Host Upgraded - File Upload won't work now

gillweb · 02-25-2008, 02:33 AM

I have had my site working for MANY years until recently my host had upgraded the server to PHP 5. Sicne then I can't get a simple file upload to work? Can anyone look at this code and tell me why it wouldn't work?

The error I am getting is "Sorry. You did not chose an image" which I am assuming means the file string isn't being passed on??? But the weirder thing is I am getting the "Ad uploaded" message also? So it's telling me there is an error but also saying it was uploaded (which it isn't).

PHP Code:


			
<?php

$adtitle = $_POST['adtitle'];

$adtype = $_POST['adtype'];

$imagetype = $_POST['imagetype'];

$adimagefile = $_POST['adimagefile'];

$url = $_POST['url'];

$groupid = $_POST['groupid'];





   $error = false;

   $errormessage = "";



   if (strlen($adtitle) < 1) {

      $error = true;

      $errormessage .= "<li><b>Sorry.  You did not enter an ad title.</b><br>\n";

   }

   if ($adtype == 'A'){

      if (strlen($code) < 1) {

         $error = true;

         $errormessage .= "<li><b>Sorry.  You did not enter an affiliate code.</b><br>\n";

      }



   }



   if ($adtype == 'O'){

      if (strlen($adimagefile) < 1) {

         $error = true;

// HERE"S MY ERROR!!!!!

         $errormessage .= "<li><b>Sorry.  You did not chose an image</b><br>\n";

      }

      if ($imagetype == "") {

         $error = true;

         $errormessage .= "<li><b>Sorry.  You did not chose an image type.</b><br>\n";

      }

      if (strlen($url) < 1) {

         $error = true;

         $errormessage .= "<li><b>Sorry.  You did enter a URL</b><br>\n";

      }



   }



?>



<h2>Save new ad</h2>

<p>



    <?php

$adid = nextid("ads", "adid");



        if($adtype == 'O'){

// HERE's MY UPLOAD

            $newfilename = "../ads/$adid.$imagetype";

            $picfile = $HTTP_POST_FILES['userfile']['tmp_name'];

            copy($adimagefile, $newfilename);

            printf("Ad uploaded<br>");

        }



        if ($error) {

           printf("<font color=red>There was a problem creating the ad!</font><br><br>Please click on back in your browser to fix the problems listed below.<br>");

           printf($errormessage);

        } else {



        if ($adtype == 'A'){

           $sql = "insert into ads (adid, code, impressions, clicks, groupid, type, name) values ($adid, '$code', 1, 1, $groupid, '$adtype', '$adtitle')";

        }



        if ($adtype == 'O'){

           $newurl = "http://";

           $newurl .= $url;

           $sql = "insert into ads (adid, image, url, impressions, clicks, groupid, type, name) values ($adid, '$imagetype', '$newurl', 1, 1, $groupid, '$adtype', '$adtitle')";

        }



           $result = mysql_query($sql ,$db);



           printf("<p>The ad has been added to the group.</p>");

           printf("<p><a href=editadgroup.php?groupid=$groupid>Back to ad group</a></p>");



        }

    ?>



</p>

abiko · 02-25-2008, 02:47 AM

Try this code:

PHP Code:


			
if($adtype == 'O'){

// HERE's MY UPLOAD

            $newfilename = "../ads/$adid.$imagetype";

            $picfile = $HTTP_POST_FILES['userfile']['tmp_name'];

    if ( move_uploaded_file( $picfile, $newfilename) ) {

            printf("Ad uploaded<br>");

    } else {

            printf( "Ad not uploaded");

   }



}

Also:

PHP Code:


			
   if (!empty($adimagefile))

or

PHP Code:


			
      if (isset($adimagefile)) {

see how this works for you.
Did you try

PHP Code:


			
print_r( $_POST );

to see is the $_POST variables are all there?
Can you tell us how does it work. is this the upload-add script or?

gillweb · 02-25-2008, 02:56 AM

this is a custom CMS backend to add advetisement banners to my website but of course now it's not working. My assumption is that PHP5 uses move_file_uploaded now instead of copy?

Ok. still didn't work. Again saying i didn't select an image? i added a var_dump(@POST); to my script and the image file is not being passed through by the script? here's my form that is submiting the info is there a problem with this maybe?

HTML Code:

<form enctype="multipart/form-data" action=savead.php method=post>
<p>
<input type=hidden name=adtype value="O">
<input type=hidden name=groupid value="2">

Title<br><input type=text name=adtitle size=50>
<br>
Upload Ad<br><input type="hidden" name="MAX_FILE_SIZE" value="100000" />
<input name="adimagefile" type="file">
<br>

image type<br><input type="radio" name="imagetype" value="jpg"> - Jpeg &nbsp;&nbsp;&nbsp;&nbsp;<input type="radio" name="imagetype" value="gif">Gif<br>
URL<br>
http://<input type=text name=url size=50>
<br>
<input type=reset value=Reset>
&nbsp;&nbsp;&nbsp;&nbsp;
<input type=submit value=Submit>
</p>
</form>

here's my $_POST variables...

