I have the following Javascript
function myFunction() {
var str = "depth10 shown";
var depth= str.match(/^depth[\d+$]/);
console.log(depth);
}
My function is trying to find if depth* is present in the string, where * is always numeric (ex: depth0, depth1, depth100) and return the numeric value in it. In the example above, depth is always returning only one digit instead of all digits. Can anyone explain why?
You're improperly utilizing a character class, you want to use a capturing group instead:
var str = 'depth10 shown or depth100 and depth1'
var re = /depth(\d+)/gi,
matches = [];
while (m = re.exec(str)) {
matches.push(m[1]);
}
console.log(matches) //=> [ '10', '100', '1' ]
Note: If you will have more than 1 "depth*" substrings in your string, you'll want to use the exec() method in a loop pushing the match result of the captured group to the results array.
Otherwise, you can use the match method here:
var r = 'depth10 shown'.match(/depth(\d+)/)
if (r)
console.log(r[1]); //=> "10"
$ Matches end of input. ie. /t$/ does not match the 't' in "eater", but does match it in "eat".
^ Matches beginning of input. ie, /^A/ does not match the 'A' in "an A", but does match the 'A' in "An E".
Try:
var str = "depth10 shown".match(/depth\d+/gi);
console.log(str[0])
Related
I need to parse a string that comes like this:
-38419-indices-foo-7119-attributes-10073-bar
Where there are numbers followed by one or more words all joined by dashes. I need to get this:
[
0 => '38419-indices-foo',
1 => '7119-attributes',
2 => '10073-bar',
]
I had thought of attempting to replace only the dash before a number with a : and then using .split(':') - how would I do this? I don't want to replace the other dashes.
Imo, the pattern is straight-forward:
\d+\D+
To even get rid of the trailing -, you could go for
(\d+\D+)(?:-|$)
Or
\d+(?:(?!-\d|$).)+
You can see it here:
var myString = "-38419-indices-foo-7119-attributes-10073-bar";
var myRegexp = /(\d+\D+)(?:-|$)/g;
var result = [];
match = myRegexp.exec(myString);
while (match != null) {
// matched text: match[0]
// match start: match.index
// capturing group n: match[n]
result.push(match[1]);
match = myRegexp.exec(myString);
}
console.log(result);
// alternative 2
let alternative_results = myString.match(/\d+(?:(?!-\d|$).)+/g);
console.log(alternative_results);
Or a demo on regex101.com.
Logic
lazy matching using quantifier .*?
Regex
.*?((\d+)\D*)(?!-)
https://regex101.com/r/WeTzF0/1
Test string
-38419-indices-foo-7119-attributes-10073-bar-333333-dfdfdfdf-dfdfdfdf-dfdfdfdfdfdf-123232323-dfsdfsfsdfdf
Matches
Further steps
You need to split from the matches and insert into your desired array.
I'm working with a string where I need to extract the first n characters up to where numbers begin. What would be the best way to do this as sometimes the string starts with a number: 7EUSA8889er898 I would need to extract 7EUSA But other string examples would be SWFX74849948, I would need to extract SWFX from that string.
Not sure how to do this with regex my limited knowledge is blocking me at this point:
^(\w{4}) that just gets me the first four characters but I don't really have a stopping point as sometimes the string could be somelongstring292894830982 which would require me to get somelongstring
Using \w will match a word character which includes characters and digits and an underscore.
You could match an optional digit [0-9]? from the start of the string ^and then match 1+ times A-Za-z
^[0-9]?[A-Za-z]+
Regex demo
const regex = /^[0-9]?[A-Za-z]+/;
[
"7EUSA8889er898",
"somelongstring292894830982",
"SWFX74849948"
].forEach(s => console.log(s.match(regex)[0]));
Can use this regex code:
(^\d+?[a-zA-Z]+)|(^\d+|[a-zA-Z]+)
I try with exmaple and good worked:
1- somelongstring292894830982 -> somelongstring
2- 7sdfsdf5456 -> 7sdfsdf
3- 875werwer54556 -> 875werwer
If you want to create function where the RegExp is parametrized by n parameter, this would be
function getStr(str,n) {
var pattern = "\\d?\\w{0,"+n+"}";
var reg = new RegExp(pattern);
var result = reg.exec(str);
if(result[0]) return result[0].substr(0,n);
}
There are answers to this but here is another way to do it.
var string1 = '7EUSA8889er898';
var string2 = 'SWFX74849948';
var Extract = function (args) {
var C = args.split(''); // Split string in array
var NI = []; // Store indexes of all numbers
// Loop through list -> if char is a number add its index
C.map(function (I) { return /^\d+$/.test(I) === true ? NI.push(C.indexOf(I)) : ''; });
// Get the items between the first and second occurence of a number
return C.slice(NI[0] === 0 ? NI[0] + 1 : 0, NI[1]).join('');
};
console.log(Extract(string1));
console.log(Extract(string2));
Output
EUSA
SWFX7
Since it's hard to tell what you are trying to match, I'd go with a general regex
^\d?\D+(?=\d)
I have a string with this format:
#someID#tn#company#somethingNew#classing#somethingElse#With
There might be unlimited #-separated words, but definitely the whole string begins with #
I have written the following regexp, though it matches it, but I cannot get each #-separated word, and what I get is the last recursion and the first (as well as the whole string). How can I get an array of every word in an element separately?
