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I try to regex for not contain some character.
I need to show /%7(.*?);/g which dose not contain "=".
I try to input
?!xx=1
and change to
?!( (.?)=(.?) )
But it dose not work.
Please help. Thanks.
//Here is my simple regex
reg = /%7((?!xx=1).*?);/g ;
//Here is my string
str = "%7aa; %7bb=11; %7cc=123; %7xx=1; %7yy; %7zz=2;"
//I need
%7aa; and %7yy;
Instead of using a negative lookahead, try using a ^ block:
const reg = /%7([^=;]+);/g;
The ([^=;]+) bit matches any non-=, the condition you're looking for, and non-;, the character at the end of your regex.
I left the capture group in since your question's regex also contains it.
const reg = /%7([^=;]+);/g;
const str = "%7aa; %7bb=11; %7cc=123; %7xx=1; %7yy; %7zz=2;"
const matches = str.match(reg);
console.log(matches);
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Can I split string by regex to match two parameters (: , ) only one time?
let str='part1:20/02/2002 part3 part4 part5';
let splittedArr=str.split(something..);
I would like the splittedArr to look like:
[part1 , 20/02/2002 , part3 part4 part5]
in js you can split on a regex
and the regex is fairly simple its ":" or " "
the or in regex is a |
let str='part1:20/02/2002 part3 part4 part5';
let splittedArr=str.split(/:| /);
console.log(splittedArr);
EDIT
I've misinterpreted the question, if I understand your question correctly;
You want to split on the first colon (:) and on the first space ( ), and capture those 3 parts.
You can achieve this with simple group matches, and ungreedy wildcards.
let str='part1:20/02/2002 part3 part4 part5';
let parts=str.match('^(.*?):(.*?) (.*)$');
console.log(parts);
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I don't know how to replace with capital letters in this input text, for example I want to put on bold the letters which they are de same on the input, but if is capital letters doesn't put on bold
this is my line of code $texto_option = $texto_option.replace($(input).val(),'<b>'+$(input).val()+'</b>');
Generate a regular expression that ignores the letters' case:
var search = $(input).val();
var re = new RegExp(search, "i");
$texto_option = $texto_option.replace(re, "<b>$&</b>");
This is an easy answer. But keep in mind that your input has to be sanitized, as some characters are control characters for regular expressions too:
search = search.replace(/[\-\[\]\/\{\}\(\)\*\+\?\.\\\^\$\|]/g, "\\$&");
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I am looking for a method to get the regex rules, for a capture group.
So if I had: /(\w) (\d)/ I would want $1 = /\w/ and $2 = /\d/.
Is there a method for this?
Provided that the regex is valid and no nested groups are used, then you can try this out.
var arr = (regex + "").match(/\(.*?\)/g).map(function(rule){
return rule.replace(/[()]/g, "");
});
Now, arr[0] will have rule of group1, arr[1] will be group2 and so on.
Although I don't know of any direct way to do this on a regex rule given by /(rule1)(rule2)/ you could build the regex rule using strings and specify your grouping in different strings.
var group1 = '([A-Za-z]+)',
group2 = '(\\d+)',
r = new RegExp(group1+group2);
r.test('hello1234');
To get the groups from a regex string you could run a regex on that regex string to extract the groups. If your regex string is "(\w+)(\d+)" then you'd have a regex to extract the groups as /(\([^\)\))+/
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So I have a string that will have basically any kind of text in it and I want to capture all the text with regular expression and replace it.
var img_url = "http://grfx.domain.com/photos/dir/school/dir/m-dir/auto_original/123456.jpeg?123456";
img_url.replace(/regexp/,newvalue);
Just trying to capture everything in the img_url var and replace it.
Thanks!
Well, if you want to replace everything, you don't need regex. Simply write:
img_url = newvalue;
If you are really determined to do this with a regular expression, this could help:
img_url = img_url.replace(/.*/, newvalue);
If you're reassigning, just use =
If you want to replace all matches in one go, use the g flag, in which case this question is a duplicate of this post
If that expression needs to be dynamic:
var expr = new RegExp(replaceVar,'gi');//globally, and case-insensitively
var newUrl = oldUrl.replace(expr,replacement);
Note that, here, all back-slashes like you'd use for \b have to be escaped, too: \\b and \\/ for forward slashes. and \\\\ for an escaped backslash
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How would I go about removing numbers and a space from the start of a string?
For example, from '13 Adam Court, Cannock' remove '13 '.
Search for
/^[\s\d]+/
Replace with the empty string. Eg:
str = str.replace(/^[\s\d]+/, '');
This will remove digits and spaces in any order from the beginning of the string. For something that removes only a number followed by spaces, see BoltClock's answer.
str.replace(/^\d+\s+/, '');
var text = '13 Adam Court, Cannock';
var match = /\d+\s/.exec(text)[0];
text.replace(match,"");
The above solution doesn't work for me. Instead I used this regex below.
var sRegExResult = "Regex Sample99";
sRegExResult.replace(/\s|[0-9]/g, '');