Develop Reference

replace a string partially with something else

lets say I have this image address like
https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b
how is it possible to replace FILE_NAME.jpg with THUMB_FILE_NAME.jpg
Note: FILE_NAME and THUMB_FILE_NAME are not static and fix.
the FILE_NAME is not fixed and I can't use string.replace method.
eventually I don't know the File_Name

Use replace
.replace(/(?<=\/)[^\/]*(?=(.jpg))/g, "THUMB_FILE_NAME")
or if you want to support multiple formats
.replace(/(?<=\/)[^\/]*(?=(.(jpg|png|jpeg)))/g, "THUMB_FILE_NAME")
Demo
var output = "https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b".replace(/(?<=\/)[^\/]*(?=(.jpg))/g, "THUMB_FILE_NAME");
console.log( output );
Explanation
(?<=\/) matches / but doesn't remember the match
[^\/]* matches till you find next /
(?=(.jpg) ensures that match ends with .jpg
To match the FILE_NAME, use
.match(/(?<=\/)[^\/]*(?=(.(jpg|png|jpeg)))/g)

var pattern = /[\w-]+\.(jpg|png|txt)/
var c = 'https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b
'
c.replace(pattern, 'YOUR_FILE_NAME.jpg')
you can add any format in the pipe operator

You can use the String's replace method.
var a = "https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b";
a = a.replace('FILE_NAME', 'THUMB_FILE_NAME');

If you know the format, you can use the split and join to replace the FILE_NAME.
let str = "https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b";
let str_pieces = str.split('/');
let str_last = str_pieces[str_pieces.length - 1];
let str_last_pieces = str_last.split('?');
str_last_pieces[0] = 'THUMB_' + str_last_pieces[0];
str_last = str_last_pieces.join('?');
str_pieces[str_pieces.length - 1] = str_last;
str = str_pieces.join('/');

Extract string between characters at the end of a URL

I have the following example url: http://example.com/this/is/the/end/
I need to extract the last piece of the url, between the last two /
There may be characters after the last / but it's always between the last two / that I need.
This is what I'm trying, I think it's pretty close but it only returns the d of end
How can I extract the full end?
Javascript
var str = 'http://example.com/this/is/the/end/';
var string = str.substring(str.lastIndexOf("/")-1,str.lastIndexOf("/"));
Here's a fiddle

Use lastIndexOf with start from index as second parameter to extract the text between the two slashes.
var str = 'http://example.com/this/is/the/end/';
var lastIndex = str.lastIndexOf('/');
var string = str.substring(str.lastIndexOf("/", lastIndex - 1) + 1, lastIndex);
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ : Get the last `/` index by starting search from `lastIndex - 1` index.
console.log(string);
You can also use string and array functions as follow.
var str = 'http://example.com/this/is/the/end/';
var string = str.split('/').slice(-2)[0];
console.log(string);
Also, regex can be used.
Regex Demo and Explanation
var str = 'http://example.com/this/is/the/end/';
var string = str.match(/\/(\w+)\/[^\/]*?$/)[1];
console.log(string);

This is a good place to use regular expressions:
Regex Live Demo
var str = 'http://example.com/this/is/the/end/';
var re = /\/([^\/]*)\/[^\/]*$/;
// \/ - look for /
// ([^\/]*) - capture zero or more characters that aren't a /
// \/ - look for last /
// [^\/]* - look for more chars that aren't /
// $ - match the end of the string
var last = re.exec(str)[1];
console.log(last); //end

You can simply split and slice
'http://example.com/this/is/the/end/'.split('/').slice(-2)[0]

Split the date! (It's a string actually)

I want to split this kind of String :
"14:30 - 19:30" or "14:30-19:30"
inside a javascript array like ["14:30", "19:30"]
so I have my variable
var stringa = "14:30 - 19:30";
var stringes = [];
Should i do it with regular expressions? I think I need an help

You can just use str.split :
var stringa = "14:30 - 19:30";
var res = str.split("-");

If you know that the only '-' present will be the delimiter, you can start by splitting on that:
let parts = input.split('-');
If you need to get rid of whitespace surrounding that, you should trim each part:
parts = parts.map(function (it) { return it.trim(); });
To validate those parts, you can use a regex:
parts = parts.filter(function (it) { return /^\d\d:\d\d$/.test(it); });
Combined:
var input = "14:30 - 19:30";
var parts = input.split('-').map(function(it) {
return it.trim();
}).filter(function(it) {
return /^\d\d:\d\d$/.test(it);
});
document.getElementById('results').textContent = JSON.stringify(parts);
<pre id="results"></pre>

Try this :
var stringa = "14:30 - 19:30";
var stringes = stringa.split("-"); // string is "14:30-19:30" this style
or
var stringes = stringa.split(" - "); // if string is "14:30 - 19:30"; style so it includes the spaces also around '-' character.
The split function breaks the strings in sub-strings based on the location of the substring you enter inside it "-"
. the first one splits it based on location of "-" and second one includes the spaces also " - ".
*also it looks more like 24 hour clock time format than data as you mentioned in your question.

var stringa = '14:30 - 19:30';
var stringes = stringa.split("-");

.split is probably the best way to go, though you will want to prep the string first. I would go with str.replace(/\s*-\s*/g, '-').split('-'). to demonstrate:
var str = "14:30 - 19:30"
var str2 = "14:30-19:30"
console.log(str.replace(/\s*-\s*/g, '-').split('-')) //outputs ["14:30", "19:30"]
console.log(str2 .replace(/\s*-\s*/g, '-').split('-')) //outputs ["14:30", "19:30"]

Don't forget that you can pass a RegExp into str.split
'14:30 - 19:30'.split(/\s*-\s*/); // ["14:30", "19:30"]
'14:30-19:30'.split(/\s*-\s*/); // ["14:30", "19:30"]

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

Remove all dots except the first one from a string

Given a string
'1.2.3.4.5'
I would like to get this output
'1.2345'
(In case there are no dots in the string, the string should be returned unchanged.)
I wrote this
function process( input ) {
var index = input.indexOf( '.' );
if ( index > -1 ) {
input = input.substr( 0, index + 1 ) +
input.slice( index ).replace( /\./g, '' );
}
return input;
}
Live demo: http://jsfiddle.net/EDTNK/1/
It works but I was hoping for a slightly more elegant solution...

