This
var re = /[^<a]b/;
var str = "<a>b";
console.log(str.match(re)[0]);
matches >b.
However, I don't understand why this pattern /[^<a>]b/ doesn't match anything. I want to capture only the "b".
The reason why /[^<a>]b/ doesn't do anything is that you are ignoring <, a, and > as individual characters, so rewriting it as /[^><a]b/ would do the same thing. I doubt this is what you want, though. Try the following:
var re = /<a>(b)/;
var str = "<a>b";
console.log(str.match(re)[1]);
This regex looks for a string that looks like <a>b first, but it captures the b with the parentheses. To access the b, simply use [1] when you call .match instead of [0], which would return the entire string (<a>b).
What you're using here is a match for a b preceded by any character that is not listed in the group. The syntax [^a-z+-] where the a-z+- is a range of characters (in this case, the range of the lowercase Latin letters, a plus sign and a minus sign). So, what your regex pattern matches is any b preceded by a character that is NOT < or a. Since > doesn't fall in that range, it matches it.
The range selector basically works the same as a list of characters that are seperated by OR pipes: [abcd] matches the same as (a|b|c|d). Range selectors just have an extra functionality of also matching that same string via [a-d], using a dash in between character ranges. Putting a ^ at the start of a range automatically turns this positive range selector into a negative one, so it will match anything BUT the characters in that range.
What you are looking for is a negative lookahead. Those can exclude something from matching longer strings. Those work in this format: (?!do not match) where do not match uses the normal regex syntax. In this case, you want to test if the preceding string does not match <a>, so just use:
(?!<a>)(.{3}|^.{0,2})b
That will match the b when it is either preceded by three characters that are not <a>, or by fewer characters that are at the start of the line.
PS: what you are probably looking for is the "negative lookbehind", which sadly isn't available in JavaScript regular expressions. The way that would work is (?<!<a>)b in other languages. Because JavaScript doesn't have negative lookbehinds, you'll have to use this alternative regex.
you could write a pattern to match anchor tag and then replace it with empty string
var str = "<a>b</a>";
str = str.replace(/((<a[\w\s=\[\]\'\"\-]*>)|</a>)/gi,'')
this will replace the following strings with 'b'
<a>b</a>
<a class='link-l3'>b</a>
to better get familiar with regEx patterns you may find this website very useful regExPal
Your code :
var re = /[^<a>]b/;
var str = "<a>b";
console.log(str.match(re));
Why [^<a>]b is not matching with anything ?
The meaning of [^<a>]b is any character except < or a or > then b .
Hear b is followed by > , so it will not match .
If you want to match b , then you need to give like this :
var re = /(?:[\<a\>])(b)/;
var str = "<a>b";
console.log(str.match(re)[1]);
DEMO And EXPLANATION
Related
I have string like this -
var string = 'callback&user_story=1&user_story=2&user_story=100&user_story=a&user_story=john';
&user_story=x (Here x can be anything) can repeat n number of times.
How to replace this '&user_story=x' with a blank value.
What would be regex for it in JS?
The regex looks like this:
var string = 'callback&user_story=1&user_story=2&user_story=100&user_story=a&user_story=john';
var re = /&user_story=.*?(?=&|$)/g
console.log(string.replace(re, ""));
Broken down:
/&user_story=.*?(?=&|$)/g
&user_story= - checks whether the match starts with "&user_story="
.*? - matches any number of any characters, but the ? makes it non-greedy, so it will find as few of these characters before finding the next part of the regex
(?=&|$) - the brackets make this a group, and ?= means it's a lookahead, i.e. it won't actually add matches to the regex match, but just checks to see they're there. It will match either another &, or the end of the string (symbolised by $).
g - is a flag which tells the regex to check the entire string, and not stop after just finding one match.
I need a regular expression for :
<<12.txt>> <<45.txt>
I have created a regular expression :
<<.+.txt>>
But this found one match in whole string but here is 2 matches:
<<12.txt>>
<<45.txt>>
if anyone have solution for this problem please help me out there
Part of the issue is that the string you've specified wouldn't match because the second > is missing in <<45.txt>.
Also, you're using the . (dot) selector, and also trying to find a period. It works, but now how you think it is.
Here's the regex you want:
var regex = /<<\d+\.txt>>/g
\d matches only numbers
\. matches an actual period
/g means global, so it won't stop at the first match
Practice Regular Expressions
https://regexr.com/43bs4
Demo
var string = "<<12.txt>> <<45.txt>>";
var regex = /<<\d+\.txt>>/g;
var matches = string.match(regex);
console.log(matches);
P.S., if you actually want to match with 1 > or 2 >>, you can with:
var regex = /<<\d+\.txt>>?/g
? optionally matches the character right before it
/<<.+.txt>>/gm
g is for global (will search through entire source)
m is for multi line search support
Trying to use this regex to verify usernames and this is what I have :
var goodUsername = /[a-zA-Z0-9_]/g;
console.log(goodUsername.test("HELO $"));
But wether or not I have $ in there it returns true. Not sure why.
