This is my code snippet:
var myString = '#EXTINF:-1 group-title="|FR| CHAINES FRANÇAISES |FR|",|FR|*****CANALSAT*****|FR|';
var group_title = /(group-title=")(\S*)["]/;
var matchgroup_title = group_title.exec(myString);
console.log(matchgroup_title);
I am not familiar to regex so I can't understand this code. Why does exec here return null?
The problem exists because of \S*. If you pop it into regex101.com, you'll see:
So, \s* will match any non-whitespace character, and a space is considered a whitespace character, so this does not match. You can simply use [^"] to check for anything that isn't another quote.
You can simplify this to:
var myString = '#EXTINF:-1 group-title="|FR| CHAINES FRANÇAISES |FR|",|FR|*****CANALSAT*****|FR|';
var group_title = /group-title="([^"]+)"/;
var matchgroup_title = group_title.exec(myString);
console.log(matchgroup_title);
(group-title=") matches group-title="
(\S*) matches zero or more non-whitespace characters, so |FR|
["] then matches a ", but there isn't one (well, there is, but there are whitespace characters first, so no match).
Related
I have a string which looks like below
str = "hey there = pola"
Now I need to check if there is equal = sign and the first word to the left of it. So this is what I do
str.match(/\w+(?= *=)/)[0]
So I get the desired result
But say I have a string like this
str = "hey there= pola so = boba"
Now I have two = signs. But the above regex will only give me the result for the first = sign.
Is there any regex that can always look for the first instance of = from the end of the string?
You can assert what is on the right is an equals sign followed by matching any char except an equals sign until the end of the string
\w+(?= *=[^=]*$)
In parts:
\w+
(?= Positive lookahead
*= Match 0+ occurrences of a space followed by =
[^=]* Match 0+ occurrences of = ( Use [^=\r\n]* to not cross line breaks)
$ End of string
) Close lookahead
Regex demo
const regex = /\w+(?= *=[^=]*$)/;
const str = `hey there= pola so = boba`;
console.log(str.match(regex)[0]);
Without using a lookahead, you could use a capturing group:
^.*\b(\w+) *=[^=]*$
Regex demo
const regex = /^.*\b(\w+) *=[^=]*$/m;
const str = `hey there= pola so = boba`;
console.log(str.match(regex)[1]);
I'm not much expert on regex but for you requirement I think split and pop should work
let str = "hey there= pola so = boba";
let endres = str.split('=').pop(); // gives the last element in the split array
Hope this helps.
I want to replace {r-group1} with "REPLACED" but leave the , where it is.
So, the string
var string = "{r-group1, }foo bar"
should output: "REPLACED, foo bar"
Using a negative lookahead, I tried adding a preceding (?![,]) group to leave the comma alone:
var replaced = string.replace(^(?:(?![,]){r-group1\})+$, 'REPLACED');
But it returns the same string. There are no matches to replace.
The same goes for a preceding comma:
var string = "foo bar{r-, group1}"
This should output: "foo bar, REPLACED"
You could do the replacement without a lookahead. You could match the curly braces and the content that comes before and after it except a comma using a negated character class [^,}]+ and capture the comma with optional whitespace chars in a capturing group.
In the replacement use the capturing groups $1REPLACED$2
Credits to #Nick for the updated pattern.
{r-(,?\s*)[^,}]+(,?\s*)}
Regex demo
const regex = /{r-(,?\s*)[^,}]+(,?\s*)}/g;
const str = `{r-group1, }foo bar`;
const subst = `$1REPLACED$2`;
const result = str.replace(regex, subst);
console.log(result);
match a path for a specific word and a / and any characters that follow.
For example.
const str = 'cars/ford';
const isCars = str.match('cars');
What I want to do is make sure it matches cars and has a slash and characters after the / then return true or false.
