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@NogDogOct 30.2006 — #
In this section you are assigning the [b]results[/b] of a mysql_query() to the variable $gift_entry, then trying to use that result set pointer as a query in a second mysql_query() command.@NogDogOct 30.2006 — #You have two mysql_query() commands. The first one executes the query which is given as a string literal in its parameter list, and assigns the return value to the variable $gift_entry. That value is going to either be a special variable type in PHP known as a "resource" (used to reference the query result in related commands) or FALSE if it failed.
You then call mysql_query() again, this time using the variable $gift_entry as the query parameter, but $gift_entry is NOT a query. What you probably want to do is:
MySQL: Query Was Empty
Anything wrong with my code, I keep getting “Query Was Empty”?
[code=php]
<?php
//DATABASE CONNECTION
include ‘includes/mysql/connection.php’;
//SELECT DATABAE
$dbname = uw_wipgfhprogram;
//CONNECT TO MYSQL DATABASE
$dbconn = mysql_connect ( $dbhost, $dbuser, $dbpass );
//CHECK IF CONNECTED
if (! $dbconn ) { die( ‘Could Not Connect: ‘ . mysql_error()); }
//SELECT DATABASE
mysql_select_db( $dbname, $dbconn );
//EXMAMPLE USER
$cn = ‘Christopher We.’;
//SELECT QUERY
$info = mysql_query( “SELECT * FROM `master_list` WHERE `cn` = ‘$cn'”, $dbconn );
//TIME/DATE
$g1_dt_signup = date( “Y-m-d H-i-s” );
$g1_dt_signup_output = date( “m/d/y” );
//INSERT QUERY
$gift_entry = mysql_query( “INSERT TO `master_list`
(
`g1_st`,
`g1_dt_signup`,
`g1_dt_signup_output`
) VALUES (
‘taken’,
‘$g1_dt_signup’
‘$g1_dt_signup_output’
)”, $dbconn );
//INSERT RESULTS
$result = mysql_query( $gift_entry ) or die( ‘Error: ‘ . mysql_error());
//CLOSE CONNECTION
mysql_close($dbconn) ?>
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@NogDogOct 30.2006 — #[code=php]
//INSERT QUERY
$gift_entry = mysql_query( "INSERT TO master_list
(
g1_st,
g1_dt_signup,
g1_dt_signup_output
) VALUES (
'taken',
'$g1_dt_signup'
'$g1_dt_signup_output'
)", $dbconn );
//INSERT RESULTS
$result = mysql_query( $gift_entry ) or die( 'Error: ' . mysql_error());
[/code]In this section you are assigning the [b]results[/b] of a mysql_query() to the variable $gift_entry, then trying to use that result set pointer as a query in a second mysql_query() command.
0
@NogDogOct 30.2006 — #You have two mysql_query() commands. The first one executes the query which is given as a string literal in its parameter list, and assigns the return value to the variable $gift_entry. That value is going to either be a special variable type in PHP known as a "resource" (used to reference the query result in related commands) or FALSE if it failed.You then call mysql_query() again, this time using the variable $gift_entry as the query parameter, but $gift_entry is NOT a query. What you probably want to do is:
[code=php]
//INSERT QUERY
$gift_entry = "INSERT TO master_list
(
g1_st,
g1_dt_signup,
g1_dt_signup_output
) VALUES (
'taken',
'$g1_dt_signup'
'$g1_dt_signup_output'
)";
$result = mysql_query( $gift_entry ) or die( 'Error: ' . mysql_error());
[/code]