed by a space. The length of our match is 2 (the 'a' and the space), so we consume two characters and start our next search at index 2.
Character 2 ('a') is neither a space nor the start of the string, and therefore it doesn't match the start of our regular expression, so we consume that character (without replacing it) and move on to the next.
Character 3 is a space, followed by an 'a' followed by another space, which is a match for our regex. We replace it with an empty string, consume the length of the match (3 characters - " a ") and move on to index 6.
Character 6 ('a') is neither a space nor the start of the string, and therefore it doesn't match the start of our regular expression, so we consume that character (without replacing it) and move on to the next.
Now we're at the end of the string, so we're done.
The reason why the regex #caeth suggested (/(^|\s+)a(?=\s|$)/g) works is because of the ?= quantifier. From the MDN Regexp Documentation:
Matches x only if x is followed by y. For example, /Jack(?=Sprat)/ matches "Jack" only if it is followed by "Sprat". /Jack(?=Sprat|Frost)/ matches "Jack" only if it is followed by "Sprat" or "Frost". However, neither "Sprat" nor "Frost" is part of the match results.
So, in this case, the ?= quantifier checks to see if the following character is a space, without actually consuming that character.

(^|\s)a(?=\s|$)
Try this.Replace by $1.See demo.
https://regex101.com/r/gQ3kS4/3

Use this instead:
"a a a a".replace(/(^|\s*)a(\s|$)/g, '$1')
With "* this you replace all the "a" occurrences
Greetings

Or you can just split the string up, filter it and glue it back:
"a ba sl lf a df a a df r a".split(/\s+/).filter(function (x) { return x != "a" }).join(" ")
>>> "ba sl lf df df r"
"a a a a".split(/\s+/).filter(function (x) { return x != "a" }).join(" ")
>>> ""
Or in ECMAScript 6:
"a ba sl lf a df a a df r a".split(/\s+/).filter(x => x != "a").join(" ")
>>> "ba sl lf df df r"
"a a a a".split(/\s+/).filter(x => x != "a").join(" ")
>>> ""
I assume that there is no leading and trailing spaces. You can change the filter to x && x != 'a' if you want to remove the assumption.

Escape string for use in Javascript regex [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Is there a RegExp.escape function in Javascript?
I am trying to build a javascript regex based on user input:
function FindString(input) {
var reg = new RegExp('' + input + '');
// [snip] perform search
}
But the regex will not work correctly when the user input contains a ? or * because they are interpreted as regex specials. In fact, if the user puts an unbalanced ( or [ in their string, the regex isn't even valid.
What is the javascript function to correctly escape all special characters for use in regex?

Short 'n Sweet (Updated 2021)
To escape the RegExp itself:
function escapeRegExp(string) {
return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
}
To escape a replacement string:
function escapeReplacement(string) {
return string.replace(/\$/g, '$$$$');
}
Example
All escaped RegExp characters:
escapeRegExp("All of these should be escaped: \ ^ $ * + ? . ( ) | { } [ ]");
>>> "All of these should be escaped: \\ \^ \$ \* \+ \? \. \( \) \| \{ \} \[ \] "
Find & replace a string:
var haystack = "I love $x!";
var needle = "$x";
var safeNeedle = escapeRegExp(needle); // "\\$x"
var replacement = "$100 bills"
var safeReplacement = escapeReplacement(replacement); // "$$100 bills"
haystack.replace(
new RegExp(safeNeedle, 'g'),
escapeReplacement(safeReplacement),
);
// "I love $100 bills!"
(NOTE: the above is not the original answer; it was edited to show the one from MDN. This means it does not match what you will find in the code in the below npm, and does not match what is shown in the below long answer. The comments are also now confusing. My recommendation: use the above, or get it from MDN, and ignore the rest of this answer. -Darren,Nov 2019)
Install
Available on npm as escape-string-regexp
npm install --save escape-string-regexp
Note
See MDN: Javascript Guide: Regular Expressions
Other symbols (~`!## ...) MAY be escaped without consequence, but are not required to be.
.
.
.
.
Test Case: A typical url
escapeRegExp("/path/to/resource.html?search=query");
>>> "\/path\/to\/resource\.html\?search=query"
The Long Answer
If you're going to use the function above at least link to this stack overflow post in your code's documentation so that it doesn't look like crazy hard-to-test voodoo.
var escapeRegExp;
(function () {
// Referring to the table here:
// https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/regexp
// these characters should be escaped
// \ ^ $ * + ? . ( ) | { } [ ]
// These characters only have special meaning inside of brackets
// they do not need to be escaped, but they MAY be escaped
// without any adverse effects (to the best of my knowledge and casual testing)
// : ! , =
// my test "~!##$%^&*(){}[]`/=?+\|-_;:'\",<.>".match(/[\#]/g)
var specials = [
// order matters for these
"-"
, "["
, "]"
// order doesn't matter for any of these
, "/"
, "{"
, "}"
, "("
, ")"
, "*"
, "+"
, "?"
, "."
, "\\"
, "^"
, "$"
, "|"
]
// I choose to escape every character with '\'
// even though only some strictly require it when inside of []
, regex = RegExp('[' + specials.join('\\') + ']', 'g')
;
escapeRegExp = function (str) {
return str.replace(regex, "\\$&");
};
// test escapeRegExp("/path/to/res?search=this.that")
}());