conditionally delete numbers from string

Question

I have two types of Strings like the following :

78:24207 PERF keytouchListAction ProtocolSelect 04.00 : 1502876877199 1502876879180

or

78:27247 PERF create tab state : 1502876879180

I'm looking for a regex to delete the numbers at the start of the string 78:24207

to produice something like this :

PERF keytouchListAction ProtocolSelect 04.00 : 1502876877199 1502876879180

or

PERF create tab state : 1502876879180

and then if the string containt two numbers after the : take only the first number :

PERF keytouchListAction ProtocolSelect 04.00 : 1502876877199 1502876879180

becomes :

PERF keytouchListAction ProtocolSelect 04.00 : 1502876877199

I've tried this replace(/^\d+\.\s*/, ''); for the first pattern but does't seem to work.

and this pattern for the second problem replace(\:.*$) but nothing changed in my string.

Any hints on what I'm doing wrong ?

Use something like regex101.com for regex testing, as creating the regex ^(\d+\:\d+\s) was quite simple. --- You may not even need the capture group... — evolutionxbox
– evolutionxbox, Commented Aug 16, 2017 at 11:12

anubhava · Accepted Answer · 2017-08-16 11:30:05Z

2

You can do this in a single replace call:

var s = '78:24207 PERF keytouchListAction ProtocolSelect 04.00 : 1502876877199 1502876879180'
var r = s.replace(/^\d+:\d+\s*|(:\s*\d+)\s+\d+$/g, '$1')

console.log(r)

s = '78:27303 PERF tab state created : 1502882663195'
var r = s.replace(/^\d+:\d+\s*|(:\s*\d+)\s+\d+$/g, '$1')

console.log(r)

edited Aug 16, 2017 at 11:30

answered Aug 16, 2017 at 11:19

anubhava

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2 Comments

JSK Over a year ago

this is worked for the second problem but I still get the numbers at the start of the string 78:27303 PERF tab state created : 1502882663195

anubhava Over a year ago

78:27303 PERF tab state created : 1502882663195 is becoming PERF tab state created : 1502882663195 as you can see in my updated demo.

SamWhan · Accepted Answer · 2017-08-16 11:32:21Z

1

As an alternative: don't replace - get the wanted content with match using

[a-z].*:\s*\d+

It matches a letter and everything after it up to a colon, followed by (optional space and) a number.

document.write(
    '78:24207 PERF keytouchListAction ProtocolSelect 04.00 : 1502876877199 1502876879180'
    .match(/[a-z].*:\s*\d+/i)
);

answered Aug 16, 2017 at 11:32

SamWhan

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Comments

marvel308 · Accepted Answer · 2017-08-16 11:18:03Z

1

You can use it by first matching and then replacing like in the code snippet

function matchAndReplace(str){
	str = str.match(/\d+:\d+ ([\w\s.]*)(:\s*\d+)/g)[0];
	console.log(str.replace(/^\d+:\d+ /,''));
}
matchAndReplace("78:24207 PERF keytouchListAction ProtocolSelect 04.00 : 1502876877199 1502876879180");
matchAndReplace("78:27247 PERF create tab state : 1502876879180");

answered Aug 16, 2017 at 11:18

marvel308

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Comments

pchaigno · Accepted Answer · 2017-08-16 11:19:46Z

1

For the first pattern you need:

replace(/^\d+:\d+\s+/, '');

For the second pattern:

replace(/(\s+\d+)\s+\d+$/, '$1');

answered Aug 16, 2017 at 11:19

pchaigno

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Comments

Matus · Accepted Answer · 2017-08-16 11:24:01Z

1

Do you care about the pattern? If not you can just replace first 9 characters using this:

 var str = "78:27247 PERF create tab state : 1502876879180";
 var res = str.replace(/^.{9}/, "");
 console.log(res);

answered Aug 16, 2017 at 11:24

Matus

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