Image uploader

dinno · #1 20-Aug-2009, 07:57

Hello everyone I have a problem with my upload when uploading I get this this message.

File has been uploadet: tastetur.jpg

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(billede,tekst) VALUES='tastetur.jpg','jjkk'' at line 1

My code looks like this

PHP Code:


<?php 
if (isset($_FILES['filnavn'])) {print "Fil der er blevet overført: {$_FILES['filnavn']['name']}<p>\n";
$tekst=$_POST['tekst'];    
    
$query = "INSERT INTO dagenstilbud SET(billede,tekst) VALUES='".$_FILES['filnavn']['name']."','$tekst'";
    $Result1 = mysql_query($query, $cms) or die(mysql_error());
    
}

{
$tempfile = $_FILES['filnavn']['tmp_name'];
$destination = "dagenstilbud/{$_FILES['filnavn']['name']}";
copy($tempfile, $destination);
}
?>
   <form method="POST" enctype="multipart/form-data" name="form1" id="form1"> 
   
<table width="100%" border="0" cellspacing="0" cellpadding="2">
       <tr>
         <td width="10%">&nbsp;</td>
         <td width="90%">
         <label>Indsæt reklamebillede</label></td>
       </tr>
       <tr>
         <td colspan="2">
          <p align="left">Fil der skal overføres:
             <label for="filnavn"></label>
             <input type="file" name="filnavn" id="filnavn" /><br />
             <label>tekst
             <input type="text" name="tekst" id="tekst" />
             </label>
          <label for="navn"></label>
          <label for="type"></label>
          </p>
        </td>
    </tr>
       <tr>
         <td colspan="2">
          <input type="submit" name="button" id="button" value="Overfør" /></td>
       </tr>
     </table>
    <p>
    <label></label>
    <span class="style9">
    <label></label>
    </span></p>
  <label></label>
  <p>
    <label></label>
  </p>
  
    
   
   <input name="hiddenField" type="hidden" id="hiddenField" value="<?php echo $row_rsmitbilled['id']; ?>" />
</form>

I really hope that there is one who can tell me what goes wrong.
Thank you in advance
dinno

TurboPT · #2 20-Aug-2009, 21:53

You have crossed the syntax of an INSERT statement with part of an UPDATE statement, but you are close...

The generic form for an INSERT or UPDATE is:

Code:

INSERT INTO table( column_list ) VALUES( column_values );

UPDATE table SET col1=value, col2=value, coln=value WHERE <condition>

So, in your SQL, the 'SET' needs removal:

PHP Code:


$query = "INSERT INTO dagenstilbud SET(billede,tekst) VALUES='".$_FILES['filnavn']['name']."','$tekst'";

For an INSERT should be formed like so:

PHP Code:


$query = "INSERT INTO dagenstilbud(billede,tekst) VALUES('".$_FILES['filnavn']['name']."','$tekst')";

dinno · #3 21-Aug-2009, 05:40

hey 1000 thanks, it works almost as it should, the upload the image to a folder and text from the entry fields in to the database but does not takes
the filename to the database ... how can it be?
dinno

TurboPT · #4 21-Aug-2009, 05:45

Good.

Sorry that I forgot to mention to also remove the = with the VALUES, but I had it correct in the example.

Anyway, echo the query string for output and make sure it looks like what is expected. Maybe the value is blank?

dinno · #5 21-Aug-2009, 06:42

Hello again have the right value was empty I did like this now and it works thanks smile again it looks like this now ...

<?php
if (isset($_FILES['filnavn'])) {print "Fil der er blevet overført: {$_FILES['filnavn']['name']}<p>\n";
$tekst=$_POST['tekst'];
$webadr=$_POST['webadresse'];
$userid=$_POST['userid'];
$filnavn=$_FILES['filnavn']['name'];
$query = "INSERT INTO dagenstilbud (billede, tekst, webadresse, mainid) VALUES ('$filnavn','$tekst','$webadr','$userid')";

$Result1 = mysql_query($query, $cms) or die(mysql_error());

}

{
$tempfile = $_FILES['filnavn']['tmp_name'];
$destination = "dagenstilbud/{$_FILES['filnavn']['name']}";
copy($tempfile, $destination);
}
?>

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