After coming to the shocking realization that regular expressions in JavaScript are somewhat different from the ones in PCE, I am stuck with the following.
In php I extract a number after x:
(?x)[0-9]+
In JavaScript the same regex doesn't work, due to invalid group resulting from the capturing parenthesis difference.
So I am trying to achieve the same trivial functionality, but I keep getting both the x and the number:
(?:x)([0-9]+)
How do I capture the number after x without including x?
This works too:
/(?:x)([0-9]+)/.test('YOUR_STRING');
Then, the value you want is:
RegExp.$1 // group 1
You can try the following regex: (?!x)[0-9]+
fiddle here: https://jsfiddle.net/xy6x938e/1/
This is assuming that you are now looking for an x followed by a number, it uses a capture group to capture just the numbers section.
var myString = "x12345";
var myRegexp = /x([0-9]+)/g;
var match = myRegexp.exec(myString);
var myString2 = "z12345";
var match2 = myRegexp.exec(myString2);
if(match != null && match.length > 1){
alert('match1:' + match[1]);
}
else{
alert('no match 1');
}
if(match2 != null && match2.length > 1){
alert('match2:' + match2[1]);
}
else{
alert('no match 2');
}
(\d+) try this!
i have tested on this tool with x12345
http://www.regular-expressions.info/javascriptexample.html
How do I capture the number after x without including x?
In fact, you just want to extract a sequence of digits after a fixed string/known pattern.
Your PCRE (PHP) regex, (?x)[0-9]+, is wrong becaue (?x) is an inline version of a PCRE_EXTENDED VERBOSE/COMMENTS flag (see "Pattern Modifiers"). It does not do anything meaningful in this case, (?x)[0-9]+ is equal to [0-9]+ or \d+.
You can use
console.log("x15 x25".match(/(?<=x)\d+/g));
You can also use a capturing group and then extract Group 1 value after a match is obtained:
const match = /x(\d+)/.exec("x15");
if (match) {
console.log(match[1]); // Getting the first match
}
// All matches
const matches = Array.from("x15,x25".matchAll(/x(\d+)/g), x=>x[1]);
console.log(matches);
You still can use exclusive pattern (?!...)
So, for your example it will be /(?!x)[0-9]+/. Give a try to the following:
/(?!x)\d+/.exec('x123')
// => ["123"]
I need to search for a word in text. For this I used this regex:
var re =/duration='\d+'/ig;
var i = text.match(re);
This gives me an array of matches like "duration='300'", "duration='400'",...
I need to get only numbers. without duration=''
You can use a capturing group:
var re = /duration='(\d+)'/ig;
var match = re.exec(text);
while (match != null) {
// matched text: match[1]
match = re.exec(text);
}
Tim's answer works well (and I'm not sure why the OP says it is not what he/she wants). That said, here is another way to do it using the String.replace() method with a callback function replacement value:
function getDurations(text) {
var re =/duration='(\d+)'/ig;
var i = [];
text.replace(re, function(m0, m1){i.push(m1); return '';});
return i;
}
Note that this technique requires no loop and is quite efficient getting the job done in a single statement.
I am having some trouble with my regex in javascript.
I have the following code, that I think should match, but it doesn't.
var rgx = new RegExp("{\d+:(\d+)}");
if (rgx.test("{0:00000}") == true) {
alert("match");
}
else
{
alert("no match");
}
I am unsure if I should use test() here. I really want to catch the group, in my regex but exec() seems to give me the same result.
So what am I doing wrong?
The problem is that you need to escape the \ character in your regex:
var rgx = new RegExp("{\\d+:(\\d+)}");
Alternatively, you can use the literal syntax:
var rgx = /{\d+:(\d+)}/;
To capture the results, you should also use the .match function as opposed to test or exec. It will return null if it doesn't match and an array of at least one element if it does match.
There are multiple issues with the regex:
var rgx = new RegExp("{\d+:(\d+)}");
First (first noted by syazdani), you must string-escape the backslashes:
var rgx = new RegExp("{\\d+:(\\d+)}");
or better yet use a regex literal:
var rgx = /{\d+:(\d+)}/
Second, { and } have a special meaning in regex and should be escaped:
var rgx = /\{\d+:(\d+)\}/
Third, as noted by Ian, you might want to ensure the entire string is matched:
var rgx = /^\{\d+:(\d+)\}$/
RegExp#test returns a boolean true/false whether the string matches.
