RE: [PHP-DB] Re: session variable in select query showing picture from database
| From: | Fortuno, Adam | Date: | Thu, 12 Feb 2009 22:28:34 +0000 |
| Subject: | RE: [PHP-DB] Re: session variable in select query showing picture from database | ||
| References: | 1 2 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-45949@lists.php.net to get a copy of this message | ||
Mika,
Echo out the dynamically created SQL statement ie., $query = "SELECT *
FROM MyTable WHERE ID = ${ID}"; ECHO $query;" Let us see what is
actually being passed.
P.S. I couldn't agree more with the poster that said, don't pass user
input directly to a SQL statement.
-----Original Message-----
From: Mika Jaaksi [mailto:mika.jaaksi@gmail.com]
Sent: Thursday, February 12, 2009 5:02 PM
To: php-db@lists.php.net
Subject: [PHP-DB] Re: session variable in select query showing picture
from database
*Answer to Rick:
in your code below it looks like you're simply hard-coding your
"$band_id" value (as "11") -- so of course it's going to work.
*Yes, I did that because one of you helpers asked me to try that.
I'll try to be clearer on whom I'm answering to...