I have following string var foo = require('foo/bar'); which I want to change to var foo = require('../../foo/bar');.
So far I wrote following code which would do this:
var search = 'foo/bar';
var replace = '../../foo/bar';
var regex = new RegExp('require[(](\"|\')' + search + '(\"|\')[)]', 'g');
var source = 'var foo = require(\'foo/bar\')';
source.replace(regex, 'require(\'' + replace + '\')');
However as you see this is very inefficient. What could I do to make this regex a bit shorter. For example:
only have to replace the path and not to rewrite require with require(\'' + replace + '\')'
make (\"|\') and [(] shorter
anything else?
Best,
Bo
Edit:
I am doing this replacement on complete javascript source files.
Edit2:
What is necessary to change var foobar = require('foo/bar'); to var foobar = require('../../foo/baz/bar') when foo/ and ../../foo/baz/ should be variable?
You can use this code:
var search = 'foo/bar';
var source = "var foo = require('foo/bar')";
var regex = new RegExp('(require)\\((["\'])(' + search + ')\\2\\)', 'g');
var repl = source.replace(regex, "$1('../../$3')");
//=> var foo = require('../../foo/bar')
Here is an alternative :
var str = "var foo = require('foo/bar');";
str = str.split(/(')/);
str.splice(2, 0, '../../');
str = str.join('');
Even simpler :
"var foo = require('foo/bar');".replace(/'([^']+)'/, "'../../$1'")
I wonder if you could do the replacement before wrapping your path inside the var xxx = require() statement.
Optimizing code that starts by undoing what has been done just before seems a bit inefficient.
Besides, it seems you already know the target path (foo/bar in your example), but it is not even necessary to use that.
The regexp here boils down to splitting the string after the opening quote and inserting the ../../ prefix.
You could do that with
source = source.split(/["']/,2).join('"../../')+'");';
The regexp is a constant, so you could put it into a variable to have it compiled once and for all.
That would probably be more efficient than generating and recompiling the regexp from your example each time.
But anyway, I wonder how many such strings you would have to replace to start noticing a performance hit.
Related
I have this string:
var chain = "providerId=12$familyId=123&brandId=1122112$officeId=21&";
I need to do a method that erases a certain word with regular expressions.
Example:
var word = "familyId";
var newChain = deleteParam(chain, word);
console.log(newChain);
Result = "providerId=12$brandId=1122112$officeId=21&";
Delete : familyId=123&
I tried to do the method in the following way, but it does not work:
function deleteParam(chain, word) {
var exp = new RegExp(param, "=[0-9]&");
var str = chain.replace(exp, ""); // Delete
return str
}
Please, I need your help, I can not make this method work, because I do not understand well how to build regular expressions.
Excuse me all, English is not my native language
thank you very much to all.
You can use something like this new RegExp(param + "=\[^&]+")
Here is an example:
var chain = "providerId=12$familyId=123&brandId=1122112$officeId=21&";
var toRemove = "familyId";
var pattern = new RegExp(toRemove + "=[^&]*&");
var newChain = chain.replace(pattern, "");
console.log(newChain);
If you're looking to process a GET request's search parameters is better to use a pattern like this (&|\?)*parameterName=[^&]*
I have a RegExp that I would like to make dynamic and create in a string. I want to change this:
var result:Object = value.match(/John\/(.*?) /);
to this:
var firstName:String = "John";
var result:Object = value.match(firstName + "\/(.*?) "); // this doesn't work
I'm using ActionScript but I think what would work in JavaScript would work as well here.
In Javascript you can create a new instance of the RegExp class:
var firstName:String = "John";
var result:Object = value.match(new RegExp(firstName + "\/(.*?) "));
When you use value.match(firstName + "\/(.*?) "); the first parameter to the match function is a string, but it should be a regular expression object.
In Javascript, how can I trim a string by a number of characters from the end, append another string, and re-append the initially cut-off string again?
In particular, I have filename.png and want to turn it into filename-thumbnail.png.
