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regex validating if string ends with specific set of words [duplicate]

I'm creating a javascript regex to match queries in a search engine string. I am having a problem with alternation. I have the following regex:
.*baidu.com.*[/?].*wd{1}=
I want to be able to match strings that have the string 'word' or 'qw' in addition to 'wd', but everything I try is unsuccessful. I thought I would be able to do something like the following:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
but it does not seem to work.

replace [wd|word|qw] with (wd|word|qw) or (?:wd|word|qw).
[] denotes character sets, () denotes logical groupings.

Your expression:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
does need a few changes, including [wd|word|qw] to (wd|word|qw) and getting rid of the redundant {1}, like so:
.*baidu.com.*[/?].*(wd|word|qw)=
But you also need to understand that the first part of your expression (.*baidu.com.*[/?].*) will match baidu.com hello what spelling/handle????????? or hbaidu-com/ or even something like lkas----jhdf lkja$##!3hdsfbaidugcomlaksjhdf.[($?lakshf, because the dot (.) matches any character except newlines... to match a literal dot, you have to escape it with a backslash (like \.)
There are several approaches you could take to match things in a URL, but we could help you more if you tell us what you are trying to do or accomplish - perhaps regex is not the best solution or (EDIT) only part of the best solution?

Combining 2 regexes, one with exact match using OR operator

I am trying to combine:
^[a-zA-Z.][a-zA-Z'\\- .]*$
with
(\W|^)first\sname(\W|$)
which should check for the exact phrase, first name, if that is correct. It should match either the first regex OR the second exact match. I tried this, but appears invalid:
^(([a-zA-Z.][a-zA-Z'\\- .]*$)|((\W|^)first\sname(\W|$))
This is in javascript btw.

Combining regular expressions generally can be done simply in the following way:
Regex1 + Regex2 = (Regex1|Regex2)
^[a-zA-Z.][a-zA-Z'\\- .]*$
+ (\W|^)first\sname(\W|$) =
(^[a-zA-Z.][a-zA-Z'\\- .]*$|(\W|^)first\sname(\W|$))
Because some SO users have a hard time understand the math analogy, here's a full word explanation.
If you have a regex with content REGEX1 and a second regex with content REGEX2 and you want to combine them in the way that was described by OP in his question, a simple way to do this without optimization is the following.
(REGEX1|REGEX2)
Where you surround both regular expressions with parenthesis and divide the two with |.
Your regex would be the following:
(^[a-zA-Z.][a-zA-Z'\\- .]*$|(\W|^)first\sname(\W|$))
Your first regex has an error in it, though, that makes it invalid. Try this instead.
(^[a-zA-Z.][a-zA-Z'\- .]*$|(\W|^)first\sname(\W|$))
You had \\ in the second character class where you wanted \

The problem is that the first regex is messed up. You don't need to double escape characters. Therefore
\\-
Will match an ascii character between \(92) and (32). Remove one of the slashes.
Reference

Unable to find a string matching a regex pattern

While trying to submit a form a javascript regex validation always proves to be false for a string.
Regex:- ^(([a-zA-Z]:)|(\\\\{2}\\w+)\\$?)(\\\\(\\w[\\w].*))+(.jpeg|.JPEG|.jpg|.JPG)$
I have tried following strings against it
abc.jpg,
abc:.jpg,
a:.jpg,
a:asdas.jpg,
What string could possible match this regex ?

This regex won't match against anything because of that $? in the middle of the string.
Apparently using the optional modifier ? on the end string symbol $ is not correct (if you paste it on https://regex101.com/ it will give you an error indeed). If the javascript parser ignores the error and keeps the regex as it is this still means you are going to match an end string in the middle of a string which is supposed to continue.
Unescaped it was supposed to match a \$ (dollar symbol) but as it is written it won't work.
If you want your string to be accepted at any cost you can probably use Firebug or a similar developer tool and edit the string inside the javascript code (this, assuming there's no server side check too and assuming it's not wrong aswell). If you ignore the $? then a matching string will be \\\\w\\\\ww.jpg (but since the . is unescaped even \\\\w\\\\ww%jpg is a match)
Of course, I wrote this answer assuming the escaping is indeed the one you showed in the question. If you need to find a matching pattern for the correctly escaped one ^(([a-zA-Z]:)|(\\{2}\w+)\$?)(\\(\w[\w].*))+(\.jpeg|\.JPEG|\.jpg|\.JPG)$ then you can use this tool to find one http://fent.github.io/randexp.js/ (though it will find weird matches). A matching pattern is c:\zz.jpg

If you are just looking for a regular expression to match what you got there, go ahead and test this out:
(\w+:?\w*\.[jpe?gJPE?G]+,)
That should match exactly what you are looking for. Remove the optional comma at the end if you feel like it, of course.

