Develop Reference

Replace all “?” by “&” except first

I’d would to replace all “?” by “&” except the first one by javascript. I found some regular expressions but they didn’t work.
I have something like:
home/?a=1
home/?a=1?b=2
home/?a=1?b=2?c=3
And I would like:
home/?a=1
home/?a=1&b=2
home/?a=1&b=2&c=3
Someone know how to I can do it?
Thanks!

I don't think it's possible with regex but you can split the string and then join it back together, manually replacing the first occurance:
var split = 'home/?a=1?b=2'.split('?'); // [ 'home/', 'a=1', 'b=2' ]
var replaced = split[0] + '?' + split.slice(1).join('&') // 'home/?a=1&b=2'
console.log(replaced);

You could match from the start of the string not a question mark using a negated character class [^?]+ followed by matching a question mark and capture that in the first capturing group. In the second capturing group capture the rest of the string.
Use replace and pass a function as the second parameter where you return the first capturing group followed by the second capturing group where all the question marks are replaced by &
let strings = [
"home/?a=1",
"home/?a=1?b=2",
"home/?a=1?b=2?c=3"
];
strings.forEach((str) => {
let result = str.replace(/(^[^?]+\?)(.*)/, function(match, group1, group2) {
return group1 + group2.replace(/\?/g, '&')
});
console.log(result);
});

You can split it by "?" and then rewrap the array:
var string = "home/?a=1?b=2";
var str = string.split('?');
var new = str[0] + '?'; // text before first '?' and first '?'
for( var x = 1; x < str.length; x++ ) {
new = new + str[x];
if( x != ( str.length - 1 ) ) new = new + '&'; //to avoid place a '&' after the string
}

You can use /([^\/])\?/ as pattern in regex that match any ? character that isn't after / character.
var str = str.replace(/([^\/])\?/g, "$1&");
var str = "home/?a=1\nhome/?a=1?b=2\nhome/?a=1?b=2?c=3\n".replace(/([^\/])\?/g, "$1&");
console.log(str);

Extract word between '=' and '('

I have the following string
234234=AWORDHERE('sdf.'aa')
where I need to extract AWORDHERE.
Sometimes there can be space in between.
234234= AWORDHERE('sdf.'aa')
Can I do this with a regular expression?
Or should I do it manually by finding indexes?
The datasets are huge, so it's important to do it as fast as possible.

Try this regex:
\d+=\s?(\w+)\(
Check Demo
in Javascript it would like that:
var myString = "234234=AWORDHERE('sdf.'aa')";// or 234234= AWORDHERE('sdf.'aa')
var myRegexp = /\d+=\s?(\w+)\(/g;
var match = myRegexp.exec(myString);
console.log(match[1]); // AWORDHERE

You could do this at least three ways. You need to benchmark to see what's fastest.
Substring w/ indexes
function extract(from) {
var ixEq = from.indexOf("=");
var ixParen = from.indexOf("(");
return from.substring(ixEq + 1, ixParen);
}
.
Splits
function extract(from) {
var spEq = from.split("=");
var spParen = spEq[1].split("(");
return spParen[0];
}
Regex (demo)
Here is some sample regex you could use
/[^=]+=([^(]+).*/g
This says
[^=]+ - One or more character which is not an =
= - The = itself
( - creates a matching group so you can access your match in code
[^(]+ - One or more character which is not a (
) - closes the matching group
.* - Matches the rest of the line
the /g on the end tells it to perform the match on all lines.

Using look around you can search for string preceded by = and followed by ( as following.
Regex: (?<==)[A-Z ]+(?=\()
Explanation:
(?<==) checks if [A-Z ] is preceded by an =.
[A-Z ]+ matches your pattern.
(?=\() checks if matched pattern is followed by a (.
Regex101 Demo

var str = "234234= AWORDHERE('sdf.'aa')";
var regexp = /.*=\s+(\w+)\(.*\)/g;
var match = regexp.exec(str);
alert( match[1] );

I made my solution for this just a little more general than you asked for, but I don't think it takes much more time to execute. I didn't measure. If you need greater efficiency than this provides, comment and I or someone else can help you with that.
Here's what I did, using the command prompt of node:
> var s = "234234= AWORDHERE('sdf.'aa')"
undefined
> var a = s.match(/(\w+)=\s*(\w+)\s*\(.*/)
undefined
> a
[ '234234= AWORDHERE(\'sdf.\'aa\')',
'234234',
'AWORDHERE',
index: 0,
input: '234234= AWORDHERE(\'sdf.\'aa\')' ]
>
As you can see, this matches the number before the = in a[1], and it matches the AWORDHERE name as you requested in a[2]. This will work with any number (including zero) spaces before and/or after the =.

Javascript Remove strings in beginning and end

base on the following string
...here..
..there...
.their.here.
How can i remove the . on the beginning and end of string like the trim that removes all spaces, using javascript
the output should be
here
there
their.here

These are the reasons why the RegEx for this task is /(^\.+|\.+$)/mg:
Inside /()/ is where you write the pattern of the substring you want to find in the string:
/(ol)/ This will find the substring ol in the string.
var x = "colt".replace(/(ol)/, 'a'); will give you x == "cat";
The ^\.+|\.+$ in /()/ is separated into 2 parts by the symbol | [means or]
^\.+ and \.+$
^\.+ means to find as many . as possible at the start.
^ means at the start; \ is to escape the character; adding + behind a character means to match any string containing one or more that character
\.+$ means to find as many . as possible at the end.
$ means at the end.
The m behind /()/ is used to specify that if the string has newline or carriage return characters, the ^ and $ operators will now match against a newline boundary, instead of a string boundary.
The g behind /()/ is used to perform a global match: so it find all matches rather than stopping after the first match.
To learn more about RegEx you can check out this guide.

