We Keep Coding

Using RegExp to substring a string at the position of a special character

Suppose I have a sting like this: ABC5DEF/G or it might be ABC5DEF-15 or even just ABC5DEF, it could be shorter AB7F, or AB7FG/H.
I need to create a javascript variable that contains the substring only up to the '/' or the '-'. I would really like to use an array of values to break at. I thought maybe to try something like this.
...
var srcMark = array( '/', '-' );
var whereAt = new RegExp(srcMark.join('|')).test.str;
alert("whereAt= "+whereAt);
...
But this returns an error: ReferenceError: Can't find variable: array
I suspect I'm defining my array incorrectly but trying a number of other things I've been no more successful.
What am I doing wrong?

Arrays aren't defined like that in JavaScript, the easiest way to define it would be with:
var srcMark = ['/','-'];
Additionally, test is a function so it must be called as such:
whereAt = new RegExp(srcMark.join('|')).test(str);
Note that test won't actually tell you where, as your variable suggests, it will return true or false. If you want to find where the character is, use String.prototype.search:
str.search(new RegExp(srcMark.join('|'));
Hope that helps.

You need to use the split method:
var srcMark = Array.join(['-','/'],'|'); // "-|/" or
var regEx = new RegExp(srcMark,'g'); // /-|\//g
var substring = "222-22".split(regEx)[0] // "222"
"ABC5DEF/G".split(regEx)[0] // "ABC5DEF"

From whatever i could understand from your question, using this RegExp /[/-]/ in split() function will work.
EDIT:
For splitting the string at all special characters you can use new RegExp(/[^a-zA-Z0-9]/) in split() function.
var arr = "ABC5DEF/G";
var ans = arr.split(/[/-]/);
console.log(ans[0]);
arr = "ABC5DEF-15";
ans = arr.split(/[/-]/);
console.log(ans[0]);
// For all special characters
arr = "AB7FG/H";
ans = arr.split(new RegExp(/[^a-zA-Z0-9]/));
console.log(ans[0]);

You can use regex with String.split.
It will look something like that:
var result = ['ABC5DEF/G',
'ABC5DEF-15',
'ABC5DEF',
'AB7F',
'AB7FG/H'
].map((item) => item.split(/\W+/));
console.log(result);
That will create an Array with all the parts of the string, so each item[0] will contain the text till the / or - or nothing.

If you want the position of the special character (non-alpha-numeric) you can use a Regular Expression that matches any character that is not a word character from the basic Latin alphabet. Equivalent to [^A-Za-z0-9_], that is: \W
var pattern = /\W/;
var text = 'ABC5DEF/G';
var match = pattern.exec(text);
var position = match.index;
console.log('character: ', match[0]);
console.log('position: ', position);

get all but last occurrences of a pattern in javascript

Given the following patterns:
"profile[foreclosure_defenses_attributes][0][some_text]"
"something[something_else_attributes][0][hello_attributes][0][other_stuff]"
I am able to extract the last part using non-capturing groups:
var regex = /(?:\w+(\[\w+\]\[\d+\])+)(\[\w+\])/;
str = "profile[foreclosure_defenses_attributes][0][properties_attributes][0][other_stuff]";
match = regex.exec(str);
["profile[foreclosure_defenses_attributes][0][properties_attributes][0][other_stuff]", "[properties_attributes][0]", "[other_stuff]"]
However, I want to be able to get everything but the last part. In other words, everything but [some_text] or [other_stuff].
I cannot figure out how to do this with noncapturing groups. How else can I achieve this?

Something like?
shorter, and matches from the back if you can have more of the [] items.
var regex = /(.*)(?:\[\w+\])$/;
var a = "something[something_else_attributes][0][hello_attributes][0][other_stuff11][other_stuff22][other_stuff33][other_stuff44]".match(regex)[1];
a;
or using replace, though less performant.
var regex = /(.*)(?:\[\w+\])$/;
var a = "something[something_else_attributes][0][hello_attributes][0][other_stuff11][other_stuff22][other_stuff33][other_stuff44]".replace(regex, function(_,$1){ return $1});
a;

If those really are your strings:
var regex = /(.*)\[/;

Find last string after delimiter: Javascript

I have a url with many delimiters '/'.
I want to find the string after the last delimiter. How can I write a javascript code?
for eg if my url is
localhost/sample/message/invitation/create/email
I want to display 'email' as my output.

var last = input.split("/").pop();
Simples!

