[RESOLVED] Rusty PHP user need some direction with query

dmacman1962

Hi All,

I am rusty with my PHP, I have a query that I am using to get all the products from a table that are "Video" products.

The challenge I am having is it only displays the first item, there are about 6 items.

I know it is a logic problem, ie I don't have enough logic right now! 🙂

Here is my code:

$sql2 = "Select * from tblProduct Where prodType='Video' ORDER BY prodName ASC";  

					$result2 = mysql_query($sql2);
						while($row = mysql_result($result2)){
						$prodName = $row['prodName'];
						echo $prodName;
						echo "<br><br";
					}

Thanks in advance,

Don

leatherback

Hi,

The echo SHOULD produce the right list.

this:
$prodName = $row['prodName'];
should be:
$prodName .= $row['prodName'];
if you want all the names in that one var.

dmacman1962

Thanks for the response Letherback.

Your suggestion does make all the items into one array, but I am not savy enough to figure out how to break them back apart into single "displayable" items.

What I have is 6 records in my database, and want to display 6 items, each on a seperate line.

I tried to use this technique, which I have seen before:

echo "<h2>Video</h2>"; 
					$sql2 = "Select * from tblProduct Where prodType='Video' ORDER BY prodName ASC";  

					$result2 = mysql_query($sql2);
						while($row = mysql_fetch_assoc($result2))				
						$VideoName .= $row['prodName'];{
						echo $VideoName[$i];
						echo "<br><br";
						 $i++;
						 }

..to break out each element of the arrya, but that did not produce what I am looking for.

Eventually, here is what I am trying to accomplish...

Apples
Bananas
Pears
etc.

Thanks for the logic/coding.

Don

leatherback

$sql2 = "Select * from tblProduct Where prodType='Video' ORDER BY prodName ASC";  

	$result2 = mysql_query($sql2);
		while($row = mysql_result($result2))
                        {
			echo $row['prodName'] . "<br />"; // this or:
                         $output .= $row['prodName'] . "<br /> 
                         ";// this
			}
// this is the second solution, which allows you to run the script in a different location then where the output is needed.
echo $ouput;

dmacman1962

HI Leatherback,

I tried both suggestions, and here is the error I get:

"Warning: Wrong parameter count for mysql_result() in /home/inthecla/public_html/noFrames/products/index_don.php on line 51"

which is this line:

while($row = mysql_result($result2))

I am not sure why?

Thanks,

Don

leatherback

Hi Don,

Sorry for that./ I do not know where that came from! Guess I was coding some other pages for myself..

That line should read:

while($row = mysql_fetch_array($result2))