PHP find character by pattern in string and replace it

Question

Here is a string:

$str = "Discount 5.5@, Update T@";

I want to replace symbol '@' to '%' but only if it has numbers (int or dec) before:

"Discount 5.5%, Update T@"

I use:

preg_replace("/[0-9.]*@/", "%", $str);

But it just removes any number before % sign, what I do worng?

The . in [] means almost any character with some modifiers all. Its not \. — Rolice
– Rolice, Commented May 21, 2014 at 9:36

Rolice · Accepted Answer · 2014-08-07 21:30:19Z

2

See my comment above. I would suggest a bit different expression:

preg_replace("/(\d+(?:\.\d+)?)(\s*)\@/", "$1$2%", $str);

This expression will match 5@, but 5.5@ as well (with or without floating point). However it will not match 5.@.

\d means numbers, equal to [0-9], the + (plus) means 1 or more, but not 0 occurrences. The second expression starting with with ?: (which means not to match as group) means to find . (dot) immediately after the first number sequence and to be followed by numbers - whole zero or one time (not to match 5.15.25@).

We then check for a spaces (0 or more times), then we turn them back after replacing with $2.

edited Aug 7, 2014 at 21:30

answered May 21, 2014 at 9:38

Rolice

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2 Comments

Ken Tang Over a year ago

Great! How about if 5.5@ or 5.5 @ space between?

Rolice Over a year ago

@KenTang, if you dedicate one week to test and learn regular expressions you will see it is not so hard. :)

Deadooshka · Accepted Answer · 2014-05-21 09:40:27Z

2

$str = preg_replace('/(?<=\d)\@/', '%', $str);

answered May 21, 2014 at 9:40

Deadooshka

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Comments

Daan · Accepted Answer · 2014-05-21 09:38:20Z

0

First check the string if there are numbers in it. Than do a simple str_replace.

$str = 'Discount 5.5@, Update T@';
preg_match_all('!\d+!', $str, $matches);
if(!empty($matches)) { 
   str_replace('@','%', $str);
}

answered May 21, 2014 at 9:38

Daan

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3 Comments

xate Over a year ago

Why are you checking it before with preg_match_all(); and dont use preg_replace right away?

Daan Over a year ago

@Xatenev because of the OP's title

xate Over a year ago

Well a preg_replace(); would fit. in his answer hes already using peg_replace() too

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