will not update

[gevans]

echo "$consname2";
$sql=("UPDATE general SET image= '$consname2'  WHERE id= $id");
$query = mysql_query($sql)or die(mysql_error());

 

look at the second line

 

$sql="UPDATE general SET image='$consname2' WHERE id= '$id'";

[kev wood]

the column i am trying to update already holds an image and i am trying to overwrite this image with this statement if that helps.

[gevans]

wrote a thread seconds before you, please take not ^^^

[kev wood]

gevans:

 

changed that line back (was just doing some trial an error in between post) here is what i got out this time

 

mage/thumb320.jpg 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/thumb320.jpg' WHERE id= ''
Affected rows: 

[gevans]

Have you set the $id variable, the error is telling you that it is not set, $id = NULL

[kev wood]

took that stray line out and we are back where we started. not getting updated and not giving me errors.

[gevans]

Took what stray line out??

 

Please put the ful code you're currently using in a post withing code tags

[kev wood]

i am echoing out the id when the page first gets loaded and it is echoing out the same number i have put in on the previous page.  so it must hold some value if it is being echoed when the page is first loaded

[gevans]

Took what stray line out??

 

Please put the ful code you're currently using in a post withing code tags

[kev wood]

here is the code i am using (ignore the stry line one removed then looked what it said in the error message)

 

echo "$consname2";
$sql="UPDATE general SET image='$consname2' WHERE id= '$id'";
$query = mysql_query($sql)or die(mysql_error());

echo 'SQL: ' . $sql . '<br />Affected rows: ' . mysql_affected_rows($query);

 

here is what is out putted

 

image/thumb122.jpg 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/thumb122.jpg' WHERE id= ''
Affected rows: 

 

the id is not showing because it has ' ' around it is it not.

[kev wood]

does it matter that the column already has data stored in it.  i thought this would just overwrite what ever was already stored in the column or should i delete what is in there already and then run the update.

[Maq]

Please, echo out $sql!

 

$sql="UPDATE general SET image='$consname2' WHERE id= '$id'";
echo $sql;
die();
$query = mysql_query($sql)or die(mysql_error());

[gevans]

No it would still show,

 

Change the code to this so w can see what id is

 

echo "$consname2<br/>";
echo "$id<br/>";
$sql="UPDATE general SET image='$consname2' WHERE id='$id'";
$query = mysql_query($sql)or die(mysql_error());
               
echo 'SQL: ' . $sql . '<br />Affected rows: ' . mysql_affected_rows($query);

[Maq]

does it matter that the column already has data stored in it.  i thought this would just overwrite what ever was already stored in the column or should i delete what is in there already and then run the update.

 

Yes, UPDATE, updates what's in the column.  It would still show up in the affected rows.

[xangelo]

If both variables are being set properly (and you're sure of this) don't add quotation marks around the id value. Quotations are for strings, integers don't require them.

 

$sql="UPDATE general SET image='$consname2' WHERE id=$id";

 

If both are being set, that should be your sql statement.

[kev wood]

here is the out put from that section of code

 

image/thumb628.jpg

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 170 SQL: UPDATE general SET image='image/thumb628.jpg' WHERE id=''
Affected rows: 

 

so it looks like the value for id is getting lost somewhere along the way?

[kev wood]

if i take the quotation marks from around the id i get this

 

You have an error in your SQL syntax near '' at line 1

[gevans]

That's your problem

 

POST ALL YOUR CODE AND WE CAN SEE WHERE $ID HAS GONE

[Adam]

$id = (int) $_POST['id'];
echo $id;

 

Try using that instead of:

 

$id = $_POST['id'];
      
      echo "$id";

[Maq]

This thread is getting ridiculously long for this type of problem.  Please do what I said 3 times now.  Echo out the $sql so we can see the actual statement you're trying to execute..........................

[gevans]

He's echoed out the sql id='' is there, his id isnt set.

 

he just need to put all the code up so we can see where he's getting the id from but he wont listen

[kev wood]

i dont understand where it can be going i have just added more articles to the db so i could change the id number i was trying to update.