(?:^\#\w*)(?:(\#\w*)+) //I know I have ruled out second capturing group with ?: , though doesn't make much difference.
And here is my Javascript code:
var reg = /(?:^\#\w*)(?:(\#\w*)+)/g;
var x = null;
while(x = reg.exec("#someID#tn#company#somethingNew#classing#somethingElse#With"))
{
console.log(x);
}
And here is the result (Firebug, console):
["#someID#tn#company#somet...sing#somethingElse#With", "#With"]
0
"#someID#tn#company#somet...sing#somethingElse#With"
1
"#With"
index
0
input
"#someID#tn#company#somet...sing#somethingElse#With"
EDIT :
I want an output like this with regular expression if possible:
["#someID", "#tn", #company", "#somethingNew", "#classing", "#somethingElse", "#With"]
NOTE that I want a RegExp solution. I know about String.split() and String operations.
You can use:
var s = '#someID#tn#company#somethingNew#classing#somethingElse#With'
if (s.substr(0, 1) == "#")
tok = s.substr(1).split('#');
//=> ["someID", "tn", "company", "somethingNew", "classing", "somethingElse", "With"]
You could try this regex also,
((?:#|#)\w+)
DEMO
Explanation:
() Capturing groups. Anything inside this capturing group would be captured.
(?:) It just matches the strings but won't capture anything.
#|# Literal # or # symbol.
\w+ Followed by one or more word characters.
OR
> "#someID#tn#company#somethingNew#classing#somethingElse#With".split(/\b(?=#|#)/g);
[ '#someID',
'#tn',
'#company',
'#somethingNew',
'#classing',
'#somethingElse',
'#With' ]
It will be easier without regExp:
var str = "#someID#tn#company#somethingNew#classing#somethingElse#With";
var strSplit = str.split("#");
for(var i = 1; i < strSplit.length; i++) {
strSplit[i] = "#" + strSplit[i];
}
console.log(strSplit);
// ["#someID", "#tn", "#company", "#somethingNew", "#classing", "#somethingElse", "#With"]
I want to capture the "1" and "2" in "http://test.com/1/2". Here is my regexp /(?:\/([0-9]+))/g.
The problem is that I only get ["/1", "/2"]. According to http://regex101.com/r/uC2bW5 I have to get "1" and "1".
I'm running my RegExp in JS.
You have a couple of options:
Use a while loop over RegExp.prototype.exec:
var regex = /(?:\/([0-9]+))/g,
string = "http://test.com/1/2",
matches = [];
while (match = regex.exec(string)) {
matches.push(match[1]);
}
Use replace as suggested by elclanrs:
var regex = /(?:\/([0-9]+))/g,
string = "http://test.com/1/2",
matches = [];
string.replace(regex, function() {
matches.push(arguments[1]);
});
In Javascript your "match" has always an element with index 0, that contains the WHOLE pattern match. So in your case, this index 0 is /1 and /2 for the second match.
If you want to get your DEFINED first Matchgroup (the one that does not include the /), you'll find it inside the Match-Array Entry with index 1.
This index 0 cannot be removed and has nothing to do with the outer matching group you defined as non-matching by using ?:
Imagine Javascript wrapps your whole regex into an additional set of brackets.
I.e. the String Hello World and the Regex /Hell(o) World/ will result in :
[0 => Hello World, 1 => o]
I have a strings "add_dinner", "add_meeting", "add_fuel_surcharge" and I want to get characters that are preceded by "add_" (dinner, meeting, fuel_surcharge).
[^a][^d]{2}[^_]\w+
I have tried this one, but it only works for "add_dinner"
[^add_]\w+
This one works for "add_fuel_surcharge", but takes "inner" from "add_dinner"
Help me to understand please.
Use capturing groups:
/^add_(\w+)$/
Check the returned array to see the result.
Since JavaScript doesn't support lookbehind assertions, you need to use a capturing group:
var myregexp = /add_(\w+)/;
var match = myregexp.exec(subject);
if (match != null) {
result = match[1];
}
[^add_] is a character class that matches a single character except a, d or _. When applied to add_dinner, the first character it matches is i, and \w+ then matches nner.
The [^...] construct matches any single character except the ones listed. So [^add_] matches any single character other than "a", "d" or "_".
If you want to retrieve the bit after the _ you can do this:
/add_(\w+_)/
Where the parentheses "capture" the part of the expression inside. So to get the actual text from a string:
var s = "add_meeting";
var result = s.match(/add_(\w+)/)[1];
This assumes the string will match such that you can directly get the second element in the returned array that will be the "meeting" part that matched (\w+).
If there's a possibility that you'll be testing a string that won't match you need to test that the result of match() is not null.
(Or, possibly easier to understand: result = "add_meeting".split("_")[1];)
You can filter _ string by JavaScript for loop ,
var str = ['add_dinner', 'add_meeting', 'add_fuel_surcharge'];
var filterString = [];
for(var i = 0; i < str.length; i ++){
if(str[i].indexOf("_")>-1){
filterString.push(str[i].substring(str[i].indexOf("_") + 1, str[i].length));
}
}
alert(filterString.join(", "));