I've written a regular expression in PHP to allow strings that are alpha-numeric with any punctuation except & or @. Essentially, I need to allow anything on a standard American keyboard with the exception of those two characters. It took me a while to come up with the following regex, which seems to be doing what I need:
if (ereg("[^]A-Za-z0-9\[!\"#$%'()*+,./:;<=>?^_`{|}~\-]", $test_string)) {
// error message goes here
}
Which brings me to my question... is there a better, simpler, or more efficient way?
Have a look at character ranges:
@[!-%'-?A-~]+@
This will exclude the characters & (\0x26) and @ (0x40).
Looking at an ASCII Table,you can see how this works:
The exclamation mark is the first character in the ASCII set, that is not whitespace. It will then match everything up to and including the % character, which immediately precedes the ampersand. Then the next range until the @ character, which lies between ? and A. After that, we match everything unto the end of the standard ASCII character set which is a ~.
Update
To make things more readable, you might also consider to do this in two steps:
At first, filter anything outside of the default ASCII range.
@[!-~]+@
In a second step, filter your undesired characters, or simply do a str_pos on the characters.
At the end, you can compare it with what you started to see whether it contained any undesired characters.
Instead, you could also use a regex such as this for the second step.
/[^@&]+/
The steps are interchangeable and doing a str_pos on @ or & as a first step, to identify bad characters, may be better performance wise.
What about this:
[^&@]
with preg_match
$str = 'a';
var_dump(preg_match('~^[^&@]+$~', $str)); // true
$str = '&';
var_dump(preg_match('~^[^&@]+$~', $str)); // false
$str = '!';
var_dump(preg_match('~^[^&@]+$~', $str)); // true
I think rather than testing for all the alpha numeric characters you can simply check for @ and & and use a not?
$reg = '/@|&/';
if(!preg_match($reg, "YOUR STRING CAN GO HERE")){
// your code goes here
}
