I'm sorry if it is a confusing question. I was trying to find a way to do this but couldn't find it so, if it is a repeated question, my apologies!
I have a text something like this: something:"askjnqwe234"
I want to be able to get askjnqwe234 using a RegExp. You can notice I want to omit the quotes. I was trying this using /[^"]+(?=(" ")|"$)/g but it returns an array. I want a RegExt to return a single string, not an array.
I don't know if it's possible but I do not want to specify the position of the array; something like this:
var x = string.match(/[^"]+(?=(" ")|"$)/g)[0];
Thanks!
Try:
/"([^"]*)"/g
in English: look for " the match and record anything that isn't " till you see another "".
match and exec always return an array or null, so, assuming you have a single double-quoted value and no newlines in the string, you could use
var x;
var str = 'something:"askjnqwe234"';
x = str.replace( /^[^"]*"|".*/g, '' );
// "askjnqwe234"
Or, if you may have other quoted values in the string
x = str.replace( /.*?something:"([^"]*)".*/, '$1' );
where $1 refers to the substring captured by the sub-pattern [^"]* between the ().
Further explanation on request.
Notwithstanding the above, I recommend that you tolerate the array indexing and just use match.
You can capture the information inside quotes like this, assuming it matches:
var x = string.match(/something:"([^"]*)"/)[1];
The memory capture at index 1 is the part inside the double quotes.
If you're not sure it will match:
var match = string.match(/something:"([^"]*)"/);
if (match) {
// use match[1] here
}
Related
For example, I have a string "esolri.gbn43sh.earbnf", and I want to remove every character after the last dot(i.e. "esolri.gbn43sh"). How can I do so with regular expression?
I could of course use non-RegExp way to do it, for example:
"esolri.gbn43sh.earbnf".slice("esolri.gbn43sh.earbnf".lastIndexOf(".")+1);
But I want a regular expression.
I tried /\..*?/, but that remove the first dot instead.
I am using Javascript. Any help is much appreciated.
I would use standard js rather than regex for this one, as it will be easier for others to understand your code
var str = 'esolri.gbn43sh.earbnf'
console.log(
str.slice(str.lastIndexOf('.') + 1)
)
Pattern Matching
Match a dot followed by non-dots until the end of string
let re = /\.[^.]*$/;
Use this with String.prototype.replace to achieve the desired output
'foo.bar.baz'.replace(re, ''); // 'foo.bar'
Other choices
You may find it is more efficient to do a simple substring search for the last . and then use a string slicing method on this index.
let str = 'foo.bar.baz',
i = str.lastIndexOf('.');
if (i !== -1) // i = -1 means no match
str = str.slice(0, i); // "foo.bar"
I have a string that can be a comma separated list of \w, such as:
abc123
abc123,def456,ghi789
I am trying to find a JavaScript regexp that will return ['abc123'] (first case) or ['abc123', 'def456', 'ghi789'] (without the comma).
I tried:
^(\w+,?)+$ -- Nope, as only the last repeating pattern will be matched, 789
^(?:(\w+),?)+$ -- Same story. I am using non-capturing bracket. However, the capturing just doesn't seem to happen for the repeated word
Is what I am trying to do even possible with regexp? I tried pretty much every combination of grouping, using capturing and non-capturing brackets, and still not managed to get this happening...
If you want to discard the whole input when there is something wrong, the simplest way is to validate, then split:
if (/^\w+(,\w+)*$/.test(input)) {
var values = input.split(',');
// Process the values here
}
If you want to allow empty value, change \w+ to \w*.
Trying to match and validate at the same time with single regex requires emulation of \G feature, which assert the position of the last match. Why is \G required? Since it prevents the engine from retrying the match at the next position and bypass your validation. Remember than ECMA Script regex doesn't have look-behind, so you can't differentiate between the position of an invalid character and the character(s) after it:
something,=bad,orisit,cor&rupt
^^ ^^
When you can't differentiate between the 2 positions, you can't rely on the engine to do a match-all operation alone. While it is possible to use a while loop with RegExp.exec and assert the position of last match yourself, why would you do so when there is a cleaner option?
If you want to savage whatever available, torazaburo's answer is a viable option.
Live demo
Try this regex :
'/([^,]+)/'
Alternatively, strings in javascript have a split method that can split a string based on a delimeter:
s.split(',')
Split on the comma first, then filter out results that do not match:
str.split(',').filter(function(s) { return /^\w+$/.test(s); })
This regex pattern separates numerical value in new line which contains special character such as .,,,# and so on.
var val = [1234,1213.1212, 1.3, 1.4]
var re = /[0-9]*[0-9]/gi;
var str = "abc123,def456, asda12, 1a2ass, yy8,ghi789";
var re = /[a-z]{3}\d{3}/g;
var list = str.match(re);
document.write("<BR> list.length: " + list.length);
for(var i=0; i < list.length; i++) {
document.write("<BR>list(" + i + "): " + list[i]);
}
This will get only "abc123" code style in the list and nothing else.
May be you can use split function
var st = "abc123,def456,ghi789";
var res = st.split(',');
I'm trying to build a regular expression that parses a string and skips things in brackets.
Something like
string = "A bc defg hi [hi] jkl mnop.";
The .match() should return "hi" but not [hi]. I've spent 5 hours running through RE's but I'm throwing in the towel.
Also this is for javascript or jquery if that matters.
Any help is appreciated. Also I'm working on getting my questions formatted correctly : )
EDIT:
Ok I just had a eureka moment and figured out that the original RegExp I was using actually did work. But when I was replaces the matches with the [matches] it simply replaced the first match in the string... over and over. I thought this was my regex refusing to skip the brackets but after much time of trying almost all of the solutions below, I realized that I was derping Hardcore.
