Re: Re: [PHP] How do I convert from perl to php?
| From: | PHP List | Date: | Tue, 20 Nov 2001 22:38:21 +0000 |
| Subject: | Re: Re: [PHP] How do I convert from perl to php? | ||
| References: | 1 2 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-75309@lists.php.net to get a copy of this message | ||
How about like this:
$required = array("field1","field2","field3");
for ($i=0;$i<count($required);$i++)
{
$temp = $required[$i];
if (trim($$temp) == "")
{
echo "You Missed One";
continue; (break?)
}
}
Something like that, I havn't acually tried it.
----- Original Message -----
From: <digital1@olypen.com>
To: <php-general@lists.php.net>
Sent: Tuesday, November 20, 2001 2:21 PM
Subject: Re: Re: [PHP] How do I convert from perl to php?
> So I would have to write a seperate if condition for each form input field
i wanted to require? that doesn't make for a very dynamic script...
>
> > if(!(isset($name) && isset($address) && isset($phone)) {
> > echo "You left one empty.";
> > }
> >
> > On Tue, 20 Nov 2001 digital1@olypen.com wrote:
> >
> > > I am a perl user trying to convert to php
> > >
> > > how would i turn this perl into php?
> > >
> > > use CGI;
> > >
> > > $name = param(name);
> > > $address = param(address);
> > > $phone = param(phone);
> > >
> > > @required = qw( name address phone );
> > >
> > > foreach $key($required)
> > > {
> > > if (!$$key) { &out("You left one empty."); }
> > > }
> > >
> > > ??
> >
>
>
> --
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