How can I write regular expression to remove double dash -- into single dash - and if string start or end with dash replace with empty string.
var oldString = "abc--xyz--"
var filtered = oldStringt.replace(???????);
Sample Input >>>> Output
abc--xyz-- >>>>> abc-xyz
abc---xyz-123 >>>>> abc-xyz-123
--abc-xyz-123 >>>>> abc-xyz-123
How about chaining replaces:
str.replace(/[-]+/g, '-').replace(/[-]+$/g, '').replace(/^[-]+/g, '')
Fiddle here.
oldString.replace(/-+/g,"-").replace(/(^-+)|(-+$)/g,"");
Here is a single regexp that should work:
oldString.replace(/^-+|-+$|(-)+/g, '$1')
Tests: http://jsfiddle.net/kd9g3/
Now, I know you specifically asked for regexp, but many replaces like this one can be done using arrays as well (and sometimes they are faster):
oldString.split(/-+/).filter(function(e){return !!e}).join('-')
Related
My string:
AA,$,DESCRIPTION(Sink, clinical),$
Wanted matches:
AA
$
DESCRIPTION(Sink, clinical)
$
My regex sofar:
\+d|[\w$:0-9`<>=&;?\|\!\#\+\%\-\s\*\(\)\.ÅÄÖåäö]+
This gives
AA
$
DESCRIPTION(Sink
clinical)
I want to keep matches between ()
https://regex101.com/r/MqFUmk/3
Here's my attempt at the regex
\+d|[\w$:0-9`<>=&;?\|\!\#\+\%\-\s\*\.ÅÄÖåäö]+(\(.+\))?
I removed the parentheses from within the [ ] characters, and allowed capture elsewhere. It seems to satisfy the regex101 link you posted.
Depending on how arbitrary your input is, this regex might not be suitable for more complex strings.
Alternatively, here's an answer which could be more robust than mine, but may only work in Ruby.
((?>[^,(]+|(\((?>[^()]+|\g<-1>)*\)))+)
That one seems to work for me?
([^,\(\)]*(?:\([^\(\)]*\))?[^,\(\)]*)(?:,|$)
https://regex101.com/r/hLyJm5/2
Hope this helps!
Personally, I would first replace all commas within parentheses () with a character that will never occur (in my case I used # since I don't see it within your inclusions) and then I would split them by commas to keep it sweet and simple.
myStr = "AA,$,DESCRIPTION(Sink, clinical),$"; //Initial string
myStr = myStr.replace(/(\([^,]+),([^\)]+\))/g, "$1#$2"); //Replace , within parentheses with #
myArr = myStr.split(',').map(function(s) { return s.replace('#', ','); }); //Split string on ,
//myArr -> ["AA","$","DESCRIPTION(Sink, clinical)","$"]
optionally, if you're using ES6, you can change that last line to:
myArr = myStr.split(',').map(s => s.replace('#', ',')); //Yay Arrow Functions!
Note: If you have nested parentheses, this answer will need a modification
At last take an aproximation of what you need:
\w+(?:\(.*\))|\w+|\$
https://regex101.com/r/MqFUmk/4
I have a textArea. I am trying to split each string from a paragraph, which has proper grammar based punctuation delimiters like ,.!? or more if any.
I am trying to achieve this using Javascript. I am trying to get all such strings in that using the regular expression as in this answer
But here, in javascript for me it's not working. Here's my code snippet for more clarity
$('#split').click(function(){
var textAreaContent = $('#textArea').val();
//split the string i.e.., textArea content
var splittedArray = textAreaContent.split("\\W+");
alert("Splitted Array is "+splittedArray);
var lengthOfsplittedArray = splittedArray.length;
alert('lengthOfText '+lengthOfsplittedArray);
});
Since its unable to split, its always showing length as 1. What could be the apt regular expression here.
The regular expression shouldn't differ between Java and JavaScript, but the .split() method in Java accepts a regular expression string. If you want to use a regular expression in JavaScript, you need to create one...like so:
.split(/\W+/)
DEMO: http://jsfiddle.net/s3B5J/
Notice the / and / to create a regular expression literal. The Java version needed two "\" because it was enclosed in a string.
Reference:
https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Regular_Expressions
You can try this
textAreaContent.split(/\W+/);
\W+ : Matches any character that is not a word character (alphanumeric & underscore).
so it counts except alphanumerics and underscore! if you dont need to split " " (space) then you can use;
var splittedArray = textAreaContent.split("/\n+/");
I've a string done like this: "http://something.org/dom/My_happy_dog_%28is%29cool!"
How can I remove all the initial domain, the multiple underscore and the percentage stuff?
For now I'm just doing some multiple replace, like
str = str.replace("http://something.org/dom/","");
str = str.replace("_%28"," ");
and go on, but it's really ugly.. any help?
Thanks!
EDIT:
the exact input would be "My happy dog is cool!" so I would like to get rid of the initial address and remove the underscores and percentage and put the spaces in the right place!
The problem is that trying to put a regex on Chrome "something goes wrong". Is it a problem of Chrome or my regex?
I'd suggest:
var str = "http://something.org/dom/My_happy_dog_%28is%29cool!";
str.substring(str.lastIndexOf('/')+1).replace(/(_)|(%\d{2,})/g,' ');
JS Fiddle demo.
The reason I took this approach is that RegEx is fairly expensive, and is often tricky to fine tune to the point where edge-cases become less troublesome; so I opted to use simple string manipulation to reduce the RegEx work.
Effectively the above creates a substring of the given str variable, from the index point of the lastIndexOf('/') (which does exactly what you'd expect) and adding 1 to that so the substring is from the point after the / not before it.
