How to select text between two string javascript?

Question

I have this string:

var str = "Toto PARKER rejected the task.<br><br>Approval comments:<br>11-07-2017 20:11:29 - James EY (Comments)<br>NO!!<br><br>";

I would like a function to return "NO!!"

To do this i using:

<p>Click the button to display the array values after the split.</p>

<button onclick="myFunction()">Try it</button>   
<p id="demo"></p>

<script>
function myFunction() {
    var str = "Toto PARKER rejected the task.<br><br>Approval comments:<br>11-07-2017 20:11:29 - James EY (Comments)<br>NO!!<br><br>";
    var res = str.match("(<br\s*[\/]?>)[1](.*)(<br\s*[\/]?><br\s*[\/]?>)");
    document.getElementById("demo").innerHTML = res[0];          
}
</script>

I have some difficulties with my regex because the return is:

 11-07-2017 20:11:29 - James EY (Comments) NO!! 

So, how to start my regex with the last   (to return  NO!! ) and after how to not return  ?

Why don't you split with br tag and remove null data from array and take last using pop method of array. If not removing null values then last but one from the array — Priya
– Priya, Commented Jul 12, 2017 at 9:19
I assume it always would be br tags, so you can use / (.*?) /g — Milan Chheda
– Milan Chheda, Commented Jul 12, 2017 at 9:25

Mistalis · Accepted Answer · 2017-07-12 09:24:56Z

1

Here is a solution without a regex. I splitted your string on  , then removed the null values, and display the last element from the result:

function myFunction() {
    var str = "Toto PARKER rejected the task.<br><br>Approval comments:<br>11-07-2017 20:11:29 - James EY (Comments)<br>NO!!<br><br>";
    var res = str.split('<br>').filter(n => n);
    document.getElementById("demo").innerHTML = res.pop();
}

<p>Click the button to display the array values after the split.</p>
<button onclick="myFunction()">Try it</button>
<p id="demo"></p>

answered Jul 12, 2017 at 9:24

Mistalis

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Comments

Frank Wisniewski · Accepted Answer · 2017-07-12 09:30:01Z

0

A short form...

console.log(
  str.split('<br>').filter( e => e ).pop()
);

answered Jul 12, 2017 at 9:30

Frank Wisniewski

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