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(", "));
Related
I want a Javascript regex or with any possible solution,
For a given string finds all the substrings that start with a particular string and end with a particular character. The returned set of subStrings can be an Array.
this string can also have nested within parenthesis.
var str = "myfunc(1,2) and myfunc(3,4) or (myfunc(5,6) and func(7,8))";
starting char = "myfunc" ending char = ")" . here ending character should be first matching closing paranthesis.
output: function with arguments.
[myfunc(1,2),
myfunc(3,4),
myfunc(5,6),
func(7,8)]
I have tried with this. but, its returning null always.
var str = "myfunc(1,2) and myfunc(3,4) or (myfunc(5,6) and func(7,8))";
var re = /\myfunc.*?\)/ig
var match;
while ((match = re.exec(str)) != null){
console.log(match);
}
Can you help here?
I tested your regex and it seems to work fine:
let input = "myfunc(1,2) and myfunc(3,4) or (myfunc(5,6) and func(7,8))"
let pattern = /myfunc.*?\)/ig
// there is no need to use \m since it does nothing, and NO you dont need it even if you use 'm' at the beginning.
console.log(input.match(pattern))
//[ "myfunc(1,2)", "myfunc(3,4)", "myfunc(5,6)" ]
If you use (?:my|)func\(.+?\) you will be able to catch 'func(7,8)' too.
(?:my|)
( start of group
?: non capturing group
my| matches either 'my' or null, this will match either myfunc or func
) end of group
Test the regex here: https://regex101.com/r/3ujbdA/1
Need to extract values from a string using regex(for perf reasons).
Cases might be as follows:
RED,100
RED,"100"
RED,"100,"
RED,"100\"ABC\"200"
The resulting separated [label, value] array should be:
['RED','100']
['RED','100']
['RED','100,']
['RED','100"ABC"200']
I looked into solutions and a popular library even, just splits the entire string to get the values,
e.g. 'RED,100'.split(/,/) might just do the thing.
But I was trying to make a regex with comma, which splits only if that comma is not enclosed within a quotes type value.
This isnt a standard CSV behaviour might be. But its very easy for end-user to enter values.
enter label,value. Do whatever inside value, if thats surrounded by quotes. If you wanna contain quotes, use a backslash.
Any help is appreciated.
You can use this regex that takes care of escaped quotes in string:
/"[^"\\]*(?:\\.[^"\\]*)*"|[^,"]+/g
RegEx Explanation:
": Match a literal opening quote
[^"\\]*: Match 0 or more of any character that is not \ and not a quote
(?:\\.[^"\\]*)*: Followed by escaped character and another non-quote, non-\. Match 0 or more of this combination to get through all escaped characters
": Match closing quote
|: OR (alternation)
[^,"]+: Match 1+ of non-quote, non-comma string
RegEx Demo
const regex = /"[^"\\]*(?:\\.[^"\\]*)*"|[^,"]+/g;
const arr = [`RED,100`, `RED,"100"`, `RED,"100,"`,
`RED,"100\\"ABC\\"200"`];
let m;
for (var i = 0; i < arr.length; i++) {
var str = arr[i];
var result = [];
while ((m = regex.exec(str)) !== null) {
result.push(m[0]);
}
console.log("Input:", str, ":: Result =>", result);
}
You could use String#match and take only the groups.
var array = ['RED,100', 'RED,"100"', 'RED,"100,"', 'RED,"100\"ABC\"200"'];
console.log(array.map(s => s.match(/^([^,]+),(.*)$/).slice(1)))
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 the following string: pass[1][2011-08-21][total_passes]
How would I extract the items between the square brackets into an array? I tried
match(/\[(.*?)\]/);
var s = 'pass[1][2011-08-21][total_passes]';
var result = s.match(/\[(.*?)\]/);
console.log(result);
but this only returns [1].
Not sure how to do this.. Thanks in advance.
You are almost there, you just need a global match (note the /g flag):
match(/\[(.*?)\]/g);
Example: http://jsfiddle.net/kobi/Rbdj4/
If you want something that only captures the group (from MDN):
var s = "pass[1][2011-08-21][total_passes]";
var matches = [];
var pattern = /\[(.*?)\]/g;
var match;
while ((match = pattern.exec(s)) != null)
{
matches.push(match[1]);
}
Example: http://jsfiddle.net/kobi/6a7XN/
Another option (which I usually prefer), is abusing the replace callback:
var matches = [];
s.replace(/\[(.*?)\]/g, function(g0,g1){matches.push(g1);})
Example: http://jsfiddle.net/kobi/6CEzP/
var s = 'pass[1][2011-08-21][total_passes]';
r = s.match(/\[([^\]]*)\]/g);
r ; //# => [ '[1]', '[2011-08-21]', '[total_passes]' ]
example proving the edge case of unbalanced [];
var s = 'pass[1]]][2011-08-21][total_passes]';
r = s.match(/\[([^\]]*)\]/g);
r; //# => [ '[1]', '[2011-08-21]', '[total_passes]' ]
add the global flag to your regex , and iterate the array returned .
match(/\[(.*?)\]/g)
I'm not sure if you can get this directly into an array. But the following code should work to find all occurences and then process them:
var string = "pass[1][2011-08-21][total_passes]";
var regex = /\[([^\]]*)\]/g;
while (match = regex.exec(string)) {
alert(match[1]);
}
Please note: i really think you need the character class [^\]] here. Otherwise in my test the expression would match the hole string because ] is also matches by .*.
'pass[1][2011-08-21][total_passes]'.match(/\[.+?\]/g); // ["[1]","[2011-08-21]","[total_passes]"]
Explanation
\[ # match the opening [
Note: \ before [ tells that do NOT consider as a grouping symbol.
.+? # Accept one or more character but NOT greedy
\] # match the closing ] and again do NOT consider as a grouping symbol
/g # do NOT stop after the first match. Do it for the whole input string.
You can play with other combinations of the regular expression
https://regex101.com/r/IYDkNi/1
[C#]
string str1 = " pass[1][2011-08-21][total_passes]";
string matching = #"\[(.*?)\]";
Regex reg = new Regex(matching);
MatchCollection matches = reg.Matches(str1);
you can use foreach for matched strings.