asked on

PHP - select - where statement

HI experts,

I'm having troubles getting my code to select a specific piece of information.

Here is my code:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>Test 1</title>
</head>
<?php
$con = mysql_connect("localhost","warehouse21","tld3");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("warehouse", $con);

$result = mysql_query("SELECT * FROM product 
WHERE productCode='HT400DVA'");

while($row = mysql_fetch_array($result))
  {
  <?php echo $row['productCode'];?>
  <?php echo $row['name'];?>
  <?php echo $row['price'];?>
 "<br />";
  }
?> 
<?php }
mysql_close($conn); ?>
</table></body></html>

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I'm trying to get my code to display the following.

Product Code | Name | Price
HT400DVA Bike $50

Thanks in advance,

Rick

Francisco Igor

Do you want a structured table or simple lines?
First, you are trying to put some php tags within another php code...

//the title line
echo "Product Code | Name | Price";
echo "<br />";

//the data lines
while($row = mysql_fetch_array($result))
{
echo $row['productCode'];
echo "|";
echo $row['name'];
echo "|";
echo $row['price'];
echo "<br />";
}

magadesign_sviluppo

you code seams ok, what error do you get?

ASKER CERTIFIED SOLUTION

Zyloch

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Ricky Nguyen

ASKER

Thanks for the fast replies!

@Zyloch

I tried your code and its brought me one step closer I'm getting the following error:

PHP Notice: Undefined variable: conn on line 35 PHP Warning: mysql_close() expects parameter 1 to be resource, null given on line 35

The product code, name and price are showing up now, but that error shows up with them.

Vampireofdarkness

You have $conn instead of $con in your mysql_close. Don't give me points for this.

Ricky Nguyen

ASKER

Thanks for the help and Thanks for the correction :)