Please WHO can help me?
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 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($db_handle);
}
else {
print "Database NOT Found";
}
?>
The code above gives me a favorable result. When I run the code with EASYPHP I get the following result
Connection to the Server opened.Database Found
*** This is where the problem now comes up. When I add this code i get an error message.
PHP Code:
[while ($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 Code:
<?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.