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I need a regex that detects words inside a string. My regex works fine when the class consists of a single word, but not for multiple words.
function repl(string){
return string.replace(/<span class="[a-z\s\-\b \b]*">/ , (n)=>{
return <Symbol/>
}
}
Works:
repl('<span class='hi'>') /==> <Symbol/>
Does not work:
repl('<span class='hi hi'>')
You can simplify your expression to: /<span class=["'][a-z_\-][\w\s\-]*["']>/i
Make sure your HTML element attributes are double-quoted.
Note: Please be aware that the closing </span> tag is mandatory and should never be ommited.
const repl = (str) =>
str.replace(/<span class=["'][a-z_\-][\w\-\s]*["']>/i, () => '<Symbol />');
console.log(repl('<span class="hi">'));
console.log(repl("<span class='hi hi'>"));
console.log(repl('<span class="one two three ">'));
Related
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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 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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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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function validate(username) {
var reg = /^\w+([-+.]\w+)*#\w+([-.]\w+)*$/;
if(reg.test(username)) {
alert("is correct");
return true;
}
else {
return false;
}
}
Pattern requires an #
This pattern ^\w+([-+.]\w+)*#\w+([-.]\w+)*$ requires an # in your input.
It matches a#a but not someusername.
If you want to build a username regex, I suggest you can with something simple like:
^[-.\w]{2,20}$
and tweak from there.
The ^ anchor asserts that we are at the beginning of the string
[-.\w] matches one word character (letters, digits, underscores), dash or period
{2,20} matches two to 20 of these characters
The $ anchor asserts that we are at the end of the string
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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, '');