Please WHO can help me?

[Gladstone365]

 

I have this code that i am learning that I got from a PHP tut.

 

When I write the code up to a certain point I get a favorable result. When I now add the code to get the data from my database i get an error message. Please tell me what should I be doing or not doing? Here is the code:-

 

<?PHP

 

$user_name = "root";

$password = "";

//$database="checking";

$database = "address_book";

$server = "127.0.0.1";

 

/*mysql_connect($server, $user_name, $password);

$db_found = mysql_select_db($database);

*/

$db_handle = mysql_connect($server, $user_name, $password);

 

$db_found = mysql_select_db($database, $db_handle);

 

print "Connection to the Server opened.";

 

if ($db_found) {

 

print "Database Found";

mysql_close($db_handle);

}

else {

 

print "Database NOT Found";

}

?>

The code above gives me a favorable result. When I run the code with EASYPHP I get the following result

Connection to the Server opened.Database Found

 

*** This is where the problem now comes up. When I add this code i get an error message.

 

[while ($db_field = mysql_fetch_assoc($result)) {

print $db_field['ID'] . "<BR>";

print $db_field['First_Name'] . "<BR>";

print $db_field['Surname'] . "<BR>";

print $db_field['Address'] . "<BR>";

}

I get the following error message[Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\Program Files\EasyPHP 2.0b1\www\docs\prac\webdesign\connect_test.php on line 25

 

Following is the full script and code that gives me the error message above

 

 

 

<?PHP

 

$user_name = "root";

$password = "";

//$database=checking";

$database = "testdb";

$server = "127.0.0.1";

 

/*mysql_connect($server, $user_name, $password);

$db_found = mysql_select_db($database);

*/

$db_handle = mysql_connect($server, $user_name, $password);

 

$db_found = mysql_select_db($database, $db_handle);

 

print "Connection to the Server opened.";

 

if ($db_found) {

 

print "Database Found";

//

$SQL = "SELECT * FROM tb_address_book";

$result = mysql_query($SQL);

 

while ($db_field = mysql_fetch_assoc($result)) {

print $db_field['ID] . "<BR>";

print $db_field['First_Name'] . "<BR>";

print $db_field['Surname'] . "<BR>";

print $db_field['Address'] . "<BR>";

}

mysql_close($db_handle);

 

}

else {

 

print "Database NOT Found";

}

?>

I now get the following message

 

[Connection to the Server opened.Database Found

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\Program Files\EasyPHP 2.0b1\www\docs\prac\webdesign\connect_test.php on line 25

 

Please who can help me? I am just learning PHP. Thanks.

 

[NathanLedet]

where is this tutorial?

 

As a noob myself, I often forget the simplest of things... database connections and queries being one of those many things.  This tutorial I find very helpful in reminding me of some basic things:

http://www.php-mysql-tutorial.com

[Psycho]

Your query is failing - add some error handling

 

$result = mysql_query($SQL) or die("Error:<br>".mysql_error()."<br>Query:<br>$SQL");