When .replace was working its magic it was on the first match, so I quite simply added a space to the end of the result word as follows:
var result = string.match(regex);
var modifiedResult = '[' + result[0].toString() + ']';
string.replace(result[0].toString() + ' ', modifiedResult + ' ');
This got it to stop targeting the original word in the string and stop adding a new set of brackets to it with every match. Thank you all for your help. I am going to give answer credit to the post that prodded me in the right direction.
preprocess the target string by removing everything between brackets before trying to match your RE
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*\]/, "")
then apply your RE to tmpstring
correction: made the match for brackets eager per nhahtd comment below, and also, made the RE global
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*?\]/g, "")
You don't necessarily need regex for this. Simply use string manipulation:
var arr = string.split("[");
var final = arr[0] + arr[1].split("]")[1];
If there are multiple bracketed expressions, use a loop:
while (string.indexOf("[") != -1){
var arr = string.split("[");
string = arr[0] + arr.slice(1).join("[").split("]").slice(1).join("]");
}
Using only Regular Expressions, you can use:
hi(?!])
as an example.
Look here about negative lookahead: http://www.regular-expressions.info/lookaround.html
Unfortunately, javascript does not support negative lookbehind.
I used http://regexpal.com/ to test, abcd[hi]jkhilmnop as test data, hi(?!]) as the regex to find. It matched 'hi' without matching '[hi]'. Basically it matched the 'hi' so long as there was not a following ']' character.
This of course, can be expanded if needed. This has a benefit of not requiring any pre-processing for the string.
r"\[(.*)\]"
Just play arounds with this if you wanto to use regular expressions.
What do yo uwant to do with it? If you want to selectively replace parts like "hi" except when it's "[hi]", then I often use a system where I match what I want to avoid first and then what I want to watch; if it matches what I want to avoid then I return the match, otherwise I return the processed match.
Like this:
return string.replace(/(\[\w+\])|(\w+)/g, function(all, m1, m2) {return m1 || m2.toUpperCase()});
which, with the given string, returns:
"A BC DEFG HI [hi] JKL MNOP."
Thus: it replaces every word with uppercase (m1 is empty), except if the word is between square brackets (m1 is not empty).
This builds an array of all the strings contained in [ ]:
var regex = /\[([^\]]*)\]/;
var string = "A bc defg hi [hi] [jkl] mnop.";
var results=[], result;
while(result = regex.exec(string))
results.push(result[1]);
edit
To answer to the question, this regex returns the string less all is in [ ], and trim whitespaces:
"A bc defg [hi] mnop [jkl].".replace(/(\s{0,1})\[[^\]]*\](\s{0,1})/g,'$1')
Instead of skipping the match you can probably try something different - match everything but do not capture the string within square brackets (inclusive) with something like this:
var r = /(?:\[.*?[^\[\]]\])|(.)/g;
var result;
var str = [];
while((result = r.exec(s)) !== null){
if(result[1] !== undefined){ //true if [string] matched but not captured
str.push(result[1]);
}
}
console.log(str.join(''));
The last line will print parts of the string which do not match the [string] pattern. For example, when called with the input "A [bc] [defg] hi [hi] j[kl]u m[no]p." the code prints "A hi ju mp." with whitespaces intact.
You can try different things with this code e.g. replacing etc.
I wanted to get all strings inside a parentheses pair. for example, after applying regex on
"fun('xyz'); fun('abcd'); fun('abcd.ef') { temp('no'); "
output should be
['xyz','abcd', 'abcd.ef'].
I tried many option but was not able to get desired result.
one option is
/fun\((.*?)\)/gi.exec("fun('xyz'); fun('abcd'); fun('abcd.ef')").
Store the regex in a variable, and run it in a loop...
var re = /fun\((.*?)\)/gi,
string = "fun('xyz'); fun('abcd'); fun('abcd.ef')",
matches = [],
match;
while(match = re.exec(string))
matches.push(match[1]);
Note that this only works for global regex. If you omit the g, you'll have an infinite loop.
Also note that it'll give an undesired result if there a ) between the quotation marks.
You can use this code will almost do the job:
"fun('xyz'); fun('abcd'); fun('abcd.ef')".match(/'.*?'/gi);
You'll get ["'xyz'", "'abcd'", "'abcd.ef'"] which contains extra ' around the string.
The easiest way to find what you need is to use this RegExp: /[\w.]+(?=')/g
var string = "fun('xyz'); fun('abcd'); fun('abcd.ef')";
string.match(/[\w.]+(?=')/g); // ['xyz','abcd', 'abcd.ef']
It will work with alphanumeric characters and point, you will need to change [\w.]+ to add more symbols.
I am doing it wrong. I know.
I want to assign the matched text that is the result of a regex to a string var.
basically the regex is supposed to pull out anything in between two colons
so blah:xx:blahdeeblah
would result in xx
var matchedString= $(current).match('[^.:]+):(.*?):([^.:]+');
alert(matchedString);
I am looking to get this to put the xx in my matchedString variable.
I checked the jquery docs and they say that match should return an array. (string char array?)
When I run this nothing happens, No errors in the console but I tested the regex and it works outside of js. I am starting to think I am just doing the regex wrong or I am completely not getting how the match function works altogether
I checked the jquery docs and they say that match should return an array.
No such method exists for jQuery. match is a standard javascript method of a string. So using your example, this might be
var str = "blah:xx:blahdeeblah";
var matchedString = str.match(/([^.:]+):(.*?):([^.:]+)/);
alert(matchedString[2]);
// -> "xx"
However, you really don't need a regular expression for this. You can use another string method, split() to divide the string into an array of strings using a separator:
var str = "blah:xx:blahdeeblah";
var matchedString = str.split(":"); // split on the : character
alert(matchedString[1]);
// -> "xx"
String.match
String.split