Re: Eval error
| From: | Mark Maggelet | Date: | Wed, 15 Aug 2001 20:03:57 +0000 |
| Subject: | Re: Eval error | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-62899@lists.php.net to get a copy of this message | ||
lots of things:
1) variables in single-quoted strings aren't evaluated so you don't
need to escape the $ in $email
2) php uses . not + for concatenation.
3) try it like this:
eval('$email = $email_'.$i.';');
or
eval("\$email = \$email_$i;");
4) you can save yourself the trouble by using arrays as field names:
<input name="email[1]">
5) RTFM! that's what its there for.
On Wed, 15 Aug 2001 15:15:09 -0300, Felipe Coury
(fcoury@creation.com.br) wrote:
>Hi,
>
>I am a beginner in PHP and I am trying to do the following: I have a
>form in
>a page that has 3 fields: email_1, email_2 and email_3. I am trying
>to send
>e-mail to those people, if the fields are filled. Relevant part of
>code:
><?php
>for ($i = 1; $i <= 6; $i++) {
> $eval = '\$email = \$email_' + $i + ';';
> eval( $eval );
> echo $email;
>}
>?>
>
>The code complains about an error in line eval( $eval );:
>
>Parse error: parse error in
>/home/httpd/htdocs/hthcombr/cgi-local/envia.php(19) : eval()'d code
>on line
>1
>
>Can anyone please help me?
>
>Regards,
>
>Felipe Coury
>
>
>--
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