I am looking for something along the lines of:
var sImage = "filename.png";
var sAppend = "-thumbnail";
var sThumbnail = magicHere(sImage, sAppend);
You can use .slice, which accepts negative indexes:
function insert(str, sub, pos) {
return str.slice(0, pos) + sub + str.slice(pos);
// "filename" + "-thumbnail" + ".png"
}
Usage:
insert("filename.png", "-thumbnail", -4); // insert at 4th from end
Try using a regular expression (Good documentation can be found at https://developer.mozilla.org/en/JavaScript/Guide/Regular_Expressions)
I haven't tested but try something like:
var re = /(.*)\.png$/;
var str = "filename.png";
var newstr = str.replace(re, "$1-thumbnail.png");
console.log(newstr);
I would use a regular expression to find the various parts of the filename and then rearrange and add strings as needed from there.
Something like this:
var file='filename.png';
var re1='((?:[a-z][a-z0-9_]*))';
var re2='.*?';
var re3='((?:[a-z][a-z0-9_]*))';
var p = new RegExp(re1+re2+re3,["i"]);
var m = p.exec(file);
if (m != null) {
var fileName=m[1];
var fileExtension=m[2];
}
That would give you your file's name in fileName and file's extension in fileExtension. From there you could append or prepend anything you want.
var newFile = fileName + '-thumbnail' + '.' + fileExtension;
Perhaps simpler than regular expressions, you could use lastindexof (see http://www.w3schools.com/jsref/jsref_lastindexof.asp) to find the file extension (look for the period - this allows for longer file extensions like .html), then use slice as suggested by pimvdb.
You could use a regular expression and do something like this:
var sImage = "filename.png";
var sAppend = "-thumbnail$1";
var rExtension = /(\.[\w\d]+)$/;
var sThumbnail = sImage.replace(rExtension, sAppend);
rExtension is a regular expression which looks for the extension, capturing it into $1. You'll see that $1 appears inside of sAppend, which means "put the extension here".
EDIT: This solution will work with any file extension of any length. See it in action here: http://jsfiddle.net/h4Qsv/
Is there a way to remove everything after a certain character or just choose everything up to that character? I'm getting the value from an href and up to the "?", and it's always going to be a different amount of characters.
Like this
/Controller/Action?id=11112&value=4444
I want the href to be /Controller/Action only, so I want to remove everything after the "?".
I'm using this now:
$('.Delete').click(function (e) {
e.preventDefault();
var id = $(this).parents('tr:first').attr('id');
var url = $(this).attr('href');
console.log(url);
}
You can also use the split() function. This seems to be the easiest one that comes to my mind :).
url.split('?')[0]
jsFiddle Demo
One advantage is this method will work even if there is no ? in the string - it will return the whole string.
var s = '/Controller/Action?id=11112&value=4444';
s = s.substring(0, s.indexOf('?'));
document.write(s);
Sample here
I should also mention that native string functions are much faster than regular expressions, which should only really be used when necessary (this isn't one of those cases).
Updated code to account for no '?':
var s = '/Controller/Action';
var n = s.indexOf('?');
s = s.substring(0, n != -1 ? n : s.length);
document.write(s);
Sample here
var href = "/Controller/Action?id=11112&value=4444";
href = href.replace(/\?.*/,'');
href ; //# => /Controller/Action
This will work if it finds a '?' and if it doesn't
May be very late party :p
You can use a back reference $'
$' - Inserts the portion of the string that follows the matched substring.
let str = "/Controller/Action?id=11112&value=4444"
let output = str.replace(/\?.*/g,"$'")
console.log(output)
It works for me very nicely:
var x = '/Controller/Action?id=11112&value=4444';
var remove_after= x.indexOf('?');
var result = x.substring(0, remove_after);
alert(result);
If you also want to keep "?" and just remove everything after that particular character, you can do:
var str = "/Controller/Action?id=11112&value=4444",
stripped = str.substring(0, str.indexOf('?') + '?'.length);
// output: /Controller/Action?
You can also use the split() method which, to me, is the easiest method for achieving this goal.