RegExp#exec returns an array holding the match and all captured groups if the string is matched, or null if the string is not matched:
var matches = /\{\d+:(\d+)\}/.exec("{0:000000}");
if(matches){
console.log(matches[1]); //logs "000000"
}
I have a problem. I have a string - "\,str\,i,ing" and i need to split by comma before which not have slash. For my string - ["\,str\,i", "ing"]. I'm use next regex
myString.split("[^\],", 2)
but it's doesn't worked.
Well, this is ridiculous to avoid the lack of lookbehind but seems to get the correct result.
"\\,str\\,i,ing".split('').reverse().join('').split(/,(?=[^\\])/).map(function(a){
return a.split('').reverse().join('');
}).reverse();
//=> ["\,str\,i", "ing"]
Not sure about your expected output but you are specifying string not a regex, use:
var arr = "\,str\,i,ing".split(/[^\\],/, 2);
console.log(arr);
To split using regex, wrap your regex in /..../
This is not easily possible with js, because it does not support lookbehind. Even if you'd use a real regex, it would eat the last character:
> "xyz\\,xyz,xyz".split(/[^\\],/, 2)
["xyz\\,xy", "xyz"]
If you don't want the z to be eaten, I'd suggest:
var str = "....";
return str.split(",").reduce(function(res, part) {
var l = res.length;
if (l && res[l-1].substr(-1) == "\\" || l<2)
// ^ ^^ ^
// not the first was escaped limit
res[l-1] += ","+part;
else
res.push(part);
return;
}, []);
Reading between the lines, it looks like you want to split a string by , characters that are not preceded by \ characters.
It would be really great if JavaScript had a regular expression lookbehind (and negative lookbehind) pattern, but unfortunately it does not. What it does have is a lookahead ((?=) )and negative lookahead ((?!)) pattern. Make sure to review the documentation.
You can use these as a lookbehind if you reverse the string:
var str,
reverseStr,
arr,
reverseArr;
//don't forget to escape your backslashes
str = '\\,str\\,i,ing';
//reverse your string
reverseStr = str.split('').reverse().join('');
//split the array on `,`s that aren't followed by `\`
reverseArr = reverseStr.split(/,(?!\\)/);
//reverse the reversed array, and reverse each string in the array
arr = reverseArr.reverse().map(function (val) {
return val.split('').reverse().join('');
});
You picked a tough character to match- a forward slash preceding a comma is apt to disappear while you pass it around in a string, since '\,'==','...
var s= 'My dog, the one with two \\, blue \\,eyes, is asleep.';
var a= [], M, rx=/(\\?),/g;
while((M= rx.exec(s))!= null){
if(M[1]) continue;
a.push(s.substring(0, rx.lastIndex-1));
s= s.substring(rx.lastIndex);
rx.lastIndex= 0;
};
a.push(s);
/* returned value: (Array)
My dog
the one with two \, blue \,eyes
is asleep.
*/
Find something which will not be present in your original string, say "###". Replace "\\," with it. Split the resulting string by ",". Replace "###" back with "\\,".
Something like this:
<script type="text/javascript">
var s1 = "\\,str\\,i,ing";
var s2 = s1.replace(/\\,/g,"###");
console.log(s2);
var s3 = s2.split(",");
for (var i=0;i<s3.length;i++)
{
s3[i] = s3[i].replace(/###/g,"\\,");
}
console.log(s3);
</script>
See JSFiddle
I wrote a regular expression which I expect should work but it doesn't.
var regex = new RegExp('(?<=\[)[0-9]+(?=\])')
JavaScript is giving me the error：
Invalid regular expression :(/(?<=[)[0-9]+(?=])/): Invalid group
Does JavaScript not support lookahead or lookbehind?
This should work:
var regex = /\[[0-9]+\]/;
edit: with a grouping operator to target just the number:
var regex = /\[([0-9]+)\]/;
With this expression, you could do something like this:
var matches = someStringVar.match(regex);
if (null != matches) {
var num = matches[1];
}
Lookahead is supported, but not lookbehind. You can get close, with a bit of trickery.
To increment multiple numbers in the form of lets say:
var str = '/a/b/[123]/c/[4567]/[2]/69';
Try:
str.replace(/\[(\d+)\]/g, function(m, p1){
return '['+(p1*1+1)+']' }
)
//Gives you => '/a/b/[124]/c/[4568]/[3]/69'
If you're quoting a RegExp, watch out for double escaping your backslashes.