If you remove escape level, the actual regex is
^(([a-zA-Z]:)|(\\{2}\w+)\$?)(\\(\w[\w].*))+(.jpeg|.JPEG|.jpg|.JPG)$
After ^start the first pipe (([a-zA-Z]:)|(\\{2}\w+)\$?) which matches an alpha followed by a colon or two backslashes followed by one or more word characters, followed by an optional literal $. There is some needless parenthesis used inside.
The second part (\\(\w[\w].*))+ matches a backslash, followed by two word characters \w[\w] which looks weird because it's equivalent to \w\w (don't need a character class for second \w). Followed by any amount of any character. This whole thing one or more times.
In the last part (.jpeg|.JPEG|.jpg|.JPG) one probably forgot to escape the dot for matching a literal. \. should be used. This part can be reduced to \.(JPE?G|jpe?g).
It would match something like
A:\12anything.JPEG
\\1$\anything.jpg
Play with it at regex101. A better readable could be
^([a-zA-Z]:|\\{2}\w+\$?)(\\\w{2}.*)+\.(jpe?g|JPE?G)$
Also read the explanation on regex101 to understand any pattern, it's helpful!

Javascript: equivalent regex for negative lookbehind?

I want to write a regular expression that will captures all double quotes " in a string, except for those that are escaped.
For example, in the following String will return the first quote only:
"HELLO\"\"\"
but the following one will return 3 matches:
"HELLO\"\""\""
I have used the following expression, but since in JavaScript there is no negative lookbehind I am stuck:
(?<!\\)"
I have looked at similar questions but most provide a programmatic interface. I don't want to use a programmatic interface because I am using Ace editor and the simplest way to go around my problem is to define this regex.
I suppose there is no generic alternative, since I have tried the alternatives proposed to the similar questions, but non of them exactly matched my case.
Thanks for your answers!

You can use this workaround:
(^|[^\\])"
" only if preceded by any char but a \ or the beginning of the string (^).
But be careful, this matches two chars: the " AND the preceding character (unless in the start-of-the-string case). In other words, if you wan't to replace all these " by ' for example, you'll need:
theString.replace(/(^|[^\\])"/g, "$1'")

The code I assume you are trying to run:
while ( matcher = /(?<!\\)"/g.exec(theString) ) {
// do stuff. matcher[0] is double quotes (that don't follow a backslash)
}
In JavaScript, using this guide to JS lookbehinds:
while ( matcher = /(\\)?"/g.exec(theString) ) {
if (!matcher[1]) {
// do stuff. matcher[0] is double quotes (that don't follow a backslash)
}
}
This looks for double quotes (") that optionally follow a backslash (\) but then doesn't act when it actually does follow a backslash.
If you were merely trying to count the number of unescaped double-quotes, the "do stuff" line could be count++.

Javascript Regex Object Doesn't Recognize {n,m}

I'm trying to write a regular expression in JS to recognize any digit up to seven times, followed by a "-" followed by 2 digits followed by "-" followed by a single digit. This is the simple regex I have:
/\d{1,7}-\d{2}-\d/g
This should match strings like:
123-12-7
1-12-7
1234567-12-7
but not 12345678-12-1
However, the above is returning true. The regex returns true when there is any number of digit in the first group.
Does the JavaScript Regex object not support {n,m}?
Here is an example of what I am talking about.
var pattern = new RegExp(/\d{1,7}-\d{2}-\d/);
alert(pattern.test("12345678-13-1"));
http://jsfiddle.net/XTRAc/1/ live example

It matches 2345678-13-1. You need to anchor it to the beginning and end of your string:
/^\d{1,7}-\d{2}-\d$/
Note though, that (as Rocket Hazmat pointed out) you do not need to use the RegExp constructor if you use a regex literal (something without string quotes).
JSFiddle

It does support the {min,max}-syntax, but .match and .test() try to find matching substrings. You will have to include start and end anchors. Also notice that you should either use the RegExp constructor to build a regex from a string or a regex literal, but not both (see MDN: creating regexes).
/^\d{1,7}-\d{2}-\d$/
new RegExp("^\\d{1,7}-\\d{2}-\\d$") // the worse choice

You are constructing your regex incorrectly. Try this (note the anchors, which ensure the string consists of nothing but your pattern):
var pattern= /^\d{1,7}-\d{2}-\d$/;
Otherwise subsets of the existing string will match your regex.

If you need to validate entire input string, use regex pattern
/^\d{1,7}-\d{2}-\d$/
If you need to validate entire line of input string, use regex pattern
/^\d{1,7}-\d{2}-\d$/mg
If you need to find matches within input string, use regex pattern
/(?:\D|^)(\d{1,7}-\d{2}-\d)(?!\d)/g
...and use $1 as a result.

It does support the {n,m} part, the problem here is that your example matches 2345678, so you would need a way of matching the character before the first set of digits