For example:
let dummyString ="Hello Javascript: This is dummy string"
dummyString = dummyString.split(':')[0]
console.log(dummyString)
// Returns "Hello Javascript"
Source: https://thispointer.com/javascript-remove-everything-after-a-certain-character/
if you add some json syringified objects, then you need to trim the spaces too... so i add the trim() too.
let x = "/Controller/Action?id=11112&value=4444";
let result = x.trim().substring(0, x.trim().indexOf('?'));
Worked for me:
var first = regexLabelOut.replace(/,.*/g, "");
It can easly be done using JavaScript for reference see link
JS String
EDIT
it can easly done as. ;)
var url="/Controller/Action?id=11112&value=4444 ";
var parameter_Start_index=url.indexOf('?');
var action_URL = url.substring(0, parameter_Start_index);
alert('action_URL : '+action_URL);
Is it possible to do something like this?
var pattern = /some regex segment/ + /* comment here */
/another segment/;
Or do I have to use new RegExp() syntax and concatenate a string? I'd prefer to use the literal as the code is both more self-evident and concise.
Here is how to create a regular expression without using the regular expression literal syntax. This lets you do arbitary string manipulation before it becomes a regular expression object:
var segment_part = "some bit of the regexp";
var pattern = new RegExp("some regex segment" + /*comment here */
segment_part + /* that was defined just now */
"another segment");
If you have two regular expression literals, you can in fact concatenate them using this technique:
var regex1 = /foo/g;
var regex2 = /bar/y;
var flags = (regex1.flags + regex2.flags).split("").sort().join("").replace(/(.)(?=.*\1)/g, "");
var regex3 = new RegExp(expression_one.source + expression_two.source, flags);
// regex3 is now /foobar/gy
It's just more wordy than just having expression one and two being literal strings instead of literal regular expressions.
Just randomly concatenating regular expressions objects can have some adverse side effects. Use the RegExp.source instead:
var r1 = /abc/g;
var r2 = /def/;
var r3 = new RegExp(r1.source + r2.source,
(r1.global ? 'g' : '')
+ (r1.ignoreCase ? 'i' : '') +
(r1.multiline ? 'm' : ''));
console.log(r3);
var m = 'test that abcdef and abcdef has a match?'.match(r3);
console.log(m);
// m should contain 2 matches
This will also give you the ability to retain the regular expression flags from a previous RegExp using the standard RegExp flags.
jsFiddle
I don't quite agree with the "eval" option.
var xxx = /abcd/;
var yyy = /efgh/;
var zzz = new RegExp(eval(xxx)+eval(yyy));
will give "//abcd//efgh//" which is not the intended result.
Using source like
var zzz = new RegExp(xxx.source+yyy.source);
will give "/abcdefgh/" and that is correct.
Logicaly there is no need to EVALUATE, you know your EXPRESSION. You just need its SOURCE or how it is written not necessarely its value. As for the flags, you just need to use the optional argument of RegExp.
In my situation, I do run in the issue of ^ and $ being used in several expression I am trying to concatenate together! Those expressions are grammar filters used accross the program. Now I wan't to use some of them together to handle the case of PREPOSITIONS.
I may have to "slice" the sources to remove the starting and ending ^( and/or )$ :)
Cheers, Alex.
Problem If the regexp contains back-matching groups like \1.
var r = /(a|b)\1/ // Matches aa, bb but nothing else.
var p = /(c|d)\1/ // Matches cc, dd but nothing else.
Then just contatenating the sources will not work. Indeed, the combination of the two is:
var rp = /(a|b)\1(c|d)\1/
rp.test("aadd") // Returns false
The solution:
First we count the number of matching groups in the first regex, Then for each back-matching token in the second, we increment it by the number of matching groups.
function concatenate(r1, r2) {
var count = function(r, str) {
return str.match(r).length;
}
var numberGroups = /([^\\]|^)(?=\((?!\?:))/g; // Home-made regexp to count groups.
var offset = count(numberGroups, r1.source);
var escapedMatch = /[\\](?:(\d+)|.)/g; // Home-made regexp for escaped literals, greedy on numbers.
