Escaping certain characters doesn't work

Question

I'm sure this is a duplicate but I couldn't find the answer through search.

Here's my code

str_replace([' ',-,\(,\),\*,-,\|,\/], '', $params['telephone']), '0'

and that works. What I want it to be is:

str_replace([' ',-,\(,\),/,|,\*,-,\#,\], '', $params['telephone']), '0'

As soon as I add in # or \ it fails, and escaping them doesn't seem to work. Is there a different way to escape them?

And although it seemingly works, it does not seem to replace the initial space. Could that be the root of my problem?

Answer 1

You want to remove a space, -, (, ), *, -, |, / and also \ and # symbols from the string.

You need to pass an array of these chars to str_replace:

$arr = [ '-', '(', ')', '*', '-', '|', '/', '\\', '#'];
$result = str_replace($arr, '', $s);

Or, you may use a regex with preg_replace like this:

$res = preg_replace('~[- ()*|/\\\\#]~', '', $s);

See the PHP demo.

Regex details:

• ~ is a regex delimiter, required by all preg_ functions
• - must be put at the start/end of the character class, or escaped if placed anywhere else, to match literal -
• \ symbol must be matched with 2 literal \s, and that means one needs 4 backslashes to match a single literal backslash
• Inside a character class, *, (, ), | (and +, ?, ., too) lose there special status and do not need to be escaped
• If you work with Unicode strings, add u modifier after the last ~ regex delimiter: '~[- ()*|/\\\\#]~u'.

The difference between string replace and preg replace

str_replace replaces a specific occurrence of a literal string, for instance foo will only match and replace that: foo.

preg_replace will perform replacements using a regular expression, for instance /f.{2}/ will match and replace foo, but also fey, fir, fox, f12, f ) etc.