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Displaying all images in Database table using PHP

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  • 16-03-2006 4:53pm
    #1
    elbow316
    Closed Accounts Posts: 252 ✭✭


    Hi guys,

    This is giving me a lot of problems. I am able to insert in images in a table, and display ONE image. But I want to display all the images in the table, this is me giving trouble, this code at the moment is displaying the same image ten times.

    As I am incrementing the $ID variable, I want different images to be displayed, but I just keep getting the same one. I do know it is working to some extent, when I put in this statement(echo $name; ) I do get all the image names.

    CODE

    <?

    $db = mysql_connect("******", "*****","*****");

    mysql_select_db("****",$db);

    $id = 1;
    while($id<10){
    $query = "SELECT name, type, size, content FROM upload where id ='$id'";
    $result = mysql_query($query) or die('Error, query failed');
    list($name, $type, $size, $content) = mysql_fetch_array($result);


    if ($_REQUEST[gim] == $id) {


    header("Content-length: $size");
    header("Content-type: $type");


    echo $content;

    exit;
    }






    ?>

    <html>
    <title>Upload an image to a database</title><body>

    <center><img src=?gim=<? echo $id ?> width=160 height=160><br>


    </body>
    </html>

    <?
    $id++;
    }
    ?>



    _______________________________
    Edit - I had pasted in the wrong version of the code, which was calling 1 as the value, rather than the incremented $id............this version seems to be calling the correct image address(gim=2 or gim=3 and so on) but isn't displaying the relevant image


Comments

  • Options
    #2
    rsynnott
    Registered Users Posts: 4,003 ✭✭✭rsynnott


    Well, for a start, don't store images in a table unless you've got a very good reason for it.


  • Options
    #3
    frost
    Registered Users Posts: 249 ✭✭frost


    You aren't incrementing $id in your loop.


  • Options
    #4
    timeout
    Registered Users Posts: 919 ✭✭✭timeout


    You aren't incrementing $id in your loop.
    Well its in it now at the end.

    This is your problem
    where id ='$id'";
    should be 
    where id=" . $id;
    OR 
    where id='" . $id . "'";
    if you need the single quotes.


  • Options
    #5
    frost
    Registered Users Posts: 249 ✭✭frost


    timeout wrote:
    Well its in it now at the end.
    Oops, missed that at the end!
    timeout wrote:
    This is your problem

    should be 
    where id=" . $id;
    OR 
    where id='" . $id . "'";
    if you need the single quotes.

    Should also work with curly braces: 
    $query = "SELECT name, type, size, content FROM upload where id ='{$id}'";


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