var r2newSource = r2.source.replace(escapedMatch, function(match, number) { return number?"\\"+(number-0+offset):match; });
return new RegExp(r1.source+r2newSource,
(r1.global ? 'g' : '')
+ (r1.ignoreCase ? 'i' : '')
+ (r1.multiline ? 'm' : ''));
}
Test:
var rp = concatenate(r, p) // returns /(a|b)\1(c|d)\2/
rp.test("aadd") // Returns true
Providing that:
you know what you do in your regexp;
you have many regex pieces to form a pattern and they will use same flag;
you find it more readable to separate your small pattern chunks into an array;
you also want to be able to comment each part for next dev or yourself later;
you prefer to visually simplify your regex like /this/g rather than new RegExp('this', 'g');
it's ok for you to assemble the regex in an extra step rather than having it in one piece from the start;
Then you may like to write this way:
var regexParts =
[
/\b(\d+|null)\b/,// Some comments.
/\b(true|false)\b/,
/\b(new|getElementsBy(?:Tag|Class|)Name|arguments|getElementById|if|else|do|null|return|case|default|function|typeof|undefined|instanceof|this|document|window|while|for|switch|in|break|continue|length|var|(?:clear|set)(?:Timeout|Interval))(?=\W)/,
/(\$|jQuery)/,
/many more patterns/
],
regexString = regexParts.map(function(x){return x.source}).join('|'),
regexPattern = new RegExp(regexString, 'g');
you can then do something like:
string.replace(regexPattern, function()
{
var m = arguments,
Class = '';
switch(true)
{
// Numbers and 'null'.
case (Boolean)(m[1]):
m = m[1];
Class = 'number';
break;
// True or False.
case (Boolean)(m[2]):
m = m[2];
Class = 'bool';
break;
// True or False.
case (Boolean)(m[3]):
m = m[3];
Class = 'keyword';
break;
// $ or 'jQuery'.
case (Boolean)(m[4]):
m = m[4];
Class = 'dollar';
break;
// More cases...
}
return '<span class="' + Class + '">' + m + '</span>';
})
In my particular case (a code-mirror-like editor), it is much easier to perform one big regex, rather than a lot of replaces like following as each time I replace with a html tag to wrap an expression, the next pattern will be harder to target without affecting the html tag itself (and without the good lookbehind that is unfortunately not supported in javascript):
.replace(/(\b\d+|null\b)/g, '<span class="number">$1</span>')
.replace(/(\btrue|false\b)/g, '<span class="bool">$1</span>')
.replace(/\b(new|getElementsBy(?:Tag|Class|)Name|arguments|getElementById|if|else|do|null|return|case|default|function|typeof|undefined|instanceof|this|document|window|while|for|switch|in|break|continue|var|(?:clear|set)(?:Timeout|Interval))(?=\W)/g, '<span class="keyword">$1</span>')
.replace(/\$/g, '<span class="dollar">$</span>')
.replace(/([\[\](){}.:;,+\-?=])/g, '<span class="ponctuation">$1</span>')
It would be preferable to use the literal syntax as often as possible. It's shorter, more legible, and you do not need escape quotes or double-escape backlashes. From "Javascript Patterns", Stoyan Stefanov 2010.
But using New may be the only way to concatenate.
I would avoid eval. Its not safe.
You could do something like:
function concatRegex(...segments) {
return new RegExp(segments.join(''));
}
The segments would be strings (rather than regex literals) passed in as separate arguments.
You can concat regex source from both the literal and RegExp class:
var xxx = new RegExp(/abcd/);
var zzz = new RegExp(xxx.source + /efgh/.source);
Use the constructor with 2 params and avoid the problem with trailing '/':
var re_final = new RegExp("\\" + ".", "g"); // constructor can have 2 params!
console.log("...finally".replace(re_final, "!") + "\n" + re_final +
" works as expected..."); // !!!finally works as expected
// meanwhile
re_final = new RegExp("\\" + "." + "g"); // appends final '/'
console.log("... finally".replace(re_final, "!")); // ...finally
console.log(re_final, "does not work!"); // does not work
No, the literal way is not supported. You'll have to use RegExp.
the easier way to me would be concatenate the sources, ex.:
a = /\d+/
b = /\w+/
c = new RegExp(a.source + b.source)
the c value will result in:
/\d+\w+/
I prefer to use eval('your expression') because it does not add the /on each end/ that ='new RegExp' does.