php validation: no spaces in string

Question

I'm setting up a custom php form validation used in conjunction with wordpress contact form 7 plugin, I've set up a validation rule as the text field, I only want people to type 1 word (i.e. no spaces) and for that one word to begin with the letter s.
The letter s validation works just fine but while this works, the no spaces (one word) validation doesn't seem to be working?
php:

if($name == 'FirstField') {
        $FirstField = $_POST['FirstField'];

        if($FirstField != '') {
            if (!preg_match("/(^[^s]|\s)/i",$FirstField)){
                $result['valid'] = true;
            } else {
                $result['valid'] = false;
                $result['reason'][$name] = 'Invalid Entry.';
            }
        }
    }

Any suggestions would be greatly appreciated!

When I tested your regex it seems to do what you want. Can you give an example of a value that does not give your expected result? — luttkens
– luttkens, Commented Dec 19, 2013 at 20:54
As a side note, you can always test your regular expressions here... rubular.com/r/CnxN4oJub2 — Charlie74
– Charlie74, Commented Dec 19, 2013 at 21:35

Jose Carlos Ramos Carmenates · Accepted Answer · 2013-12-19 21:31:56Z

2

Test:

if($name == 'FirstField') { $FirstField = $_POST['FirstField'];

    if($FirstField != '') {
        if (!preg_match("/(^[^s]|\+s)/i",$FirstField)){
            $result['valid'] = true;
        } else {
            $result['valid'] = false;
            $result['reason'][$name] = 'Invalid Entry.';
        }
    }
}

UPDATED:

if($name == 'FirstField') { $FirstField = $_POST['FirstField'];

    if($FirstField != '' && isset($FirstField) ) {
        $FirstField = explode(' ', $FirstField); //Get only first word
        if (!preg_match("/(^[^s]|\+s)/i",$FirstField[0])){
            $result['valid'] = true;
        } else {
            $result['valid'] = false;
            $result['reason'][$name] = 'Invalid Entry.';
        }
    }
}

FIXED ERROR last:

replace:

if($FirstField != '' && isset($FirstField)

by:

if($FirstField != '' && isset($FirstField) )

I hope help you.

edited Dec 19, 2013 at 21:31

answered Dec 19, 2013 at 20:56

Jose Carlos Ramos Carmenates

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7 Comments

user1374796 Over a year ago

thanks for your suggestion, still doesn't seem to be working though. the s validation is working fine, the no spaces one though, still isn't :(

user1374796 Over a year ago

still no luck with that, there's a syntax error in there too, missing a closing bracket ;) thanks for your help though…just still no luck :(

Jose Carlos Ramos Carmenates Over a year ago

if($FirstField != '' && isset($FirstField) ) ... The problem is closet ")" the condition statement "if" ;-)

user1374796 Over a year ago

even with the syntax correction is still doesn't seem to work, again the s validation is still working, the spaces one is still a no go though, I can't figure it out :(

Jose Carlos Ramos Carmenates Over a year ago

but do not worry anyway with denial "!" in condition and works. My is working in my test site.

|

BeeLaw · Accepted Answer · 2013-12-19 21:42:57Z

0

I believe this should work:

preg_match("/^s[a-z]*$/i", $FirstField)

Assuming that your word consists of only letters.

answered Dec 19, 2013 at 21:42

BeeLaw

1

Comments

Jasper N. Brouwer · Accepted Answer · 2013-12-19 21:49:00Z

0

Starting with 's' (case insensitive), no spaces: preg_match('/^s[^ ]*$/i', $string)

Starting with 's' (case insensitive), no white-space (spaces, tabs, etc): preg_match('/^s\S*$/i', $string)

edit

You might also be interested in \w, which matches "word characters" (A-Z, a-z, 0-9, _): preg_match('/^s\w*$/i', $string)

edited Dec 19, 2013 at 21:49

answered Dec 19, 2013 at 21:43

Jasper N. Brouwer

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