preg_match does not work

[vigour]

I want to check for illegal characters, only A-Z are allowed. But my code does not work. Is this the right way to do it?

if (preg_match('/^[A-Z]/', 'QWERTY') == 1)
    {
    // contains valid characters
    }
else
    {
    // error
    }

[neilmasters]

if (preg_match('/^[A-Z]/', 'QWERTY') == 1)

You have your square brackets in the wrong position 🙂
Also preg_match returns true/false so you do not need the ==1 at the end(like substr).

Cleaned code:

if(preg_match('/[^A-Z]/', 'QWERTY'))
{
     // contains valid characters
}
else
{
     // error
} 

[johanafm]

neilmasters;10964704 wrote:

if (preg_match('/^[A-Z]/', 'QWERTY') == 1)

if(preg_match('/[^A-Z]/', 'QWERTY'))
{
     // contains valid characters
}
else
{
     // error
} 

Except that you also need to change either the comparison or the if- and else-blocks.
That is, either

if(preg_match('/[^A-Z]/', 'QWERTY') == false)

or

if(preg_match('/[^A-Z]/', 'QWERTY'))
{
     // error
}
else
{
     // contains ONLY valid characters
} 

[vigour]

Thanks for the great answers, it works now and I have learned a lot.

[johanafm]

vigour;10964793 wrote:

I want to check if there are 2 or more spaces in a row, how do I modify my code so it wont allow that? Or even better, take away all but 1 space automatically?

Use this to replace multiple whitespaces with a single one in each place. But do note that \s matches any whitespace character, such as new line, horizontal tab etc. Thus

$string = "A  B

C";
$string = preg_replace('/(\s)+/', '\1', $string);

Gives

A B

C

If you only want to allow the ordinary space character, then use that instead of \s

^/[A-Za-z0-9 ]$/

And the replacement would no longer need a back reference

$string = preg_replace('/ +/', ' ', $string);