The characters after cars/... will change so I can't match it excatly. Just need to match any characters along with the /
Would love to use regex not sure what it should be. Looking into how to achieve that via regex tutorials.
var str = "cars/ford";
var patt = new RegExp("^cars/"); //or var patt = /^cars\//
var res = patt.test(str); //true
console.log(res);
https://www.w3schools.com/js/js_regexp.asp
https://www.rexegg.com/regex-quickstart.html
You could use test() that returns true or false.
const str = "cars/ford";
const str2 = "cars/";
var isCars = (str)=>/^cars\/./i.test(str)
console.log(isCars(str));
console.log(isCars(str2));
Here is a quick regex to match "cars/" followed by any characters a-z.
(cars\/[a-z]+)
This will only match lowercase letters, so you can add the i flag to make it case insensitive.
/(cars\/[a-z]+)/i
It is a basic regular expression
var str = "cars/ford"
var result = str.match(/^cars\/(.*)$/)
console.log(result)
^ - start
cars - match exact characters
\/ - match /, the \ escapes it
(.*) - capture group, match anything
$ - end of line
Visualize it: RegExper
var url = document.referrer;
var a=document.createElement('a');
a.href=url;
var path = a.pathname;
Let's say path is this:
/cat-dog-fish/
I want to remove leading and trailing slashes, if they exist, else do nothing.
I can do this (removes trailing slash):
a.pathname.replace(/\/$/,'')
Or this (removes leading slash)
a.pathname.replace(/^\//,'')
But how do I remove both at once, in a oner, if they exist?
A regex literal like /^\/|\/$/g can be used to replace with empty string, or you may use /^\/([^]*)\// (match /, then any 0+ chars up to the last / capturing what is in-between the slashes) to replace with $1:
var s = "/cat-dog-fish/";
console.log(s.replace(/^\/|\/$/g, ''));
console.log(s.replace(/^\/([^]*)\/$/, '$1'));
Note:
^\/ - matches the start of string and a / right there
| - means OR
\/$ - matches a / at the end of string
([^]*) - is a capturing group (...) that captures 0 or more (*) any characters as [^] means not nothing.
var a="/cat-dog-fish/";
var d = a.replace(new RegExp("(^\/|\/$)",'g'),'');
console.log(d);
a.pathname.replace(new RegExp("(^\/|\/$)",'g'),'');
I'm trying to match all the words starting with # and words between 2 # (see example)
var str = "#The test# rain in #SPAIN stays mainly in the #plain";
var res = str.match(/(#)[^\s]+/gi);
The result will be ["#The", "#SPAIN", "#plain"] but it should be ["#The test#", "#SPAIN", "#plain"]
Extra: would be nice if the result would be without the #.
Does anyone has a solution for this?
You can use
/#\w+(?:(?: +\w+)*#)?/g
See the demo here
The regex matches:
# - a hash symbol
\w+ - one or more alphanumeric and underscore characters
(?:(?: +\w+)*#)? - one or zero occurrence of:
(?: +\w+)* - zero or more occurrences of one or more spaces followed with one or more word characters followed with
# - a hash symbol
NOTE: If there can be characters other than word characters (those in the [A-Za-z0-9_] range), you can replace \w with [^ #]:
/#[^ #]+(?:(?: +[^ #]+)*#)?/g
See another demo
var re = /#[^ #]+(?:(?: +[^ #]+)*#)?/g;
var str = '#The test-mode# rain in #SPAIN stays mainly in the #plain #SPAIN has #the test# and more #here';
var m = str.match(re);
if (m) {
// Using ES6 Arrow functions
m = m.map(s => s.replace(/#$/g, ''));
// ES5 Equivalent
/*m = m.map(function(s) {
return s.replace(/#$/g, '');
});*/ // getting rid of the trailing #
document.body.innerHTML = "<pre>" + JSON.stringify(m, 0, 4) + "</pre>";
}
You can also try this regex.
#(?:\b[\s\S]*?\b#|\w+)
(?: opens a non capture group for alternation
\b matches a word boundary
\w matches a word character
[\s\S] matches any character
See demo at regex101 (use with g global flag)