PHP否定前瞻正则表达式不起作用

Question

I am trying to construct a regex in my PHP application that takes a variable, such as $search , and replaces that with another variable. What I am trying to do specifically, is replace the $search value only if it is not followed directly by an = .

What I have tried so far is:

preg_replace("/($search)(?!=)/i", $replacement, $string, -1, $count);

But this has not worked. I'm sure I am just misunderstanding something, but I can not seem to determine what that is. I am fairly new to regex, and I've looked at several of the answers from previous questions and have not been able to figure it out.

If you need any other information, let me know.

EDIT

To clarify, assume the following code:

$searchTerm = "Hello";
$search = "/($searchTerm)(<?!=)/i";
$replacement = '<span style="background:red;">$0</span>';
$result = preg_replace($search, $replacement, $string, -1, $count);

where the value of $string would be something like:

$string = "hello=hola";

So the idea is that if "hello" was not followed by and '=', then it would highlight it in red, but if it is, then nothing happens. At assume that the value of $string can change. Very lame example, I know, but hopefully it clears it up a little.

Answer 1

Actually you need to escape $ character by a backslash \\ in your pattern. Also to make it more confident, you can add \\s* which means zero or more white space(s) before = character.

Use this pattern:

/\$search(?!\s*\=)/i

Online Demo

• / delimiter
• \\$ matches $ literally
• search matches that word literally
• (?! negative Lookahead which checks being both white space(s) (if exists) and = after that variable name.
• /i modifire makes the pattern insensitive (there isn't any different between az and AZ )

Note: You have to use single quotes '/pattern/' for pattern above. If you want to use double quotes " then you should escape backslashes again:

/\\$search(?!\\s*\\=)/i

Answer 2

The $ is an anchor in regex to match the end of a line, so since it's a special character you have to escape it.

You can use:

preg_replace("/(\\$search)(?!=)/i", $replacement, $string, -1, $count);

or

preg_replace('/(\$search)(?!=)/i', $replacement, $string, -1, $count);

Regex demo