There is a pretty short solution (assuming input is your string):
var output = input.split('.');
output = output.shift() + '.' + output.join('');
If input is "1.2.3.4", then output will be equal to "1.234".
See this jsfiddle for a proof. Of course you can enclose it in a function, if you find it necessary.
EDIT:
Taking into account your additional requirement (to not modify the output if there is no dot found), the solution could look like this:
var output = input.split('.');
output = output.shift() + (output.length ? '.' + output.join('') : '');
which will leave eg. "1234" (no dot found) unchanged. See this jsfiddle for updated code.

It would be a lot easier with reg exp if browsers supported look behinds.
One way with a regular expression:
function process( str ) {
return str.replace( /^([^.]*\.)(.*)$/, function ( a, b, c ) {
return b + c.replace( /\./g, '' );
});
}

You can try something like this:
str = str.replace(/\./,"#").replace(/\./g,"").replace(/#/,".");
But you have to be sure that the character # is not used in the string; or replace it accordingly.
Or this, without the above limitation:
str = str.replace(/^(.*?\.)(.*)$/, function($0, $1, $2) {
return $1 + $2.replace(/\./g,"");
});

You could also do something like this, i also don't know if this is "simpler", but it uses just indexOf, replace and substr.
var str = "7.8.9.2.3";
var strBak = str;
var firstDot = str.indexOf(".");
str = str.replace(/\./g,"");
str = str.substr(0,firstDot)+"."+str.substr(1,str.length-1);
document.write(str);
Shai.

Here is another approach:
function process(input) {
var n = 0;
return input.replace(/\./g, function() { return n++ > 0 ? '' : '.'; });
}
But one could say that this is based on side effects and therefore not really elegant.

This isn't necessarily more elegant, but it's another way to skin the cat:
var process = function (input) {
var output = input;
if (typeof input === 'string' && input !== '') {
input = input.split('.');
if (input.length > 1) {
output = [input.shift(), input.join('')].join('.');
}
}
return output;
};

Not sure what is supposed to happen if "." is the first character, I'd check for -1 in indexOf, also if you use substr once might as well use it twice.
if ( index != -1 ) {
input = input.substr( 0, index + 1 ) + input.substr(index + 1).replace( /\./g, '' );
}

var i = s.indexOf(".");
var result = s.substr(0, i+1) + s.substr(i+1).replace(/\./g, "");
Somewhat tricky. Works using the fact that indexOf returns -1 if the item is not found.

Trying to keep this as short and readable as possible, you can do the following:
JavaScript
var match = string.match(/^[^.]*\.|[^.]+/g);
string = match ? match.join('') : string;
Requires a second line of code, because if match() returns null, we'll get an exception trying to call join() on null. (Improvements welcome.)
Objective-J / Cappuccino (superset of JavaScript)
string = [string.match(/^[^.]*\.|[^.]+/g) componentsJoinedByString:''] || string;
Can do it in a single line, because its selectors (such as componentsJoinedByString:) simply return null when sent to a null value, rather than throwing an exception.
As for the regular expression, I'm matching all substrings consisting of either (a) the start of the string + any potential number of non-dot characters + a dot, or (b) any existing number of non-dot characters. When we join all matches back together, we have essentially removed any dot except the first.

var input = '14.1.2';
reversed = input.split("").reverse().join("");
reversed = reversed.replace(\.(?=.*\.), '' );
input = reversed.split("").reverse().join("");

Based on #Tadek's answer above. This function takes other locales into consideration.
For example, some locales will use a comma for the decimal separator and a period for the thousand separator (e.g. -451.161,432e-12).
First we convert anything other than 1) numbers; 2) negative sign; 3) exponent sign into a period ("-451.161.432e-12").
Next we split by period (["-451", "161", "432e-12"]) and pop out the right-most value ("432e-12"), then join with the rest ("-451161.432e-12")
(Note that I'm tossing out the thousand separators, but those could easily be added in the join step (.join(','))
var ensureDecimalSeparatorIsPeriod = function (value) {
var numericString = value.toString();
var splitByDecimal = numericString.replace(/[^\d.e-]/g, '.').split('.');
if (splitByDecimal.length < 2) {
return numericString;
}
var rightOfDecimalPlace = splitByDecimal.pop();
return splitByDecimal.join('') + '.' + rightOfDecimalPlace;
};

let str = "12.1223....1322311..";
let finStr = str.replace(/(\d*.)(.*)/, '$1') + str.replace(/(\d*.)(.*)/, '$2').replace(/\./g,'');
console.log(finStr)

const [integer, ...decimals] = '233.423.3.32.23.244.14...23'.split('.');
const result = [integer, decimals.join('')].join('.')
Same solution offered but using the spread operator.
It's a matter of opinion but I think it improves readability.