php and sql ?

[mac_shinobi]

Hi

 

I have phpmyadmin setup on my web server and have created a database called gecko with one table called feedback.

 

I am trying to create a php page that will execute a the Select field1, field2, field 3 from feedback but am just getting it returning

 

Connected Succesfully and Resource id #3

 

What does that mean and how do I fix it ?

 

   $user="gecko";
   $host='localhost:/var/run/mysqld/mysqld.sock';
   $password="password";
   $database="gecko";

$link = mysql_connect($host, $user, $password);
if (!$link) {
   die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>


$sql = 'SELECT Forename , Surname , Comment FROM feedback; ';
$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

print($result);
mysql_close($link);    
   
?>

 

I did find this :

 

PHP/MYQSL brings back wrong value (resource ID #3) - Dynamic Drive Forums

 

which basically says

 

"You cannot echo your raw query. You must first grab the information it is returning with a function such as mysql_fetch_array()"

 

Can someone correct my php code so that it displays what I have in my database table feedback.

[Jona]

something like

print_r(mysql_fetch_array($result));

 

Should do the job, to display it in a more formatted way you'll want to assign the array to a variable and then do a for loop such as:

$rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['field1'];
print $item['field2'];
}

 

Hope that helps.

 

Cheers

Jona

 

Ninja edit: To correct typo.

Edited October 19, 2008 by Jona

[Gatt]

if u have PHPMyAdmin running, then tehre should be an option somehwere to run a query, then get it do display the PHP coding for that query..

[mac_shinobi]

something like

print_r(mysql_fetch_array($result));

Should do the job, to display it in a more formatted way you'll want to assign the array to a variable and then do a for loop such as:

$rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['field1'];
print $item['field2];
}

Hope that helps.

 

Cheers

Jona

 

hi

 

ok tried that and got this error :

 

Parse error: syntax error, unexpected ')', expecting T_PAAMAYIM_NEKUDOTAYIM in /home/gecko/www/feedback.php on line 19

[Gatt]

$rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['field1'];
print $item['field2];
}  

 

Should read

 

$rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['field1'];
print $item['field2'];
}  

 

missing ' after field2

[Jona]

As above, typing error, will correct in the orignal post.

[mac_shinobi]

ok lets try again :

 

$link = mysql_connect($host, $user, $password);
if (!$link) {
   die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>


$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';
$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

$rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['Forename'];
print $item['Surname'];
print $item['Comment'];
}  


mysql_close($link);    
   
?>

 

I get that error using the above code which has the commas etc in the correct place yet if I put a dollar symbol on this part :

 

$rows = mysql_fetch_array($result);
foreach($rows as $item){
print $item['Forename'];
print $item['Surname'];
print $item['Comment'];
}  

 

As you can see in the foreach part I have put a dollar symbol aka $ before the item and then each time I run it it just returns

 

mmmmmm

 

What good is that - I have emptied my database and have not got any fields then I go to my add.php which works fine and it adds a record into the database but I have not once entered mmmmmm

[Gatt]

ignore this... Edited October 19, 2008 by Gatt

[mac_shinobi]

As you can see from my PHP above I only want it to return

 

$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';

 

The Forename, Surname and Comment that each user has made in the order they were posted.

 

What PHP do I need to do this please.

[Gatt]

This is an exert from one of my PHP scripts:

 

		 
   			$query = "SELECT * FROM dvds ORDER BY dvd_name ASC";
   			$result = mysql_query($query);
   			$numofrows = mysql_num_rows($result);

   			while ($i < $numofrows) :
         			$information = mysql_result($result, $i, "dvd_name");
         			$i++;
   			endwhile;
		?>

 

SO yours would be:

 

		 
   			$query = "SELECT Forename, Surname, Comment FROM Feedback LIMIT 0, 30";
   			$result = mysql_query($query);
   			$numofrows = mysql_num_rows($result);

   			while ($i < $numofrows) :
         			$information = mysql_result($result, $i, "forename");
         			$i++;
   			endwhile;
		?>

[mac_shinobi]

This is an exert from one of my PHP scripts:

 

            
               $query = "SELECT * FROM dvds ORDER BY dvd_name ASC";
               $result = mysql_query($query);
               $numofrows = mysql_num_rows($result);

               while ($i < $numofrows) :
                     $information = mysql_result($result, $i, "dvd_name");
                     $i++;
               endwhile;
           ?>

