I want to echo a string with the variable inside including the ($) like so:
echo "$string";
I dont want it to echo the variable for string, I want it to echo '$string' itself, and not the contents of a variable. I know I can do this by adding a '\' in front of the ($), but I want to use preg_replace to do it. I tried this and it doesnt work:
$new = preg_replace("/\$/","\\$",$text);
Use single quotes for your string declaration:
echo '$string';
Variables inside single quotes do not get expanded:
Note: Unlike the two other syntaxes, variables and escape sequences for special characters will not be expanded when they occur in single quoted strings.
Another solution would be to escape the $ like you already did within the preg_replace call:
echo "\$string";
If I understand your question correctly, you can't use preg_replace to do what you want. preg_replacegets its arguments after variable substitution takes place.
So either there's nothing for preg_replaceto replace (because the substitution already occurred), or there is no need for preg_replaceto do anything (because the dollar sign was already escaped).
As you observed, the $ is treated for variable interpolation inside double quoted strings. So your regex pattern should be constructed using single quoted strings as well. Also, the $ is a special character inside the regex-replacement string, so you need to escape it extra:
preg_replace('/\$/', '\\\\$', $text)
But there is no need for the preg_ function here. str_replace should do it:
str_replace('$', '\\$', $text)
This might clear things up for you, but as Gumbo suggests: why not use single quotes to echo?
Well I think you want to remove the text which exists in some variables. Then you can do it like
$foo = '123';
$bar = 'bqwe';
$str= preg_replace("/".$foo ."/", '', $str, 1);
$str= preg_replace("/".$bar ."/", '', $str, 1);
