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#1 chico1st

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Posted 08 August 2006 - 03:51 AM

OKay im trying to do an image upload, i have echoed all of my variables and they seem right.
here is the form with the upload:

<form action="addNews.php" method="post" enctype="multipart/form-data" name="uploadform">
<table>
<tr>
<td width="246"><input type="hidden" name="MAX_FILE_SIZE" value="2000000"><input name="userfile" type="file" class="box" id="userfile"></td>
<td width="80"><input name="upload" type="submit" class="box" id="upload" value="  Upload  "></td>
</tr>
</table>
</form>


and here is the part of the program taht shoudl enter it into my database

<?php
if(isset($_POST['upload']))
{
        $fileName = $_FILES['userfile']['name'];
        $tmpName  = $_FILES['userfile']['tmp_name'];
        $fileSize = $_FILES['userfile']['size'];
        $fileType = $_FILES['userfile']['type'];
       

echo "$fileName --- $tmpName --- $fileSize --- $fileType";// debugging

        $fp = fopen($tmpName, 'r');
        $content = fread($fp, $fileSize);
        $content = addslashes($content);
        fclose($fp);


echo "---$content-----------------------------------------------"; //debugging
       
        if(!get_magic_quotes_gpc())
        {
            $fileName = addslashes($fileName);
        }
       

echo $fileName;//debugging

        include '../../lib.php';
$dbConn = connect();

        $query = "INSERT INTO 'picture' (`image_ID`, `name`, `type`, `content`, `size`) VALUES (NULL, '$fileName',  '$fileType', '$content', '$fileSize')";


mysql_query($query, $dbConn) or die('Error, file not uploaded');


       
        echo "File $fileName uploaded";
}       
?>


i always get the:" Error, file not uploaded", i figure it must have something to do with the insert because the variables are good, and the connection to my database i use in many different programs and it works there

Thanks for any help you can give

THANKS!


#2 fenway

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Posted 08 August 2006 - 04:38 AM

Well, in general, you should be adding mysql_error() to your query failure control path.  However, in this case, I believe the error is being generated because you're quoting your table name as a string literal instead of backticking it (like the column names).
Seriously... if people don't start reading this before posting, I'm going to consider not answering at all.

#3 chico1st

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Posted 08 August 2006 - 06:08 AM

your my hero... ive been working on this for hours. ( i took some time off to watch a movie in there too)




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