Try to use the following regex
var text = '...here..\n..there...\n.their.here.';
var replaced = text.replace(/(^\.+|\.+$)/mg, '');

Here is working Demo
Use Regex /(^\.+|\.+$)/mg
^ represent at start
\.+ one or many full stops
$ represents at end
so:
var text = '...here..\n..there...\n.their.here.';
alert(text.replace(/(^\.+|\.+$)/mg, ''));

Here is an non regular expression answer which utilizes String.prototype
String.prototype.strim = function(needle){
var first_pos = 0;
var last_pos = this.length-1;
//find first non needle char position
for(var i = 0; i<this.length;i++){
if(this.charAt(i) !== needle){
first_pos = (i == 0? 0:i);
break;
}
}
//find last non needle char position
for(var i = this.length-1; i>0;i--){
if(this.charAt(i) !== needle){
last_pos = (i == this.length? this.length:i+1);
break;
}
}
return this.substring(first_pos,last_pos);
}
alert("...here..".strim('.'));
alert("..there...".strim('.'))
alert(".their.here.".strim('.'))
alert("hereagain..".strim('.'))
and see it working here : http://jsfiddle.net/cettox/VQPbp/

Slightly more code-golfy, if not readable, non-regexp prototype extension:
String.prototype.strim = function(needle) {
var out = this;
while (0 === out.indexOf(needle))
out = out.substr(needle.length);
while (out.length === out.lastIndexOf(needle) + needle.length)
out = out.slice(0,out.length-needle.length);
return out;
}
var spam = "this is a string that ends with thisthis";
alert("#" + spam.strim("this") + "#");
Fiddle-ige

Use RegEx with javaScript Replace
var res = s.replace(/(^\.+|\.+$)/mg, '');

We can use replace() method to remove the unwanted string in a string
Example:
var str = '<pre>I'm big fan of Stackoverflow</pre>'
str.replace(/<pre>/g, '').replace(/<\/pre>/g, '')
console.log(str)
output:
Check rules on RULES blotter

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);

Regular expression to parse jQuery-selector-like string

text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
regex = /(.*?)\.filter\((.*?)\)/;
matches = text.match(regex);
log(matches);
// matches[1] is '#container a'
//matchss[2] is '.top'
I expect to capture
matches[1] is '#container a'
matches[2] is '.top'
matches[3] is '.bottom'
matches[4] is '.middle'
One solution would be to split the string into #container a and rest. Then take rest and execute recursive exec to get item inside ().
Update: I am posting a solution that does work. However I am looking for a better solution. Don't really like the idea of splitting the string and then processing
Here is a solution that works.
matches = [];
var text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
var regex = /(.*?)\.filter\((.*?)\)/;
var match = regex.exec(text);
firstPart = text.substring(match.index,match[1].length);
rest = text.substring(matchLength, text.length);
matches.push(firstPart);
regex = /\.filter\((.*?)\)/g;
while ((match = regex.exec(rest)) != null) {
matches.push(match[1]);
}
log(matches);
Looking for a better solution.

This will match the single example you posted:
<html>
<body>
<script type="text/javascript">
text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
matches = text.match(/^[^.]*|\.[^.)]*(?=\))/g);
document.write(matches);
</script>
</body>
</html>
which produces:
#container a,.top,.bottom,.middle
EDIT
Here's a short explanation:
^ # match the beginning of the input
[^.]* # match any character other than '.' and repeat it zero or more times
#
| # OR
#
\. # match the character '.'
[^.)]* # match any character other than '.' and ')' and repeat it zero or more times
(?= # start positive look ahead
\) # match the character ')'
) # end positive look ahead
EDIT part II
The regex looks for two types of character sequences:
one ore more characters starting from the start of the string up to the first ., the regex: ^[^.]*
or it matches a character sequence starting with a . followed by zero or more characters other than . and ), \.[^.)]*, but must have a ) ahead of it: (?=\)). This last requirement causes .filter not to match.

You have to iterate, I think.
var head, filters = [];
text.replace(/^([^.]*)(\..*)$/, function(_, h, rem) {
head = h;
rem.replace(/\.filter\(([^)]*)\)/g, function(_, f) {
filters.push(f);
});
});
console.log("head: " + head + " filters: " + filters);
The ability to use functions as the second argument to String.replace is one of my favorite things about Javascript :-)

You need to do several matches repeatedly, starting where the last match ends (see while example at https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Global_Objects/RegExp/exec):
If your regular expression uses the "g" flag, you can use the exec method multiple times to find successive matches in the same string. When you do so, the search starts at the substring of str specified by the regular expression's lastIndex property. For example, assume you have this script:
var myRe = /ab*/g;
var str = "abbcdefabh";
var myArray;
while ((myArray = myRe.exec(str)) != null)
{
var msg = "Found " + myArray[0] + ". ";
msg += "Next match starts at " + myRe.lastIndex;
print(msg);
}
This script displays the following text:
Found abb. Next match starts at 3
Found ab. Next match starts at 9
However, this case would be better solved using a custom-built parser. Regular expressions are not an effective solution to this problem, if you ask me.

var text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
var result = text.split('.filter');
console.log(result[0]);
console.log(result[1]);
console.log(result[2]);
console.log(result[3]);

text.split() with regex does the trick.
var text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
var parts = text.split(/(\.[^.()]+)/);
var matches = [parts[0]];
for (var i = 3; i < parts.length; i += 4) {
matches.push(parts[i]);
}
console.log(matches);