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PHP Specific Selection

Hi experts,

I'm stuck on a problem and have no idea how to go about it.

Here is what I'm trying to do.

I'm trying to write an SQL statement that will extract say the name and price from the products table but it will only display the information for a specific product code that I type in.

Example
Product Code = HT400

Display
Code     |     Name      | Price
HT400          Bike          $50

I hope my question is clear.

Thanks in advance,

Rick.
0
Ronniel Allan Castanito
Asked:
Ronniel Allan Castanito
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1 Solution
 
kivan24Commented:
Something like that?
select code, name, price from product_table where code='HT400'

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0
 
Ronniel Allan CastanitoIT ManagerAuthor Commented:
Thanks for the fast reply!

Something like?

<?php
require_once("dbconn.php");
$name = $_POST["productCode"];
$sql = "select * from product ";
$sql = $sql . "where productCode = HT400'";
$rs = mysql_query($sql, $dbConn)
or die ('Problem with query' . mysql_error()); ?>
<body>
<?php if (mysql_num_rows($rs)>0){ ?>
<table width="700" border="1" cellpadding="10">
<tr><td>Product Code</td><td>Name</td><td>Price</td></tr>
<?php while ($row = mysql_fetch_array($rs)) { ?>
<tr>
<td><?php echo $row["productCode"]?></td>
<td><?php echo $row["Name"]?></td>
<td><?php echo $row["Price"]?></td>
</tr>
<?php } ?></table>
<?php }
else {?> <p>Code does not exsist with <?php echo HT400 ?>
in autoservice database.</p>
<?php } ?> </body></html>

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0
 
kivan24Commented:
line #5
$sql = $sql . "where productCode = '".$name."'";

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lime #20
else {?> <p>Code does not exsist with <?php echo $name ?>

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Ronniel Allan CastanitoIT ManagerAuthor Commented:
Thanks!,

I think I'm on the right track. I threw in that code in with the rest of 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>testing php</title>
</head>
<?php
$conn = mysql_connect("localhost", "test", "test1");
mysql_select_db("warehouse", $conn)
or die ('Database not found ' . mysql_error() );?>
<?php	
$name = $_POST["productCode"];
$sql = "select * from product ";
$sql = $sql . "where productCode = '".$name."'";
$rs = mysql_query($sql, $dbConn)
or die ('Problem with query' . mysql_error()); ?>
<body>
<?php if (mysql_num_rows($rs)>0){ ?>
<table width="700" border="1" cellpadding="10">
<tr><td>Product Code</td><td>Name</td><td>Price</td></tr>
<?php while ($row = mysql_fetch_array($rs)) { ?>
<tr>
<td><?php echo $row["productCode"]?></td>
<td><?php echo $row["name"]?></td>
<td><?php echo $row["price"]?></td>
</tr>
<?php } ?></table>
<?php }
else {?> <p>Code does not exsist with <?php echo $name ?>p>
<?php } ?> </body></html>
<?php }
mysql_close($conn); ?>
</table></body></html>

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I'm running into the problem of the database not loading up at all now.
0
 
kivan24Commented:
What error displaying?
check line #9 connection parameters $conn = mysql_connect("localhost", "test", "test1"); host user and password
0
 
Ronniel Allan CastanitoIT ManagerAuthor Commented:
The Error:

 The website cannot display the page
 HTTP 500  
   Most likely causes:
•The website is under maintenance.
•The website has a programming error.

The host user and password is right, I've been using the same while testing some other PHP stuff.
0
 
Ronniel Allan CastanitoIT ManagerAuthor Commented:
Still unresolved but have a better understanding now.

Thanks.
0
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