Cannot display multiple images per same id in one entry-Need some help

[Jankie]

I need to dispaly a user's multiple images in one entry.Right now,say if a user uploads 3 images,three entries for the same id display to match 3 images.
I only want 1 entry to display all of a user's related images like below:
Id
First Name
Last Name
Title
Image1
Image2
Image3

Currently I get;
1
First Name
Last Name
Title
Image1

1
First Name
Last Name
Title
Image2

1
First Name
Last Name
Title
Image3

My select code is as follows:

[php]
include 'Con.inc.php';
$query = "SELECT Members.first_n ame,Members.las t_name,Members. category,Member s.title,Members .message,Images .name,Images.ty pe,Images.size, Images.path FROM Members LEFT JOIN Images ON Members.member_ id=images.membe r_id";
$result = mysql_query($qu ery);

while(list($fir st_name,$last_n ame,$category,$ title,$message, $file,$Type,$Si ze,$path)= mysql_fetch_row ($result))
{
echo "first_name :$first_name <br>" .
"last_name :$last_name<br> <br>".
"category :$category <br><br>".
"title :$title <br><br>".
"message:$messa ge<br><br>".
"<img src='../FileDir/$path' alt='$first_nam e s profile photo' /><br>";
}

mysql_close($co nn);
?>
[/php]

Thank you very much for your response in advance!

[MarkoKlacar]

Hi,

if you want the output you desire, you can't read/write all the information in one single loop. I'm guessing you need at least one more loop where you go though all the images that a user with the specific userid has linked to him/her.

Hope this helps.

Cheers

[Jankie]

Thank you MarkoKlacar,for your response !
I havent done this before so I have no clue how to add a conditioned second loop in my code above

[MarkoKlacar]

Hi,

just to get an understanding of your database structure. The $path variable reads from the "Images.pat h" in the database. How is this connected to the user id?

As I see it, it's from .."Members.memb er_id=images.me mber_id", right?

Now what you could do is not select the images, but instead inside the while-loop you have execute one more query saying:

[PHP]$queryForMatchi ng = "SELECT Images.part FROM Images WHERE Images.member_i d = '$firstName'";[/PHP]

In the case above I'm assuming that the $firstName is the id used to map a user with his uploaded images.

After this you loop though the result that you receive and print out all the pictures.

Hope this helps.

Cheers

[Jankie]

thanks for your help

"As I see it, it's from .."Members.memb er_id=images.me mber_id", right?"

Yes.

member_id is what links a user to their images,not $firstname (I just missed it above)

[MarkoKlacar]

Hi Jankie,

did you solve your problem, was I clear enough in my previous post?

Cheers

[Jankie]

Sorry for not updating you
Not yet really,do you mean a nested loop ?
If i understood you correctly,I have to delete the image selection info from the select statement(then I dont need the join statement,too) .Instead,I should include your query code inside the outer loop,correct ?
or how does your code fit in my code above ?
Sorry to sound dumb,didnt do it before
and thank you very much for your follow up ! Appreciate it.

[MarkoKlacar]

Hi Jankie,

Yes I mean a nested loop. The nested loop goes through a new result array that was received from a new query where you select all the image.ids bellonging to a key that you're currently working with in the outer loop.

Hope I didn't confuse you with my mediocre description.

Let me know.

[Jankie]

MarkoKlacar
Thank you very much indeed
No,no your description is fine
yes,I got it to work with a nested loop inside the second query.
But I had to remove the join code along with image selection info from first query.

they are both linked now but just a where clause after I have stripped the table name part from the columns in the query.
[php]
"SELECT first_name,last _name,category, title,message FROM Members WERE member_id=membe r_id";
[/php]
[php]
"Select name,type,size, path FROM Images WERE member_id=membe r_id";

[/php]

I am just concerned the 2 tables are not now technically linked


Again,I do appreciate your staying with me,thanks a lot !

[MarkoKlacar]

Hi Jankie,

they are linked using the primary keys, member id in this case.

Cheers