Mika Jaaksi wrote:
I'm trying to show picture from database. Everything works until I add
variable into where part of the query.
It works with plain number. example ...WHERE id=11... ...picture is shown
on the page.
Here's the code that retrieves the picture. show_pic.php
?php
Thanks for the quick responce...
to Valentin Nedkov:
I have session_start() on another page. Session start gets band_id as a
value when user logs in.
I've tried to echo session variable on show_pic page and it works.
And I belive that I can't set default value for band_id because the picture
I
Still fighting with it...
So, these work:
$query=SELECT * FROM pic_upload;
$query=SELECT * FROM pic_upload WHERE band_id=11;
picture is shown on the other page
but when adding variable into query it doesn't show the picture on the other
page
$query=SELECT * FROM pic_upload WHERE
I tried
$query = SELECT * FROM pic_upload WHERE band_id =
'.$_SESSION['session_var'].' ;
didn't work.
And I've tried to echo session variable and it has right data in it.
I've also tried
band_id=$band_id
band_id='$band_id'
band_id=$band_id
band_id='{$band_id}'
band_id={$band_id}
Session
Okay, I added it and got this
SELECT * FROM pic_upload WHERE band_id=11
Seems to me that it's the way i should be.
For some mystical reason it still doesn't work...
: session variable in select query showing picture
from database
Still fighting with it...
So, these work:
$query=SELECT * FROM pic_upload;
$query=SELECT * FROM pic_upload WHERE band_id=11;
picture is shown on the other page
but when adding variable into query it doesn't show the picture on the
other
Sorry, but this didn't work either
$query=SELECT * FROM pic_upload WHERE band_id='${band_id}';
Thanks to everybody who has tried to help...
Don't see session_start() in your script. If you work with SESSION, you
must have it on the first lines of the file (before any output and work
with $_SESSION so it's good to put it on the first lines).
And it must be in every file which works with them (except for included
files). It should
$band_id = 11;
$query=SELECT * FROM pic_upload WHERE band_id=$band_id;
print_r($_SESSION);
gives this:
Array ( [session_var] = 11 )
and picture is shown on the page
And about the session start: I have session start on the index2.php page
when user has logged in.
Page that should show the
*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...
With these:
$band_id = $_SESSION['session_var'];
echo band_id: . $band_id;
$query=SELECT * FROM pic_upload WHERE band_id=$band_id;
echo query: . $query;
I get these:
band_id: 11
query: SELECT * FROM pic_upload WHERE band_id=11
SQL injections: Are these what I should use?
$db = new
On Fri, Feb 13, 2009 at 6:01 PM, Mika Jaaksi mika.jaa...@gmail.com wrote:
With these:
$band_id = $_SESSION['session_var'];
echo band_id: . $band_id;
$query=SELECT * FROM pic_upload WHERE band_id=$band_id;
echo query: . $query;
I get these:
band_id: 11
query: SELECT * FROM pic_upload
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