CODEWITHSUNDEEP
phpregex
Thildemar
Thildemar

Reputation: 99

php Dollar amount Regular Expression

I am have completed javascript validation of a form using Regular Expressions and am now working on redundant verification server-side using PHP.

I have copied this regular expression from my jscript code that finds dollar values, and reformed it to a PHP friendly format:

/\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/

Specifically:

if (preg_match("/\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/", $_POST["cost"])){}

While the expression works great in javascript I get :

Warning: preg_match() [function.preg-match]: Compilation failed: nothing to repeat at offset 1

when I run it in PHP. Anyone have a clue why this error is coming up?

Upvotes: 1

Views: 3463

Answers (3)

Jonathan Kuhn
Jonathan Kuhn

Reputation: 15301

you have to double escape the $.

if (preg_match("/\\\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/", $_POST["cost"])){}

just putting \$ will have php escape the $ from starting a variable. You also have to add another escaped \ (\\) in front. Or you could just use single quotes so php doesn't interpret the $ as the start of a variable.

Upvotes: 2

Emil Vikström
Emil Vikström

Reputation: 91942

Try this:

if (preg_match('/\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/', $_POST["cost"])){}

The dollar sign between " are considered as the start of a variable. Compare with these two code snippets:

1:

$var = "hello";
echo "$var world";

2:

$var = "hello";
echo '$var world';

Upvotes: 1

bancer
bancer

Reputation: 7525

From PHP documentation:

Do not use preg_match() if you only want to check if one string is contained in another string. Use strpos() or strstr() instead as they will be faster.

Upvotes: 1

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