SO yours would be:

 

            
               $query = "SELECT Forename, Surname, Comment FROM Feedback LIMIT 0, 30";
               $result = mysql_query($query);
               $numofrows = mysql_num_rows($result);

               while ($i < $numofrows) :
                     $information = mysql_result($result, $i, "forename");
                     $i++;
               endwhile;
           ?>

 

ok tried that and am getting this :

 

Connected successfully

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/gecko/www/feedback.php on line 19

 

 

The script I am using is this :

 

   $user="gecko";
   $host='localhost:/var/run/mysqld/mysqld.sock';
   $password="mypassword";
   $database="gecko";

$link = mysql_connect($host, $user, $password);
if (!$link) {
   die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>


$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';
//$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

               //$query = "SELECT Forename, Surname, Comment FROM Feedback LIMIT 0, 30";
               $result = mysql_query($sql);
               $numofrows = mysql_num_rows($result);

               while ($i < $numofrows) :
                     $information = mysql_result($result, $i, "forename");
                     $i++;
               endwhile;
           


mysql_close($link);       
?>

[mac_shinobi]

ok have this code now :

 

   $user="gecko";
   $host='localhost:/var/run/mysqld/mysqld.sock';
   $password="mypassword";
   $database="gecko";
$link = mysql_connect($host, $user, $password);
mysql_select_db("gecko", $link);
if (!$link) {
   die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>


//$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';
//$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

               $query = "SELECT Forename, Surname, Comment FROM feedback LIMIT 0, 30";
               $result = mysql_query($query, $link);
               $numofrows = mysql_num_rows($result);

               while ($i < $numofrows) :
                     $information = mysql_result($result, $i, "Forename");
                     $i++;
               endwhile;
           


mysql_close($link);       
?>

 

It just states that its connected successfully but no errors so how do I make it print each result ie Forname, Surname and Comment ??

[Gatt]

Sorry my fault - add a line to decalre what $i is

 

$i=0

 

Just after your DB details:

 

   $user="gecko";
   $host='localhost:/var/run/mysqld/mysqld.sock';
   $password="mypassword";
   $database="gecko";

$i=0;  //Declare counter here..
$link = mysql_connect($host, $user, $password);
if (!$link) {
   die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>


$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';
//$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

               //$query = "SELECT Forename, Surname, Comment FROM Feedback LIMIT 0, 30";
               $result = mysql_query($sql);
               $numofrows = mysql_num_rows($result);

               while ($i < $numofrows) :
                     $information = mysql_result($result, $i, "forename");
                     $information = mysql_result($result, $i, "Surname");
                     $information = mysql_result($result, $i, "Comment");
                     $i++;
               endwhile;
           


mysql_close($link);       
?>

 

Does it work now?

[mac_shinobi]

Nope

 

getting this error :

 

Connected successfully

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/gecko/www/feedback.php on line 21

[Gatt]

Hmm.. I can only imagine its those quote marks..

 

Here is my full php script..

 

include ("includes/db_connect.php");
?>


	 DVD Library - 
						echo gmdate("l, dS F, Y");
		?>
	


	 DVD Library....
	 
		$counter = 0;
	?>
	</pre><form method="post" action="vew_disc.php">
   		
		 
   			$query2 = "SELECT * FROM dvds ORDER BY dvd_name ASC";
   			$result2 = mysql_query($query2);
   			$numofrows2 = mysql_num_rows($result2);

   			while ($i < $numofrows2) :
         			$information = mysql_result($result2, $i, "dvd_name");
        			print "$information";
         			$i++;
   			endwhile;
		?>
 
  			
		

   		
   	</form> 	

 

Also, this is how PHPMyAdmin generated the PHP Code for another query..

 

$sql = 'SELECT Season , Episode_Title , Stardate FROM `tng` LIMIT 0, 30 ';

[PaulBM]

If you paste the SQL query...

 

SELECT Forename , Surname , Comment FROM feedback

 

into phpmyadmin's SQL page and run it do you get data back?

If the num_rows call is failing with the error you posted, it looks like the result set has an error or is null.

After $result = mysql_query($sql); check that $result is valid, before beginning to display the results.

 

For debug purposes...

 

print mysql_errno()." : ".mysql_error()."

";

 

Zero is what you're looking for.

Edited October 21, 2008 by PaulBM