JS What's the fastest way to display one specific line of a list?

Question

In my Javascript code, I get one very long line as a string. This one line only has around 65'000 letters. Example:

config=123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...

What I have to do is replace all & with an break (\\n) first and then pick only the line which starts with "path_of_code=" . This line I have to write in a variable.

The part with replace & with an break (\\n) I already get it, but the second task I didn't.

    var obj = document.getElementById('div_content');
    var contentJS= obj.value;
    var splittedResult;
    splittedResult = contentJS.replace(/&/g, '\n');

What is the fastest way to do it? Please note, the list is usually very long.

Answer 1

It sounds like you want to extract the text after &path_of_code= up until either the end of the string or the next & . That's easily done with a regular expression using a capture group, then using the value of that capture group:

var rex = /&path_of_code=([^&]+)/;
var match = rex.exec(theString);
if (match) {
    var text = match[1];
}

Live Example:

 var theString = "config=123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs..."; var rex = /&path_of_code=([^&]+)/; var match = rex.exec(theString); if (match) { var text = match[1]; console.log(text); } 

Answer 2

but the second task I didn't.

You can use filter() and startsWith()

splittedResult = splittedResult.filter(i => i.startsWith('path_of_code='));

Answer 3

Use combination of String.indexOf() and String.substr()

 var contentJS= "123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs..."; var index = contentJS.indexOf("&path_of_code"), substr = contentJS.substr(index+1), res = substr.substr(0, substr.indexOf("&")); console.log(res)