PHP Code:


			
Array
(
    [adtype] => O
    [groupid] => 2
    [adtitle] => Test Title
    [MAX_FILE_SIZE] => 100000
    [imagetype] => jpg
    [url] => www.google.com
)

No $adimagefile

Wildhoney · 02-25-2008, 03:20 AM

Uploaded items get stored in $_FILES, not $_POST.

abiko · 02-25-2008, 03:20 AM

Edit: hehe Wildhoney - same post time :)
-----
oh.
YOu don't get 'adimagefile' in the $_POST -
there should on be adtitle, imagetype and url.

The adimagefile data, e.g. the data about the uploaded image is in $_FILES['adimagefile'].
Here's an array

PHP Code:


			
$file        =        array(                'tmp_name'    =>    $_FILES['adimagefile']['name'],
                                                                    'tmp_type'    =>    $_FILES['adimagefile']['type'],
                                                                    'tmp_file'    =>    $_FILES['adimagefile']['tmp_name'],
                                                                    'tmp_size'    =>    $_FILES['adimagefile']['size']
                                                                );

And there you have :

tmp_name => the name of the file - (image1.jpg/image1.gif)
tmp_type => the type - image/jpeg, image/gif, image/png
tmpl_file => the path to the tmp file created by php in the tmp folder :)
tmp_size => the size of the image

Hope this works for you:

PHP Code:


			
<?php
$adtitle = $_POST['adtitle'];
$adtype = $_POST['adtype'];
$imagetype = $_POST['imagetype'];
$url = $_POST['url'];
$groupid = $_POST['groupid'];
$file        =        array(                'tmp_name'    =>    $_FILES['adimagefile']['name'],
                                                                    'tmp_type'    =>    $_FILES['adimagefile']['type'],
                                                                    'tmp_file'    =>    $_FILES['adimagefile']['tmp_name'],
                                                                    'tmp_size'    =>    $_FILES['adimagefile']['size']
                                                                );

   $error = false;
   $errormessage = "";

   if (strlen($adtitle) < 1) {
      $error = true;
      $errormessage .= "<li><b>Sorry.  You did not enter an ad title.</b><br>\n";
   }
     if (strlen($url) < 1) {
         $error = true;
         $errormessage .= "<li><b>Sorry.  You did enter a URL</b><br>\n";
      }

   }
    if (strlen($code) < 1) {
                 $error = true;
                 $errormessage .= "<li><b>Sorry.  You did not enter an affiliate code.</b><br>\n";
              }
              
              
if ( !$error ) {    
    echo "<h2>Save new ad</h2>
<p>";
        if ( $file['tmp_type'] == 'image/jpeg' | $file['tmp_type'] == 'image/gif') {
                            // Upload
                              $newfilename = "../ads/$adid.$imagetype";
                    $picfile = $file['tmp_file'];
                    if ( move_uploaded_file( $picfile, $newfilename) ) {
                                printf("Ad uploaded<br>");
                                // Add to the DB                                  
                                           switch ( $adtype) {
                                                case 'A':
                                                 $sql = "insert into ads (adid, code, impressions, clicks, groupid, type, name) values ($adid, '$code', 1, 1, $groupid, '$adtype', '$adtitle')";
                                                break;    
                                                
                                                // Right ad - upload and add to the DB
                                                case 'O':
                                                     $newurl = "http://";
                                           $newurl .= $url;
                                           $sql = "insert into ads (adid, image, url, impressions, clicks, groupid, type, name) values ($adid, '$imagetype', '$newurl', 1, 1, $groupid, '$adtype', '$adtitle')";                                                    
                                                break;
                                          }                                          
                                             $result = mysql_query($sql ,$db);
                                             
                                             if ( $result ) {
                                           printf("<p>The ad has been added to the group.</p>");
                                           printf("<p><a href=editadgroup.php?groupid=$groupid>Back to ad group</a></p>");
                                       } else {
                                           printf( "Error inserting into the DB");
                                       }
                            } else {
            printf( "Ad not uploaded");
                               } 
  echo "</p>";    
    } else {
                printf "<li><b>Sorry.  You did not chose an image</b><br>\n";
} 
          
} else {
      printf("<font color=red>There was a problem creating the ad!</font><br><br>Please click on back in your browser to fix the problems listed below.<br>");
           printf($errormessage);
    
}

gillweb · 02-25-2008, 03:23 AM

should this be taken out then?

PHP Code:


			
$adimagefile = $_POST['adimagefile'];

Wildhoney · 02-25-2008, 03:30 AM

It should be like the following:

php Code:

$adimagefile = $_FILES['adimagefile'];

Then you'll have like:

php Code:

$adimagefile['tmp_name'];
$adimagefile['name'];
$adimagefile['size'];

Edit: Hehe, abiko!

gillweb · 02-25-2008, 03:53 AM

thanks! Got it working.