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Skor

Trying to do my first $_get query to generate dynamic pages

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???
Yes, I'm a newbie. Trying to teach myself PHP.  Ran into a roadblock yesterday.

I've been pouring over various books and web docs to to use the _get along with a database query. The best I can get is the column names to appear. I have a very similar query on another page that works just fine, the addition of the $_get is throwing it off. My goal is to create dynamic pages based on the prodID. Looking to see where I veered off.

<?php

$host="localhost";
$user="user";
$password="pw";
$db_name ="products";
$db_table="table";

//check connection to db
mysql_connect($host, $user, $password)
or die ("Unable to connect to server");
mysql_select_db($db_name)
or die ("Couldn't retrieve database information");


$Prodid = $_GET['Prodid'];
print($Prodid); //to make sure you're getting what you wanted

$query = "SELECT Prodid, Name, Category, Price FROM testdb WHERE Prodid = '".$_GET['Prodid']."' AND Status = 'A'";
$result = mysql_query($query) or die('Query failed: ' . mysql_error());


?>
<table border ="2">

<?php


echo '<tr><td>Prodid</td><td>Name</td><td>Category</td><td>Price</td></tr>';

$cols = mysql_num_fields( $result );

while ($row = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>".$row['Item_no']."</td>";
echo "<td>".$row['Name']."</td>";
echo "<td>".$row['Category']."</td>";
echo "<td>".$row['Price']."</td>";
echo "</tr>";
}

?>
</table>

This results in the following being displayed:

netted in a table with only labels being displayed.

<table border ="2">

<tr><td>Prodid</td><td>Name</td><td>Category</td><td>Price</td></tr></table>

Thanks in advance, I'm baffled.

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Give this  a try:

[code]
<?php
$host="localhost";
$user="user";
$password="pw";
$db_name ="products";
$db_table="table";

//check connection to db
mysql_connect($host, $user, $password)
or die ("Unable to connect to server");
mysql_select_db($db_name)
or die ("Couldn't retrieve database information");

$Prodid = $_GET['Prodid'];
print($Prodid); //to make sure you're getting what you wanted
$query = "SELECT Prodid, Name, Category, Price FROM testdb WHERE Prodid = '$Prodid' AND Status = 'A'";
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
?>

<table border ="2">
  <tr>
    <td>Prodid</td>
    <td>Name</td>
    <td>Category</td>
    <td>Price</td>
  </tr>

<?php
$cols = mysql_num_fields( $result );
while ($row = mysql_fetch_array($result)) {
  $Item_no = $row['Item_no'];
  $Name = $row['name'];
  $Category = $row['Category'];
  $Price = $row['Price'];

  echo "<tr>";
  echo "<td>$Item_no</td>";
  echo "<td>$Name</td>";
  echo "<td>$Category</td>";
  echo "<td>$price</td>";
  echo "</tr>";
}
?>

</table>
[/code]

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