Regular expression replace characters with the correct number of spaces

Question

I wanna replace with spaces all characters except number, lecters, space and other characters #=<>();*,.+\\/-

eg preg_replace("/[^ #=<>();*,.+\\/-\\w]+/", " ", $string);

My problem is that when in the $string there are two or more consecutive characters to be replaced, the function replace this characters with just one space, while I need that the functions replace the two or more characters with two or more spaces.

Is there a way?

Answer 1

You should match only one character at a time. You must also escape some of the characters.

Change

 preg_replace("/[^ #=<>();*,.+/-\w]+/", " ", $string);

to

 preg_replace("/[^ #=<>();*,\\.+\\/\\-\\w]/", " ", $string);

Answer 2

• If your character class contains both forward and backward slash, you need to escape both forward and backward slashes which are present inside the character class.

I wanna replace with spaces all characters except number, lecters, space and other characters #=<>();*,.+\\/-

• \\w represent letters,numbers and also _ symbol. So avoid using \\w inside the character class.

• As another answer said, you need to remove the + after character class, which replaces one or more characters with a single space.

• And your regex should be,

   [^- #=<>();*,.+\\\\\\/0-9A-Za-z] 

DEMO

• In the demo it matches _ symbol because it isn't included in the NOT character class. In the replacement part i gave only a single space. It replaces three _ symbols with three spaces.