Is there a javascript string function that search a regex and it will start the search at the end?
If not, what is the fastest and/or cleanest way to search the index of a regex starting from the end?
example of regex:
/<\/?([a-z][a-z0-9]*)\b[^>]*>?/gi
Is there a javascript string function that search a regex and it will start the search at the end?
If not, what is the fastest and/or cleanest way to search the index of a regex starting from the end?
example of regex:
/<\/?([a-z][a-z0-9]*)\b[^>]*>?/gi
- 1 Is there a version of JavaScript's String.indexOf() that allows for regular expressions? – Andreas Commented Oct 18, 2013 at 9:26
7 Answers
Reset to default 4Maybe this can be useful and easier:
str.lastIndexOf(str.match(<your_regex_here>).pop());
Perhaps something like this is suitable for you?
Javascript
function lastIndexOfRx(string, regex) {
var match = string.match(regex);
return match ? string.lastIndexOf(match.slice(-1)) : -1;
}
var rx = /<\/?([a-z][a-z0-9]*)\b[^>]*>?/gi;
console.log(lastIndexOfRx("", rx));
console.log(lastIndexOfRx("<i>it</i><b>bo</b>", rx));
jsFiddle
And just for interest, this function vs the function that you choose to go with. jsperf
This requires that you format your regex correctly for matching exactly the pattern you want and globally (like given in your question), for example /.*(<\/?([a-z][a-z0-9]*)\b[^>]*>?)/i will not work with this function. But what you do get is a function that is clean and fast.
You may create a reverse function like:
function reverse (s) {
var o = '';
for (var i = s.length - 1; i >= 0; i--)
o += s[i];
return o;
}
and then use
var yourString = reverse("Your string goes here");
var regex = new Regex(your_expression);
var result = yourString.match(regex);
Another idea: if you want to search by word in reverse order then
function reverseWord(s) {
var o = '';
var split = s.split(' ');
for (var i = split.length - 1; i >= 0; i--)
o += split[i] + ' ';
return o;
}
var yourString = reverseWord("Your string goes here");
var regex = new Regex(your_expression);
var result = yourString.match(regex);
Andreas gave this from the ment:
https://stackoverflow./a/274094/402037
String.prototype.regexLastIndexOf = function(regex, startpos) {
regex = (regex.global) ? regex : new RegExp(regex.source, "g" + (regex.ignoreCase ? "i" : "") + (regex.multiLine ? "m" : ""));
if(typeof (startpos) == "undefined") {
startpos = this.length;
} else if(startpos < 0) {
startpos = 0;
}
var stringToWorkWith = this.substring(0, startpos + 1);
var lastIndexOf = -1;
var nextStop = 0;
while((result = regex.exec(stringToWorkWith)) != null) {
lastIndexOf = result.index;
regex.lastIndex = ++nextStop;
}
return lastIndexOf;
}
Which gives the functionality that I need, I tested my regex, and it is successful. So I'll use this
It depends what you exactly want to search for. You can use string.lastIndexOf or inside the regexp to use $ (end of the string).
Update: try the regexp
/<\/?([a-z][a-z0-9]*)\b[^>]*>?[\w\W]*$/gi
var m = text.match(/.*(<\/?([a-z][a-z0-9]*)\b[^>]*>?)/i);
if (m) {
textFound = m[1];
position = text.lastIndexOf(textFound);
}
Use .* to skip as much text as posible, capture the text found and search it with lastIndexOf
EDIT:
Well, if text is found, no need to search with lastIndexOf. m[0] contains the full coincidence (including all the initial padding), and m[1] the searched text. So position of found text is m[0].length - m[1].length
var m = text.match(/.*(<\/?([a-z][a-z0-9]*)\b[^>]*>?)/i);
if (m) {
textFound = m[1];
position = m[0].length - m[1].length;
}
Assuming you're looking for a string 'token', then you need the position of 'token' that has no other 'token' following until the end of the string.
So you should pose your regex something like that:
$token = 'token';
$re = "/(?:$token)[^(?:$token)]*$/";
This will find your 'token' where no further 'token' can be found until string end. The "(?:" grouping simply makes the group non-storing, slightly speeding up performance and saving memory.