Query NOT working

[zed420]

Hi All

Can someone please tell me what am I doing wrong in this query??? Its inserting everthing even those date that are already there. So basically its this line that's not working.

if (($result) == ($_POST['s_time'])) {

why???

Many thanks

Zed

function insert(){
$b_id = $_POST['b_id'];
$dateTime = $_POST['dateTime'];
$user_id = $_POST['user_id'];
$s_time = $_POST['s_time'];
$e_time = $_POST['e_time'];
$request_date = $_POST['request_date'];

     $query = "SELECT s_time,request_date FROM booking WHERE request_date = '$request_date'";
echo "$query";
     $result = mysql_query($query)or die(mysql_error());

if (($result) == ($_POST['s_time'])) {
	error_message("Sorry, this hour/s is already booked please choose another !");   
}else{
$query = "INSERT INTO booking  VALUES
(NULL,'$dateTime','$user_id','$s_time','$e_time','$request_date')";
	if (@mysql_query($query)) {
		comfirmed_booking();
		} else {
		echo '<p>Error adding submitted Information: ' .
		mysql_error() . '</p>';
	}//end of else
}//end of if
}//end of insert

[trq]

Take a look at the manual entry for mysql_query. I'm not sure what you think it returns, but your code makes no sense at all.

[5kyy8lu3]

Hi All

Can someone please tell me what am I doing wrong in this query??? Its inserting everthing even those date that are already there. So basically its this line that's not working.

if (($result) == ($_POST['s_time'])) {

why???

Many thanks

Zed

function insert(){
$b_id = $_POST['b_id'];
$dateTime = $_POST['dateTime'];
$user_id = $_POST['user_id'];
$s_time = $_POST['s_time'];
$e_time = $_POST['e_time'];
$request_date = $_POST['request_date'];

     $query = "SELECT s_time,request_date FROM booking WHERE request_date = '$request_date'";
echo "$query";
     $result = mysql_query($query)or die(mysql_error());

if (($result) == ($_POST['s_time'])) {
	error_message("Sorry, this hour/s is already booked please choose another !");   
}else{
$query = "INSERT INTO booking  VALUES
(NULL,'$dateTime','$user_id','$s_time','$e_time','$request_date')";
	if (@mysql_query($query)) {
		comfirmed_booking();
		} else {
		echo '<p>Error adding submitted Information: ' .
		mysql_error() . '</p>';
	}//end of else
}//end of if
}//end of insert

 

i'm not great with php but I think you need to get your results into a variable from your result set

 

like such: $row = mysql_fetch_array($result);

 

if ( $row['s_time'] == $_POST['s_time'] )

 

not sure that helps but i tried lol

[zed420]

All I'm trying to do is look into the database if there is a similar s_time as POSTED one, give an error msg else insert into databse.

Zed

[5kyy8lu3]

<?php

$query = "SELECT s_time,request_date FROM booking WHERE request_date = '$request_date'";

$result = mysql_query($query) or die(mysql_error());

$row = mysql_fetch_array($result); //you need to pull the results out so you can compare it to the request

if ( $row['s_time'] == $_POST['s_time'] )
{
error_message("Sorry, this hour/s is already booked please choose another !");   
}

else
{
$query = "INSERT INTO booking  VALUES
(NULL,'$dateTime','$user_id','$s_time','$e_time','$request_date')";

?>

[zed420]

Thanks for your reply, it still not having it

 

Zed

[trq]

All I'm trying to do is look into the database if there is a similar s_time as POSTED one, give an error msg else insert into databse.

 

Define similar.

[zed420]

Sorry for not making myself clear, I'm POSTING hrs from a dropdown form, if the hr(s_time) from 8am to 9am has been booked(already in database) and someone tries to book this hr the error MSG should come up saying that it been booked please choose another.

Zed

[zed420]

This code works but only for the first record meaning that you can not insert any more same records as the first one but after that any other record can be duplicated. why??

Thanks for your reply, it still not having it

 

Zed