Hi Guys,

I am trying to select some data taken from one table called 'uploads' in MySQL database that is held in a dropdown menu and place that data in another table called 'news' in the MySQL database in a variable called 'piicture' when I hit submit.

PHP Code:


<select name="picturemenuchoice">
<option value="" selected>select an image...</option>
<?php 
  $query = mysql_query("SELECT * FROM `uploads`"); 
  while($picture = mysql_fetch_assoc($query)){ 
?> 
  <option value="<?php echo $picture['url']; ?>"><?php
  echo $picture['name']; ?></option>
    <?php 
    } 
    ?> 
</select>

Now this works a treat, but when I submit the data the name of the image does NOT stay on the name of the image I have just chosen, it justs goes back to 'select an image', and also it does not not show the image name I have chosen when I go in and edit the selected dropdown item and again shows 'select an image'.

Now I have this code:

PHP Code:


<?php

// login stuff

include("nameoffile.php");
include("ddwsff.php");
checklogin();

$picture = "";

if(isset($_POST['Submit']))
{


    $picture = $_POST['picturemenuchoice'];
    
    if(!isset($_GET['newsid']))
    {
        $result = mysql_query("Insert into news(picture) values('$picture')");
        $msg = "New record is saved";
    }
    else
    {
        $result = mysql_query("Update news set picture='$picturemenuchoice' where newsid=".$_GET['newsid']);
        $msg = "News Record is updated";
    }
}
if(isset($_GET['newsid']))
{
    $result = mysql_query("Select * From news where newsid=".$_GET['newsid'],$link);
    $row = mysql_fetch_array($result, MYSQL_BOTH);
    $picture = $row['picturemenuchoice'];
}

if(mysql_error())
{
echo mysql_error() ."<br>\n";
}

?>

Which states what should happen when I submit. have I got it wrong anywhere when it saves so that instead of showing 'select an image' it will show the image name I have selected from the dropdown when I hit submit?

Many Many Thanks

CB

Well... have you firstly checked to see if the database is being added to?? Also, why in the php have you got a $_POST and $_GET working together?? Surely that's not possible, unless in the HTML you have <FORM ACTION="www.server.com/?var=value" METHOD="POST"> ??

The last bit where u talk about the drop down still being on "select an image..." it would be that all the time becase its the first data in the drop down and you have "SELECTED" in the option bracket.

We generally advise u posting the whole PHP script

Hi,

Thanks for the reply. Points noted!

Reason I have POST and GET working together is because when I POST the data I can then use the same form to GET the data and modify it for the database.

As for the dropdown, where it returns to 'select an image' when you hit submit, where do I change that so it stays on what I selected - I don't understand?

Many Thanks

Ok, since i'm not too sure how ur processing the PHP and stuff for the drop down, i'll use one of my own examples. This is taken from a client in a band (www.festerband.com) who can basically control their content using my CMS. The below code prints out in a drop down box for the year in the date of a future gig they wish to post on their Gigs page, but also allows them to edit the existing gigs they have posted, so we need to be able to recall the date into the drop down boxes on the page:

PHP Code:

<SELECT SIZE=1 CLASS=forminput NAME=gigs_listing_year>');

// 2005 is the current year!
$i = '2005';
// Let's allow them to book gigs 3 years in advance, so add 3 to 2005 (=2008)
$j = $i + 3;

        // So while $i is less than or equal to $j
    while ($i <= $j) {

        // Be it selected or not, we have to print the basic HTML
        echo('<OPTION VALUE=' . $i . ' ');
        /* The next if statement uses a variable which has been
            taken from the SQL DB where the gigs are stored, so if
            $i (which gets increased each loop) equals what is in the 
            database... */
        if ($gigs_listing_year == $i) {
            echo('SELECTED');
        }

         // This bits obvious
        echo('>' . $i . '</option>
        ');

         // We MUST increase $i, or we have a neverending loop!
        $i++;
            
    }

echo('
</SELECT> 


Hope that explains something?? You need to get the $gig_listing_year (change the name as appropriate) from somewhere, be it a database or from the form that has just been processed, or just from a query string - that bits up to u to figure!

Hope this helps??