hope someone can help me with this.

i am trying to update records in a mysql db with a new image but i keep gettin errors.

i am using a hidden form input to post a variable across to a new page, i can echo the variable out at the top of the page but it will not update the database using this variable.

here is the code i am using to pass the variable across to the new page.

echo "<form action=\"$page\" method=\"post\">";
echo "<input type=\"hidden\" name=\"id_num\" value=\"$id\" />";
echo "<input type=\"submit\" value=\"Upload Image\"/>";
echo "</form>";

then the code for the page that handles this data is as follows.


$id_num = $_POST['id_num'];
echo "$id_num";    //echo's out the number from the previous page
$link  =  mysql_connect(localhost, xxxxxxxx, xxxxxxx) or die("Could not connect to host.");
mysql_select_db(xxxxxxdb) or die("Could not find database.");
//define a maxim size for the uploaded images
define ("MAX_SIZE","500"); 
// define the width and height for the thumbnail
// note that theese dimmensions are considered the maximum dimmension and are not fixed, 
// because we have to keep the image ratio intact or it will be deformed
define ("WIDTH","150"); 
define ("HEIGHT","120"); 
 // this is the function that will create the thumbnail image from the uploaded image
// the resize will be done considering the width and height defined, but without deforming the image
 function make_thumb($img_name,$filename,$new_w,$new_h)
//get image extension.
//creates the new image using the appropriate function from gd library
if(!strcmp("jpg",$ext) || !strcmp("jpeg",$ext))
//gets the dimmensions of the image
 // next we will calculate the new dimmensions for the thumbnail image
// the next steps will be taken: 
// 	1. calculate the ratio by dividing the old dimmensions with the new ones
//	2. if the ratio for the width is higher, the width will remain the one define in WIDTH variable
//		and the height will be calculated so the image ratio will not change
//	3. otherwise we will use the height ratio for the image
// as a result, only one of the dimmensions will be from the fixed ones
if($ratio1>$ratio2)	{
else	{
// we create a new image with the new dimmensions
// resize the big image to the new created one
// output the created image to the file. Now we will have the thumbnail into the file named by $filename
if (!strcmp("gif",$ext))
//destroys source and destination images. 
// This function reads the extension of the file. 
// It is used to determine if the file is an image by checking the extension. 
function getExtension($str) {
 $i = strrpos($str,".");
 if (!$i) { return ""; }
 $l = strlen($str) - $i;
 $ext = substr($str,$i+1,$l);
 return $ext;
 // This variable is used as a flag. The value is initialized with 0 (meaning no error found) 
 //and it will be changed to 1 if an error occures. If the error occures the file will not be uploaded.
 // checks if the form has been submitted
 //reads the name of the file the user submitted for uploading
// if it is not empty
if ($image) 
// get the original name of the file from the clients machine
$filename = stripslashes($_FILES['cons_image']['name']);
// get the extension of the file in a lower case format
$extension = getExtension($filename);
$extension = strtolower($extension);
// if it is not a known extension, we will suppose it is an error, print an error message 
//and will not upload the file, otherwise we continue
if (($extension != "jpg")  && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))	
echo '<h1>Unknown extension!  Please use .gif, .jpg or .png files only.</h1>';
// get the size of the image in bytes
// $_FILES[\'image\'][\'tmp_name\'] is the temporary filename of the file in which 
//the uploaded file was stored on the server
//compare the size with the maxim size we defined and print error if bigger
if ($sizekb > MAX_SIZE*1024)
echo '<h1>You have exceeded the 1MB size limit!</h1>';
$rand= rand(0, 1000);
//we will give an unique name, for example a random number
//the new name will be containing the full path where will be stored (images folder)
$copied = copy($_FILES['cons_image']['tmp_name'], $consname);
$copied = copy($_FILES['cons_image']['tmp_name'], $consname2);
echo "$consname2<br/>";
$sql="UPDATE general SET image= '$consname2'  WHERE id= '$id_num'"or die(mysql_error());
$query = mysql_query($sql)or die(mysql_error());
echo 'SQL: ' . $sql . '<br />Affected rows: ' . mysql_affected_rows($query);
//we verify if the image has been uploaded, and print error instead
if (!$copied) {
echo '<h1>Copy unsuccessfull!</h1>';
// the new thumbnail image will be placed in images/thumbs/ folder
$thumb_name=$consname2	;
// call the function that will create the thumbnail. The function will get as parameters 
//the image name, the thumbnail name and the width and height desired for the thumbnail
 //If no errors registred, print the success message and show the thumbnail image created
 if(isset($_POST['Submit']) && !$errors) 
echo "<h5>Thumbnail created Successfully!</h5>";
echo '<img src="'.$thumb_name.'">';
echo "<form name=\"newad\" method=\"post\" enctype=\"multipart/form-data\"  action=\"\">";
echo "<input type=\"file\" name=\"cons_image\"  >";
echo "<input name=\"Submit\" type=\"submit\"  id=\"image1\" value=\"Upload image\" />";
echo "</form>";


when i try to upload the image i get these errors

Warning: mysql_affected_rows(): supplied argument is not a valid MySQL-Link resource in /home/acmeart/public_html/lock/cms/cons_image2_up.php on line 169 SQL: UPDATE general SET image= 'image/thumb505.jpg' WHERE id= ''
Affected rows:

any help with this would be much appreciated.

8 Years
Discussion Span
Last Post by kevin wood

$sql="UPDATE general SET image= '$consname2' WHERE id= '$id_num'"or die(mysql_error()); change to $sql="UPDATE general SET image= '$consname2' WHERE id_num= '$id_num' "or die(mysql_error()); becoz y got value in $id_num =$_POST

otherwise better option check column name your table in datase..


i have a column named id which is set to auto increment. this code should update the table depending on what id number the user chose on a previous page. the id the user chose is posted over and stored in the variable $id_num which i am able to echo out at the top of the page for some reason the valus seems to be getting lost while the runs through the creation of the thumbnail.

mysql_select_db(xxxxxxdb) or die("Could not find database.");
echo "$id_num";

check it prints or not

then try this after mysql_select_db(xxxxxxdb) or die("Could not find database."); use....

$sql=mysql_query("SELECT * FROM TABLE where id='$id_num'");
		// change table name
		 $row = mysql_fetch_array($sql);

 echo "$id_num";

the variable echo's out until the code is ran for the upload of the image then it decides to go missing and i cannot work out why this is happening.

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