The regex: (_) matches the underscores, the | just serves as an or operator and the (%\d{2,}) serves to match digit characters that occur twice in succession and follow a % sign.
The parentheses surrounding each part of the regex around the |, serve to identify matching groups, which are used to identify what parts should be replaced by the ' ' (single-space) string in the second of the arguments passed to replace().
References:
lastIndexOf().
replace().
substring().
You can use unescape to decode the percentages:
str = unescape("http://something.org/dom/My_happy_dog_%28is%29cool!")
str = str.replace("http://something.org/dom/","");
Maybe you could use a regular expression to pull out what you need, rather than getting rid of what you don't want. What is it you are trying to keep?
You can also chain them together as in:
str.replace("http://something.org/dom/", "").replace("something else", "");
You haven't defined the problem very exactly. To get rid of all stretches of characters ending in %<digit><digit> you'd say
var re = /.*%\d\d/g;
var str = str.replace(re, "");
ok, if you want to replace all that stuff I think that you would need something like this:
/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g
test
var string = "http://something.org/dom/My_happy_dog_%28is%29cool!";
string = string.replace(/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g,"");
I have a string like that:
var str = 'aaaaaa, bbbbbb, ccccc, ddddddd, eeeeee ';
My goal is to delete the last space in the string. I would use,
str.split(0,1);
But if there is no space after the last character in the string, this will delete the last character of the string instead.
I would like to use
str.replace("regex",'');
I am beginner in RegEx, any help is appreciated.
Thank you very much.
Do a google search for "javascript trim" and you will find many different solutions.
Here is a simple one:
trimmedstr = str.replace(/\s+$/, '');
When you need to remove all spaces at the end:
str.replace(/\s*$/,'');
When you need to remove one space at the end:
str.replace(/\s?$/,'');
\s means not only space but space-like characters; for example tab.
If you use jQuery, you can use the trim function also:
str = $.trim(str);
But trim removes spaces not only at the end of the string, at the beginning also.
Seems you need a trimRight function. its not available until Javascript 1.8.1. Before that you can use prototyping techniques.
String.prototype.trimRight=function(){return this.replace(/\s+$/,'');}
// Now call it on any string.
var a = "a string ";
a = a.trimRight();
See more on Trim string in JavaScript? And the compatibility list
You can use this code to remove a single trailing space:
.replace(/ $/, "");
To remove all trailing spaces:
.replace(/ +$/, "");
The $ matches the end of input in normal mode (it matches the end of a line in multiline mode).
Try the regex ( +)$ since $ in regex matches the end of the string. This will strip all whitespace from the end of the string.
Some programs have a strip function to do the same, I do not believe the stadard Javascript library has this functionality.
Regex Reference Sheet
Working example:
var str = "Hello World ";
var ans = str.replace(/(^[\s]+|[\s]+$)/g, '');
alert(str.length+" "+ ans.length);
Fast forward to 2021,
The trimEnd() function is meant exactly for this!
It will remove all whitespaces (including spaces, tabs, new line characters) from the end of the string.
According to the official docs, it is supported in every major browser. Only IE is unsupported. (And lets be honest, you shouldn't care about IE given that microsoft itself has dropped support for IE in Aug 2021!)
I'm trying to write a regex for use in javascript.
var script = "function onclick() {loadArea('areaog_og_group_og_consumedservice', '\x26roleOrd\x3d1');}";
var match = new RegExp("'[^']*(\\.[^']*)*'").exec(script);
I would like split to contain two elements:
match[0] == "'areaog_og_group_og_consumedservice'";
match[1] == "'\x26roleOrd\x3d1'";
This regex matches correctly when testing it at gskinner.com/RegExr/ but it does not work in my Javascript. This issue can be replicated by testing ir here http://www.regextester.com/.
I need the solution to work with Internet Explorer 6 and above.
Can any regex guru's help?
Judging by your regex, it looks like you're trying to match a single-quoted string that may contain escaped quotes. The correct form of that regex is:
'[^'\\]*(?:\\.[^'\\]*)*'
(If you don't need to allow for escaped quotes, /'[^']*'/ is all you need.) You also have to set the g flag if you want to get both strings. Here's the regex in its regex-literal form:
/'[^'\\]*(?:\\.[^'\\]*)*'/g
If you use the RegExp constructor instead of a regex literal, you have to double-escape the backslashes: once for the string literal and once for the regex. You also have to pass the flags (g, i, m) as a separate parameter:
var rgx = new RegExp("'[^'\\\\]*(?:\\\\.[^'\\\\]*)*'", "g");
while (result = rgx.exec(script))
print(result[0]);
The regex you're looking for is .*?('[^']*')\s*,\s*('[^']*'). The catch here is that, as usual, match[0] is the entire matched text (this is very normal) so it's not particularly useful to you. match[1] and match[2] are the two matches you're looking for.
var script = "function onclick() {loadArea('areaog_og_group_og_consumedservice', '\x26roleOrd\x3d1');}";
var parameters = /.*?('[^']*')\s*,\s*('[^']*')/.exec(script);
alert("you've done: loadArea("+parameters[1]+", "+parameters[2]+");");
The only issue I have with this is that it's somewhat inflexible. You might want to spend a little time to match function calls with 2 or 3 parameters?
EDIT
In response to you're request, here is the regex to match 1,2,3,...,n parameters. If you notice, I used a non-capturing group (the (?: ) part) to find many instances of the comma followed by the second parameter.
/.*?('[^']*')(?:\s*,\s*('[^']*'))*/
Maybe this:
'([^']*)'\s*,\s*'([^']*)'