Exclude part of string using Regular expression

Question

I have an address like

New street 4 
462005 MG Road 
Tel .: 02281 / 93-1-212  
Fax .: 02681 / 93-1148

I want string before "Tel" and remove whatever comes after "Tel"

New street 4 
462005 MG Road

What i have tried is

(?!^Tel$)(^.*$)

But its only gives first line. New street 4

How can be achieve this using regular expression without any PHP function.

Oh, without any PHP function.. missed that important update. — chris85
– chris85, Commented Dec 29, 2015 at 13:37
@Adas Please notice that I updated my answer so you won't get screwed when a street name includes the string Tel. — timgeb
– timgeb, Commented Dec 29, 2015 at 13:39

timgeb · Accepted Answer · 2015-12-29 13:37:43Z

1

Make sure to activate single line mode:

(?s).*?(?=Tel)

Demo: https://regex101.com/r/dX7kU7/1

edit:
A little improvement: this updated version should be a tad more robust and stop when a new line starts with Tel, i.e. you won't get screwed when a street name includes the substring Tel: https://regex101.com/r/dX7kU7/2

edited Dec 29, 2015 at 13:37

answered Dec 29, 2015 at 13:24

timgeb

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2 Comments

Adas Over a year ago

Bingo...exactly what i need. thank you very much @timgeb

timgeb Over a year ago

@Adas no problem, happy coding

buckley · Accepted Answer · 2015-12-29 13:24:25Z

0

This should do the trick

Replace

([\s\S]*?)(Tel[\s\S]*)

With

$1

edited Dec 29, 2015 at 13:24

answered Dec 29, 2015 at 13:22

buckley

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1 Comment

Adas Over a year ago

It gives with Tel, and i don't want that, i want only first two lines of address.

Deep · Accepted Answer · 2015-12-29 13:25:17Z

0

Be original! )))

if (false !== $pos = strpos($str, 'Tel')) {
    $str = substr($str, 0, $pos);
}

answered Dec 29, 2015 at 13:25

Deep

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