I basically only want letters, numbers and _ in usernames and that's it
It seems to work here https://regex101.com/r/nP4iG7/1
The RegEx that you use searches any match in the subject string. In your case HELO matches the criteria. If you like to apply the criteria to the whole string you should define the string begin and end using
var goodUsername = /^[a-zA-Z0-9_]+$/;
console.log(goodUsername.test("HELO $"));//false
You need to add anchors..
/^[a-zA-Z0-9_]+$/;
Anchors help to do exact matching. ^ start of the line anchor, $ end of the line anchor. And also you need to repeat the char class one or more times otherwise it would match a string which contains exactly one character.
You could search for any characters not in the list (a "negated character set"):
var badUsername = /[^a-zA-Z0-9_]/;
console.log(!badUsername.test("HELO $"));
or more simply
var badUsername = /\W/;
since \W is defined as
Matches any character that is not a word character from the basic Latin alphabet. Equivalent to [^A-Za-z0-9_].
If you prefer to do a positive match, using anchors as other answers have suggested, you can shorten your regexp by using \w:
var goodUsername = /^\w+$/;
My string can be something like A01, B02, C03, possibly AA18 in the future as well. I thought I could use a regex to get just the letters and work on my regex since I haven't done much with it. I wrote this function:
function rowOffset(sequence) {
console.log(sequence);
var matches = /^[a-zA-Z]+$/.exec(sequence);
console.log(matches);
var letter = matches[0].toUpperCase();
return letter;
}
var x = "A01";
console.log(rowOffset(x));
My matches continue to be null. Am I doing this correctly? Looking at this post, I thought the regex was correct: Regular expression for only characters a-z, A-Z
You can use String#replace to remove all non letters from input string:
var r = 'AA18'.replace(/[^a-zA-Z]+/g, '');
//=> "AA"
Your main issue is the use of the ^ and $ characters in the regex pattern. ^ indicates the beginning of the string and $ indicates the end, so you pattern is looking for a string that is ONLY a group of one or more letters, from the beginning to the end of the string.
Additionally, if you want to get each individual instance of the letters, you want to include the "global" indicator (g) at the end of your regex pattern: /[a-zA-Z]+/g. Leaving that out means that it will only find the first instance of the pattern and then stop searching . . . adding it will match all instances.
Those two updates should get you going.
EDIT:
Also, you may want to use match() rather than exec(). If you have a string of multiple values (e.g., "A01, B02, C03, AA18"), match() will return them all in an array, whereas, exec() will only match the first one. If it is only ever one value, then exec() will be fine (and you also wouldn't need the "global" flag).
If you want to use match(), you need to change your code order just a bit to:
var matches = sequence.match(/[a-zA-Z]+/g);
To return an array of separate letters remove +:
var matches = sequence.match(/[a-zA-Z]/g);
You're confused about what the goal of the other question was: he wanted to check that there were only letters in his string.
You need to remove the anchors ^$, who match respectively the beginning and end of the string:
[a-zA-Z]+
This will match the first of letters in your input string.
If there might be more (ie you want multiple matches in your single string), use
sequence.match(/[a-zA-Z]+/g)
This /[^a-z]/g solves the problem. Look at the example below.
function pangram(str) {
let regExp = /[^a-z]/g;
let letters = str.toLowerCase().replace(regExp, '');
document.getElementById('letters').innerHTML = letters;
}
pangram('GHV 2## %hfr efg uor7 489(*&^% knt lhtkjj ngnm!##$%^&*()_');
<h4 id="letters"></h4>
You can do this:
var r = 'AA18'.replace(/[\W\d_]/g, ''); // AA
Also can be done by String.prototype.split(regex).
'AA12BB34'.split(/(\d+)/); // ["AA", "12", "BB", "34", ""]
'AA12BB34'.split(/(\d+)/)[0]; // "AA"
Here regex divides the giving string by digits (\d+)
the following regex
var str = "1234,john smith,jack jone";
var match = str.match(/([^,]*,[^,]*,[^ ]*)/g);
alert(match);
returns
1234,john smith,jack
But what I am trying to get is the whole string which is
1234,john smith,jack jones
Basically my script does the job only for the first whitespace between commas but I want to do it everytime there is a white space between commas.
Can anyone help me out pls.
Your pattern excludes spaces from the last section so as soon as it encounters a space in after the third comma, that's the end of the match. You might want to try this instead:
var match = str.match(/[^,]*,[^,]*,.*/g);
This will allow anything after the second comma, including spaces or more commas (since your original pattern allowed commas after the the second).
If you'd like to match pattern only on a single, use start / end anchors (^ / $) as well as the multiline flag (m), like us this:
var match = str.match(/^[^,]*,[^,]*,.*$/mg);
You can try it out with this simple demo.
Why you're not using split ?
"1234,john smith,jack jone".split(/,/)
or
"1234,john smith,jack jone".split(",")