[RESOLVED] mysql_insert_id problem.

mrbill501

Hi
Im new to php and have spent the last 2 days trying to work something out and I'm at the point of panic now as the thing I cant work out is for a school project.
The aim of this is to add an image as a blob to a database table and to add a thumbnail sized image to another table at the same time. Both these images need to have the same ID number to make things easier down the road. At the moment I'm trying to make the two have the same ID by:

$sql = "INSERT INTO tbl_images
( image_id , image_type ,image, image_size, image_name, log_name, gallery_id)
VALUES
('', '{$size['mime']}', '{$imgData}', '{$size[3]}', '{$FILES['pix']['name']}', '{$SESSION['logname']}', '{$_POST['gallery']}')";

$var = mysql_insert_id();
echo $var;
The problem is that it is returning mysql_insert_id with a value of zero instead of the same number as the image_id (which is an auto increment field).
Any help would be really good as I cant move on till Ive done this bit.
Thanks in advance to anyone who can help me out here.
Andy

devinemke

your code as posted never actually runs the query, it only creates a string variable of the query. [man]mysql_insert_id[/man] will not return anything until you have run the query using [man]mysql_query[/man].

Houdini

DO not include an auto_incrementing field like the image_id into the list, it will automatically do that for you.

$sql = "INSERT INTO tbl_images
(  image_type ,image, image_size, image_name, log_name, gallery_id)
VALUES
( '{$size['mime']}', '{$imgData}', '{$size[3]}', '{$_FILES['pix']['name']}', '{$_SESSION['logname']}', '{$_POST['gallery']}')";

Do this instead, check that I did in fact remove the peoper field which you were not setting to anything.

mrbill501

Thanks guys, if I can get past this bit I will owe you both my life.
I've taken the image_id out of the INSERT query and have added:

$last_id = mysql_insert_id();
$query6 = "SELECT * FROM tbl_images WHERE image_id = $last_id";$result6 = mysql_query($query6) or die('Error with query 6.');
echo $result6;

When I ask it to show me the query result its giving me: Resource id #6

devinemke

mrbill501 wrote:
When I ask it to show me the query result its giving me: Resource id #6

...exactly as expected. if you want to see the actual contents of the result read up on:

[man]mysql_fetch_row[/man]
[man]mysql_fetch_array[/man]
[man]mysql_fetch_assoc[/man]
[man]mysql_fetch_object[/man]

mrbill501

Ok will do thanks. I only want to be able to see the image_id so I can make it the same in the next INSERT statement to add the smaller image in to a different table. I'm only trying to see the result at the moment so I can tell if its working or not.
At the moment I'm getting no display when I run the first INSERT statement:

$sql = "INSERT INTO tbl_images
( image_type ,image, image_size, image_name, log_name, gallery_id)
VALUES
('{$size['mime']}', '{$imgData}', '{$size[3]}', '{$FILES['pix']['name']}', '{$SESSION['logname']}', '{$_POST['gallery']}')";
$last_id = mysql_insert_id();
$query6 = "SELECT image_id FROM tbl_images WHERE image_id = $last_id";
$result6 = mysql_query($query6) or die('Error with query 6.');
$row6 = mysql_fetch_row($result6);
echo $row6;

Sorry if I'm being a bit slow, I only started trying php out the other week and its been a bit intensive.

devinemke

re-read my first post above, about never running the INSERT query. you are still not running it. also you really need to read up on arrays. they are rather crucial when dealing the database results in PHP. look at examples in the manual.

mrbill501

devinemke wrote:
re-read my first post above, about never running the INSERT query. you are still not running it.

The INSERT statement works though, when I look the tables the image is inserted into the table.

mrbill501

Thanks for helping me out man, its just started to work now. You were right. I owe you big time.