[SOLVED] mysql_query("SELECT * FROM listings WHERE ('$site' = 'active)")

[tommyda]

What i need to do is when i call the url

 

www.mysite.com/list.php?site=fakesite

 

that should set the $site var to 'fakesite'

 

then select all from database where 'fakesite' = 'active'

 

But its not pulling any results from the database

 

is it reading '$site' as the previusly posted $POST['fakesite']  ??

 

 

<?php

$site = $_POST['site'];

 

include"mysql.php";

 

// Get the data from the table

$result = mysql_query("SELECT * FROM listings WHERE ('$site' = 'active)")or die(mysql_error());

 

include'table.php';

?>

[Hooker]

Just a couple of questions:

 

* Is that your full source?

* Are you getting any errors?

 

It looks like you're pulling the data, just not using it.

 

[tommyda]

No errors just a blank screen??

 

Full source for list.php

 

<?php

$payment = $_POST['payment'];

$subcat = $_POST['subcat'];

include"mysql.php";

 

// Get all the data from the "example" table

$result = mysql_query("SELECT * FROM listings WHERE ('$payment' = 'yes')")or die(mysql_error());

 

// keeps getting the next row until there are no more to get

while($row = mysql_fetch_array( $result )) {

 

// Print out the contents of each row

echo $row['title'];

};

?>

[trq]

You really ought check your query for success before attempting to use any results.

 

<?php

if (isset($_POST['submit'])) {
  $payment = $_POST['payment'];
  $subcat = $_POST['subcat'];
  include "mysql.php";

  // Get all the data from the "example" table
  if ($result = mysql_query("SELECT title FROM listings WHERE $payment = 'yes'")) {
    if (mysql_num_rows($result)) {
      // keeps getting the next row until there are no more to get
      while ($row = mysql_fetch_assoc($result )) {
        // Print out the contents of each row 
        echo $row['title'];
      }
    }
  }
}

?>

[Hooker]

Can you show your  MYSQL table? seem's a little strange to have to define the column with a variable.

 

If you're trying to find out if the website exists in your database and if it's active this is what i'd probably do to make your code work:

* First, instead of just the site name, have the site name (in this instance the column is called "website" in the table) and another field that would be a boolen called "active".

<?php
include"mysql.php";

$site = $_POST['site'];
$site = $_POST['active'];

// Get all the data from the "example" table
$sql = "SELECT * FROM listings WHERE website=$site AND active=$active";
$result = mysql_query($sql)or die(mysql_error());

// keeps getting the next row until there are no more to get
if (mysql_num_rows($result)) {
     while($row = mysql_fetch_array( $result )) {

// Print out the contents of each row
echo $row['title'];
     }
}
?>

 

This would then select only the active websites that it finds in the database.

 

Note: to choose if you want active or inactive websites you simply change the value of "$active" to 1 or 0 respectively.