Add a variable in preg_replace

Question

I want to add a variable in preg_replace.

for Ex :-

In this blelow line of code, I want to add $text instead of "sometext"

$text = "sometext";
$lineOfValue = preg_replace("/.*sometext[\"', ]*/", "", $line);

Can any one help me?

$lineOfValue = preg_replace("/.*$text[\"', ]*/", "", $line); Where is the problem?! — Rizier123
– Rizier123, Commented Feb 23, 2015 at 8:14
Getting this error when adding like your reply : Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in D:\xampp\htdocs\wordpresstest\write.php on line 12 — Abdul Nassar M
– Abdul Nassar M, Commented Feb 23, 2015 at 8:19
Then just concatenate it: $lineOfValue = preg_replace("/.*" . $text . "[\"', ]*/", "", $line); ?! — Rizier123
– Rizier123, Commented Feb 23, 2015 at 8:23
If the contents of $text is guaranteed to be safe for use in the expression you can use "/.*{$text}[\"', ]*/" as well, wrapping the variable references in curly braces. — Ja͢ck
– Ja͢ck, Commented Feb 23, 2015 at 8:44

jones · Accepted Answer · 2015-02-23 20:57:31Z

1

In a regular expression the $ symbol signifies the end of the string being processed, therefore including a variable within a regular expression as suggested by @Rizier123 won't work.

Instead the variable should be concatenated into the regular expression as follows to avoid the $ being misinterpreted:

$text = 'sometext';
$lineOfValue = preg_replace("/.*" . $text . "[\"', ]*/", "", $line);

Also - if your $text variable may contain any character, you should consider using preg_quote() to make the string safe for use within a regular expression:

$text = 'sometext';
$lineOfValue = preg_replace("/.*" . preg_quote($text, '/') . "[\"', ]*/", "", $line);

edited Feb 23, 2015 at 20:57

answered Feb 23, 2015 at 8:30

jones

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2 Comments

Ja͢ck Over a year ago

The expression delimiter should also be escaped.

jones Over a year ago

Good call - I've edited the example to show the use of preg_quote()'s optional second parameter

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