Wamp phpmyadmin

**keyboard** · 08-03-2011, 11:13 PM

Also it's coming up with this

( ! ) Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\Canberra Amatuer Productions\member.php on line 88
Call Stack
# Time Memory Function Location
1 0.0082 383600 {main}( ) ..\member.php:0
2 0.0146 393384 mysql_fetch_array ( ) ..\member.php:88

On this page

PHP Code:


<?php 



 mysql_connect("localhost", "****", "****") or die(mysql_error()); 

 mysql_select_db("****") or die(mysql_error()); 

 



 if(isset($_COOKIE['ID_my_site'])) 

 { 

     $username = $_COOKIE['ID_my_site']; 

     $pass = $_COOKIE['Key_my_site']; 

          $check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error()); 

     while($info = mysql_fetch_array( $check ))      

         { 

 



         if ($pass != $info['password']) 

             {             header("Location: login.php"); 

             } 

 

     

     else 

             { 

               



             } 

         } 

         } 

 else 

 



 {             

 header("Location: login.php"); 

 } 

 





$links = "link2.php";









?>



<?php 



 mysql_connect("localhost", "****", "****") or die('could not connect to database'); 

 mysql_select_db("****") or die('Could not select database');



if (isset ($_POST['submit'])) 

{

    $comment = nl2br(mysql_escape_string (trim ($_POST['comment']))); 



$quantity = $_POST['quantity'];

$item = $_POST['item'];



$cheese =  "$username  - $comment  ";







$sql = mysql_query ("INSERT INTO comments (id,comments) VALUES ('0','".$cheese."')");

    

    echo 'Your comment has been entered successfully!';

    echo '<a href="member.php">Please click here</a>';

        

} 

    

else 

{





?>





<html>

<head>

<body style="background-color:lightgreen">

</head>

<body>

<?php include("$links"); 

echo "Hey ".$username;

?>  

<br /><br />

<p><b>COMMENTS</b></p>

<?php

    // POST data wasnt entered, so display the comments and comment form 

    // view comments from database

       $sql = mysql_query ("SELECT * FROM comments");

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

       echo $row['comments'].'<br />';

}







    echo '<br /><br />

    <form action="member.php" method="post">

    Comments:<br />

    <textarea name="comment" cols="40" rows="7"></textarea>

    <input type="submit" value="Submit" name="submit">

    </form>';

}

?>

</body>

</html>

Any help would be great

**JShor** · 08-04-2011, 12:16 AM

There are actually two php.ini files included with WAMP (I never understood why).

I found this out the hard way. There was one in the Apache directory, and there's one in the PHP installation directory. I'm not sure which one is the active one, but try modifying both of them.

**keyboard** · 08-04-2011, 01:40 AM

Yep, changed the second php.ini and the notices stopped but the warning is still coming up and the comments box won't work.

The error is

( ! ) Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\Canberra Amatuer Productions\member.php on line 88
Call Stack
# Time Memory Function Location
1 0.0082 383600 {main}( ) ..\member.php:0
2 0.0146 393384 mysql_fetch_array ( ) ..\member.php:88

**JShor** · 08-04-2011, 04:01 AM

In the future, you should be posting these errors under the PHP discussion board to get more attention. Please provide us with the code for member.php so we can diagnose the problem.

**keyboard** · 08-04-2011, 07:46 AM

The code above is member.php

**keyboard** · 08-04-2011, 07:47 AM

Originally Posted by keyboard1333

PHP Code:


<?php 



 mysql_connect("localhost", "****", "****") or die(mysql_error()); 

 mysql_select_db("****") or die(mysql_error()); 

 



 if(isset($_COOKIE['ID_my_site'])) 

 { 

     $username = $_COOKIE['ID_my_site']; 

     $pass = $_COOKIE['Key_my_site']; 

          $check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error()); 

     while($info = mysql_fetch_array( $check ))      

         { 

 



         if ($pass != $info['password']) 

             {             header("Location: login.php"); 

             } 

 

     

     else 

             { 

               



             } 

         } 

         } 

 else 

 



 {             

 header("Location: login.php"); 

 } 

 





$links = "link2.php";









?>



<?php 



 mysql_connect("localhost", "****", "****") or die('could not connect to database'); 

 mysql_select_db("****") or die('Could not select database');



if (isset ($_POST['submit'])) 

{

    $comment = nl2br(mysql_escape_string (trim ($_POST['comment']))); 



$quantity = $_POST['quantity'];

$item = $_POST['item'];



$cheese =  "$username  - $comment  ";







$sql = mysql_query ("INSERT INTO comments (id,comments) VALUES ('0','".$cheese."')");

    

    echo 'Your comment has been entered successfully!';

    echo '<a href="member.php">Please click here</a>';

        

} 

    

else 

{





?>





<html>

<head>

<body style="background-color:lightgreen">

</head>

<body>

<?php include("$links"); 

echo "Hey ".$username;

?>  

<br /><br />

<p><b>COMMENTS</b></p>

<?php

    // POST data wasnt entered, so display the comments and comment form 

    // view comments from database

       $sql = mysql_query ("SELECT * FROM comments");

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

       echo $row['comments'].'<br />';

}







    echo '<br /><br />

    <form action="member.php" method="post">

    Comments:<br />

    <textarea name="comment" cols="40" rows="7"></textarea>

    <input type="submit" value="Submit" name="submit">

    </form>';

}

?>

</body>

</html>

Any help would be great

That's it again

**JShor** · 08-04-2011, 04:06 PM

Sorry, I overlooked that other post and didn't see the code posted before.

Try changing line 88 to the following:

PHP Code:


$sql = mysql_query ("SELECT * FROM comments") or die(mysql_error());;

mysql_error() will tell us if MySQL is not executing the code properly and what error it is giving if $sql's query fails. When you get the error, post it here.

Also, you're reconnecting to MySQL twice in your script. You really only should be connecting once. Connecting multiple times when a connection is already open puts a lot of stress on your DB.

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