I need to replace less or greater than(< >) characters, but keep any html tags(simple tags will do like <b>text</b> without arguments).
So the input below:
<b>> 0.5 < 0.4</b> - <>
Should be:
<b>> 0.5 < 0.4</b> - <>
All I managed to find and edit now is this expr:
<\/?[a-z][a-z0-9]*[^>\s]*>|([<>])
It groups the < and > characters but it also matches tags witch I don't need to replace
UPD: Thanks to @Sree Kumar, here's the final functions:
String.prototype.replaceAt = function (index, char) {
let arr = this.split('');
arr[index] = char;
return arr.join('');
};
String.prototype.escape = function () {
let p = /(?:<[a-zA-Z]+>)|(?:<\/[a-zA-Z]+>)|(?<lt><)|(?<gt>>)/g,
result = this,
match = p.exec(result);
while (match !== null) {
if (match.groups.lt !== undefined) {
result = result.replaceAt(match.index, '<');
}
else if (match.groups.gt !== undefined) {
result = result.replaceAt(match.index, '>');
}
match = p.exec(result);
}
return result;
};
Try this regex:
<(?!\/?\w+>)|(?<!<\w+|<\/\w+)>
Explanation
<(?!\/?\w+>) finds all '<' symbols (except in tags)(?<!<\w+|<\/\w+)> finds all '>' symbols (except in tags)You can use them separately:
let str = '<b>> 0.5 < 0.4</b> - <>';
let lessThen = /<(?!\/?\w+>)/g;
let greaterThen = /(?<!<\w+|<\/\w+)>/g;
str = str.replace(lessThen, '<');
str = str.replace(greaterThen, '>');
console.log(str); // <b>> 0.5 < 0.4</b> - <>NB! It only finds symbols '<' and '>' between tags. It doesn't check that html is valid. For text like that <a></b> it will not find any matches.
Here is a way to do it using named groups. That is, name your desired group and look for it. It may be null or undefined at times because it didn't match. Hence, you will have to add the null check.
Notice (?<B>...) surrounding the "desired" group. Also, notice the null check in the 5th line.
let p = /(?:<[a-zA-Z]+>)|(?:<\/[a-zA-Z]+>)|(?<B>[<>])/g
let input = '<b>> 0.5 < 0.4</b> - <>';
let match = p.exec( input );
while( match !== null) {
if( match.groups.B !== undefined ) console.log( match.groups.B );
match = p.exec( input )
}