 

gone through the page selected a different id number from the list, went to the page in question and the correct id number is being echoed out when the page first loads up.  i choose the file i want uploaded click the submit button which has no action so it is processed on the page it is on and the id goes missing.

 

 

the id can only be going on this page, as i said the correct number is being displayed at the top of the page.

 

here is the code for this page

 

<?php



	echo "$id"; //this echo's out the correct id number sent from the previous page.

	set_time_limit(0); 

	$link  =  mysql_connect(localhost, construction, constest) or die("Could not connect to host.");
	mysql_select_db(constructiondb) 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.
		$ext=getExtension($img_name);
		//creates the new image using the appropriate function from gd library
		if(!strcmp("jpg",$ext) || !strcmp("jpeg",$ext))
			$src_img=imagecreatefromjpeg($img_name);

		if(!strcmp("png",$ext))
			$src_img=imagecreatefrompng($img_name);

		if(!strcmp("gif",$ext))
			$src_img=imagecreatefromgif($img_name);

			//gets the dimmensions of the image
		$old_x=imageSX($src_img);
		$old_y=imageSY($src_img);

		 // 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
		$ratio1=$old_x/$new_w;
		$ratio2=$old_y/$new_h;
		if($ratio1>$ratio2)	{
			$thumb_w=$new_w;
			$thumb_h=$old_y/$ratio1;
		}
		else	{
			$thumb_h=$new_h;
			$thumb_w=$old_x/$ratio2;
		}

		// we create a new image with the new dimmensions
		$dst_img=ImageCreateTrueColor($thumb_w,$thumb_h);
		// resize the big image to the new created one
		imagecopyresampled($dst_img,$src_img,0,0,0,0,$thumb_w,$thumb_h,$old_x,$old_y); 

		// output the created image to the file. Now we will have the thumbnail into the file named by $filename
		if(!strcmp("png",$ext))
			imagepng($dst_img,$filename); 
		else
			imagejpeg($dst_img,$filename);

		if (!strcmp("gif",$ext))
			imagegif($dst_img,$filename); 

		//destroys source and destination images. 
		imagedestroy($dst_img); 
		imagedestroy($src_img); 
	 }

	 // 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.
	 $errors=0;
	 // checks if the form has been submitted
	 if(isset($_POST['Submit']))
	 {
	 //reads the name of the file the user submitted for uploading
		$image=$_FILES['cons_image']['name'];
		// 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>';
				$errors=1;
			}
			else
			{
				// 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
				$size=getimagesize($_FILES['cons_image']['tmp_name']);
				$sizekb=filesize($_FILES['cons_image']['tmp_name']);

				//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>';
					$errors=1;
				}


				$rand= rand(0, 1000);
				//we will give an unique name, for example a random number
				$image_name=$rand.'.'.$extension;
				//the new name will be containing the full path where will be stored (images folder)
				$consname="image/".$image_name;
				$consname2="image/thumb".$image_name;
				$copied = copy($_FILES['cons_image']['tmp_name'], $consname);
				$copied = copy($_FILES['cons_image']['tmp_name'], $consname2);
				$id = $_POST['id2'];
				echo "$consname2<br/>";
				echo "$id<br/>";
				$sql="UPDATE general SET image='$consname2' WHERE id=$id";
				$query = mysql_query($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>';
					$errors=1;
				}
				else
				{
					// 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
					$thumb=make_thumb($consname,$thumb_name,WIDTH,HEIGHT);
				}
			}	
		}
	 }

	  //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 $id;
	 }

	 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>";



?>

[gevans]

You can't just echo $id

 

You need to assign a number to it sent from the previous page

 

so $id = $_REQUEST['id'];

[Adam]

His printed the sQL out:

 

SQL: UPDATE general SET image='image/thumb628.jpg' WHERE id=''

 

you're trying to get the ID form $_POST, when you don't have it in your form:

 

       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>";

 

You need to add an input named ID to be able to get the ID from $_POST.

[Maq]

Yeah, show us the page where you get

 

$_POST['id2'];

 

I bet if you echo out $_POST['id2']; it's empty.