Splitting on a regex that matches spaces or hyphens and taking the last element
var lw = function(v) {
return (""+v).replace(/[\s-]+$/,'').split(/[\s-]/).pop();
};
lw('This is a test.'); // returns 'test.'
lw('localhost/sample/message/invitation/create/email,'); // returns 'email,'

var url="localhost/sample/message/invitation/create/email";
url.split("/").pop()
or
var last=$(url.split("/")).last();

Usng simple regex
var str = "localhost/sample/message/invitation/create/email";
var last = str.match(/[^/]*$/)[0]";
Above regex return all character after last "/"

Javascript RegExp match & Multiple backreferences

I'm having trouble trying to use multiple back references in a javascript match so far I've got: -
function newIlluminate() {
var string = "the time is a quarter to two";
var param = "time";
var re = new RegExp("(" + param + ")", "i");
var test = new RegExp("(time)(quarter)(the)", "i");
var matches = string.match(test);
$("#debug").text(matches[1]);
}
newIlluminate();
#Debug when matching the Regex 're' prints 'time' which is the value of param.
I've seen match examples where multiple back references are used by wrapping the match in parenthesis however my match for (time)(quarter)... is returning null.
Where am I going wrong? Any help would be greatly appreciated!

Your regex is literally looking for timequarterthe and splitting the match (if it finds one) into the three backreferences.
I think you mean this:
var test = /time|quarter|the/ig;

Your regex test simply doesn't match the string (as it does not contain the substring timequarterthe). I guess you want alternation:
var test = /time|quarter|the/ig; // does not even need a capturing group
var matches = string.match(test);
$("#debug").text(matches!=null ? matches.join(", ") : "did not match");

Regex to grab strings between square brackets

I have the following string: pass[1][2011-08-21][total_passes]
How would I extract the items between the square brackets into an array? I tried
match(/\[(.*?)\]/);
var s = 'pass[1][2011-08-21][total_passes]';
var result = s.match(/\[(.*?)\]/);
console.log(result);
but this only returns [1].
Not sure how to do this.. Thanks in advance.

You are almost there, you just need a global match (note the /g flag):
match(/\[(.*?)\]/g);
Example: http://jsfiddle.net/kobi/Rbdj4/
If you want something that only captures the group (from MDN):
var s = "pass[1][2011-08-21][total_passes]";
var matches = [];
var pattern = /\[(.*?)\]/g;
var match;
while ((match = pattern.exec(s)) != null)
{
matches.push(match[1]);
}
Example: http://jsfiddle.net/kobi/6a7XN/
Another option (which I usually prefer), is abusing the replace callback:
var matches = [];
s.replace(/\[(.*?)\]/g, function(g0,g1){matches.push(g1);})
Example: http://jsfiddle.net/kobi/6CEzP/

var s = 'pass[1][2011-08-21][total_passes]';
r = s.match(/\[([^\]]*)\]/g);
r ; //# => [ '[1]', '[2011-08-21]', '[total_passes]' ]
example proving the edge case of unbalanced [];
var s = 'pass[1]]][2011-08-21][total_passes]';
r = s.match(/\[([^\]]*)\]/g);
r; //# => [ '[1]', '[2011-08-21]', '[total_passes]' ]

add the global flag to your regex , and iterate the array returned .
match(/\[(.*?)\]/g)

I'm not sure if you can get this directly into an array. But the following code should work to find all occurences and then process them:
var string = "pass[1][2011-08-21][total_passes]";
var regex = /\[([^\]]*)\]/g;
while (match = regex.exec(string)) {
alert(match[1]);
}
Please note: i really think you need the character class [^\]] here. Otherwise in my test the expression would match the hole string because ] is also matches by .*.

'pass[1][2011-08-21][total_passes]'.match(/\[.+?\]/g); // ["[1]","[2011-08-21]","[total_passes]"]
Explanation
\[ # match the opening [
Note: \ before [ tells that do NOT consider as a grouping symbol.
.+? # Accept one or more character but NOT greedy
\] # match the closing ] and again do NOT consider as a grouping symbol
/g # do NOT stop after the first match. Do it for the whole input string.
You can play with other combinations of the regular expression
https://regex101.com/r/IYDkNi/1

[C#]
string str1 = " pass[1][2011-08-21][total_passes]";
string matching = #"\[(.*?)\]";
Regex reg = new Regex(matching);
MatchCollection matches = reg.Matches(str1);
you can use foreach for matched strings.