Develop Reference

Use regex to remove "public://" from string

Im lost in part of this.
I want to remove the public:// in every link of an image like public://china-taxi_4.jpg
I have tried this but returns null:
var _img = 'public://china-taxi_4.jpg';
var regex = /(public:)(\/\w+)/;
var matches = _img.match(regex);
console.log(matches);
Hope you can help.

I want to remove the 'public://' in every link of an image.
> var img = 'public://china-taxi_4.jpg';
> img.replace(/public:\/\/(?=\S+?\.jpg(?:\s|$))/, "")
'china-taxi_4.jpg'
It removes the word public:// only in the strings which ends with .jpg

You are removing a literal string, not a regular expression. So try:
var _img = 'public://china-taxi_4.jpg';
var result = _img.replace("public://","");
console.log(result);
Regexes are for matching complex expressions.

I think you're missing a slash, try:
var _img = 'public://china-taxi_4.jpg';
var regex = /(public:)(\/\/\w+)/;
var matches = _img.match(regex);
console.log(matches);

From Mozilla Developer Network String.prototype.replace():
Example: Defining the regular expression in replace()
In the following example, the regular expression is defined in
replace() and includes the ignore case flag.
var str = 'Twas the night before Xmas...';
var newstr = str.replace(/xmas/i, 'Christmas');
console.log(newstr);
This prints:
'Twas the night before Christmas...'
To match the beginning of a string, use ^
To escape characters that have special meaning in regexp like : and / so that regexp will match these literally, prepend \
This suggests:
var _img = 'public://china-taxi_4.jpg';
var newimg = _img.replace(/^public\:\/\//i, '');
Tested and working in chrome browser console window.
Note: This answer also matches an earlier comment by #dystroy, so I have marked it CW.

var re = /(public:)(\/\/[\w-.]+)/g;
See demo.
http://regex101.com/r/rA7aS3/9

Passing Parameter into function match

I am using the function match for a search engine, so whenever a user types a search-string I take that string and use the match function on an array containing country names, but it doesn't seem to work.
For example if I do :
var string = "algeria";
var res = string.match(/alge/g); //alge is what the user would have typed in the search bar
alert(res);
I get a string res = "alge": //thus verifying that alge exists in algeria
But if I do this, it returns null, why? and how can I make it work?
var regex = "/alge/g";
var string = "algeria";
var res = string.match(regex);
alert(res);

To make a regex from a string, you need to create a RegExp object:
var regex = new RegExp("alge", "g");
(Beware that unless your users will be typing actual regular expressions, you'll need to escape any characters that have special meaning within regular expressions - see Is there a RegExp.escape function in Javascript? for ways to do this.)

You don't need quotes around the regex:
var regex = /alge/g;

Remove the quotes around the regex.
var regex = /alge/g;
var string = "algeria";
var res = string.match(regex);
alert(res);

found the answer, the match function takes a regex object so have to do
var regex = new RegExp(string, "g");
var res = text.match(regex);
This works fine

Regex working fine in C# but not in Javascript

I have the following javascript code:
var markdown = "I have \(x=1\) and \(y=2\) and even \[z=3\]"
var latexRegex = new RegExp("\\\[.*\\\]|\\\(.*\\\)");
var matches = latexRegex.exec(markdown);
alert(matches[0]);
matches has only matches[0] = "x=1 and y=2" and should be:
matches[0] = "\(x=1\)"
matches[1] = "\(y=2\)"
matches[2] = "\[z=3\]"
But this regex works fine in C#.
Any idea why this happens?
Thank You,
Miguel

Specify g flag to match multiple times.
Use String.match instead of RegExp.exec.
Using regular expression literal (/.../), you don't need to escape \.
* matches greedily. Use non-greedy version: *?
var markdown = "I have \(x=1\) and \(y=2\) and even \[z=3\]"
var latexRegex = /\[.*?\]|\(.*?\)/g;
var matches = markdown.match(latexRegex);
matches // => ["(x=1)", "(y=2)", "[z=3]"]

Try non-greedy: \\\[.*?\\\]|\\\(.*?\\\). You need to also use a loop if using the .exec() method like so:
var res, matches = [], string = 'I have \(x=1\) and \(y=2\) and even \[z=3\]';
var exp = new RegExp('\\\[.*?\\\]|\\\(.*?\\\)', 'g');
while (res = exp.exec(string)) {
matches.push(res[0]);
}
console.log(matches);

Try using the match function instead of the exec function. exec only returns the first string it finds, match returns them all, if the global flag is set.
var markdown = "I have \(x=1\) and \(y=2\) and even \[z=3\]";
var latexRegex = new RegExp("\\\[.*\\\]|\\\(.*\\\)", "g");
var matches = markdown.match(latexRegex);
alert(matches[0]);
alert(matches[1]);
If you don't want to get \(x=1\) and \(y=2\) as a match, you will need to use non-greedy operators (*?) instead of greedy operators (*). Your RegExp will become:
var latexRegex = new RegExp("\\\[.*?\\\]|\\\(.*?\\\)");

regex: any string between two slashes first of them is prefixed with a defined string

I'd like to get the talker name of some mp3s files paths such as the following:
/assets/audio/James_Lee/001.mp3
/assets/audio/Marc_Smith/001.mp3
/aasets/audio/blahblah/001.mp3
In the previous example we note that each talker name is surrounded by two slashes where the first of them is prefixed with the word audio. I need a pattern that matches names like the example above using javascript.
I tried at http://regexpal.com/ :
audio/.*/
but it only matches *audio/The_name/* where I need *The_name* only. The other thing I don't know how could I use such patterns with javascript replace().

This will get your the name: (?<=\/assets\/audio\/).*(?=\/)
Here's the regex in use: http://regexr.com?34747
Considering Javascript, you could do this:
var string = "/assets/audio/James_Lee/001.mp3";
var name = string.replace(/^.*\/audio\/|\/[\d]+\..*$/g, '');

Try this:
var str = "/assets/audio/James_Lee/001.mp3\n/assets/audio/Marc_Smith/001.mp3";
var pattern = /audio\/(.+?)\//g;
var match;
var matches = [];
while ((match = pattern.exec(str)) !== null){
matches.push(match[1]);
}
console.log(matches);
// If you want a string with only the names, you can re-combine the matches
str = matches.join('\n');

how about this?
str.replace(/.*audio\/([^\/]*)\/.*/,"$1")

Shouldn't this RegExp work?

testString = "something://something/task?type=Checkin";
patt = new RegExp("something\/(\w*)\?");
match = patt.exec(testString);
document.querySelector('#resultRegexp').innerHTML = match[1];
I want to capture task So shouldn't this RegExp work?
I am grabbing any alphanumeric character up until the question mark... and capturing it.
http://jsfiddle.net/h4yhc/2/

You would need to escape the slash in regex literals, and the backslash in string literals which you create regexes from:
var patt = /something\/(\w*)\?/g;
// or
var patt = new RegExp("something/(\\w*)\\?", 'g');
I strongly recommend the first version, it is more readable.

I think this would be enough: (\w*)\?, since / is not captured by \w and the only ? in the string is after your target string.

This is what you need:
patt = new RegExp(".*/(\\w*)\\?");
http://jsfiddle.net/FJcfd/

try with this: var pat = /something:\/\/(?:[^\/]+\/)+(\w+)\?(\w+=\w+)/;
it can match string such as:
something://something/task?type=Checkin
something://something/foo/task?type=Checkin
something://something/foo/bar/task1?type3=Checkin4