[SOLVED] No Double entry? Please help.

[kendallkamikaze]

I've been working on trying to fix this code for the last 2 hours, I want it to make it so that it will not let you enter two times. this is what i have so far:

 

<?php  
  
include 'header.php'; 
  
$horseid = $_POST['horseid'];
$userid = $_POST['userid']; 
$classtype = $_POST['classtype'];
$classname = $_POST['classname'];


$sql="INSERT INTO winterfest (horseid, userid, classname, classtype)
VALUES
('$_POST[horseid]','$id','$_POST[classname]','$_POST[classtype]')";
if (!mysql_query($sql))
  {
  die('You cannot enter that show because you have already that horse in one of its type.');
}

echo "Success! You have entered Winterfest!"; 
  
?>

[kendallkamikaze]

oops wrong thread 

[kendallkamikaze]

<?php  
  
include 'header.php'; 
  
$horseid = $_POST['horseid'];
$userid = $_POST['userid']; 
$classtype = $_POST['classtype'];
$classname = $_POST['classname'];


$sql="INSERT INTO winterfest (horseid, userid, classname, classtype)
VALUES
('$_POST[horseid]','$id','$_POST[classname]','$_POST[classtype]')";
if (!mysql_query($sql))
  {
  die('You cannot enter that show because you have already that horse in one of its type.');
}

echo "Success! You have entered Winterfest!"; 
  
?>

 

I've been working on trying to fix this code for the last 2 hours, I want it to make it so that it will not let you enter two times.

[suma237]

use REPLACE instead on INSERT.

[kendallkamikaze]

<?php  
  
include 'header.php'; 
  
$horseid = $_POST['horseid'];
$userid = $_POST['userid']; 
$classtype = $_POST['classtype'];
$classname = $_POST['classname'];


$sql="REPLACE INTO winterfest (horseid, userid, classname, classtype)
VALUES
('$_POST[horseid]','$id','$_POST[classname]','$_POST[classtype]')";
if (!mysql_query($sql))
  {
  die('You cannot enter that show because you have already that horse in one of its type.');
}

echo "Success! You have entered Winterfest!"; 
  
?>

didnt work

[Yesideez]

First use SELECT to see if they have that type of horse and if they don't use INSERT to add one.

[Maq]

Not tested but should work.  If the user has 2 or more records than it won't let them insert, are there any other restrictions?

 

include 'header.php'; 
  
$horseid = $_POST['horseid'];
$userid = $_POST['userid']; 
$classtype = $_POST['classtype'];
$classname = $_POST['classname'];

$check_query = "SELECT * FROM winterfest WHERE id = $userid";
$num_rows = mysql_num_rows($check_query) or die(mysql_error());
if($num_rows > 1)
{
echo "You have enter 2 or more times.";
}
else {
$sql="INSERT INTO winterfest (horseid, userid, classname, classtype) VALUES ('$horseid','$userid','$classname','$classtype')";
mysql_query($sql) or die(mysql_query());
echo "Success! You have entered Winterfest!"; 
}
?>

[revraz]

I think he wants it so you can't enter twice, if so, he would need to change

 

if($num_rows > 1)

 

to either

 

if($num_rows => 1)

 

or

 

if($num_rows > 0)

 

 

[Maq]

Yes, she certainly did.  Thanks revraz 

[kendallkamikaze]

still no luck :/

[gevans]

Does your table have a primary key?

[kendallkamikaze]

yes.

 

id int(10) No auto_increment

[gevans]

Can you double check that it is a primary key, as replace needs a PK to work.

 

Also your query is using POST data when you've already assigned it to variables.

 

And in the query the var $id hasn't been assigned anywhere

[revraz]

Dupe post?

 

http://www.phpfreaks.com/forums/index.php/topic,230036.0.html

[kendallkamikaze]

yes, sorry i had posted it in the wrong thread. I didnt even realize people were posting on it O.O

 

With the code MAQ provided, seems like it would work but I'm getting:

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/collegh9/public_html/one-stride/winterfest_enter.php on line 10

 

<?php  
include 'header.php'; 
  
$horseid = $_POST['horseid'];
$userid = $_POST['userid']; 
$classtype = $_POST['classtype'];
$classname = $_POST['classname'];

$check_query = "SELECT * FROM winterfest WHERE id = $userid";
$num_rows = mysql_num_rows($check_query) or die(mysql_error());
if($num_rows => 1)
{
echo "You have enter 2 or more times.";
}
else {
$sql="INSERT INTO winterfest (horseid, userid, classname, classtype) VALUES ('$horseid','$userid','$classname','$classtype')";
mysql_query($sql) or die(mysql_query());
echo "Success! You have entered Winterfest!"; 
}
?>

[revraz]

echo $check_query, sounds like $userid is empty or maybe it's a string?

 

If so, put single quotes around $userid

 

 

$check_query = "SELECT * FROM winterfest WHERE id = '$userid'";

 

[PFMaBiSmAd]

Actually, greater than or equal is >=

 

=> has special significance and is not a comparison operator.

 

Upon further review, the posted code has a few other problems. There is no mysql_query() in it to execute the SELECT query and the or die() statement would need to be part of that mysql_query().

 

Replace this line -

 

$num_rows = mysql_num_rows($check_query) or die(mysql_error());

 

With these -

$result = mysql_query($check_query) or die(mysql_error());
$num_rows = mysql_num_rows($result));

[Maq]

Sorry thank you for the corrections guys, I don't have an environment to test this in right now ;/

 

@PFMaBiSmAd

You have an extra ')' in the second line you corrected.

 

My revised code (that should work  ):

 

include 'header.php'; 
  
$horseid = $_POST['horseid'];
$userid = $_POST['userid']; 
$classtype = $_POST['classtype'];
$classname = $_POST['classname'];

$check_query = "SELECT * FROM winterfest WHERE id = $userid";
$result = mysql_query($check_query) or die(mysql_error())
$num_rows = mysql_num_rows($check_query);
if($num_rows >= 1)
{
   echo "You have enter 2 or more times.";
}
else {
   $sql="INSERT INTO winterfest (horseid, userid, classname, classtype) VALUES ('$horseid','$userid','$classname','$classtype')";
   mysql_query($sql) or die(mysql_query());
   echo "Success! You have entered Winterfest!"; 
}
?>

[kendallkamikaze]

with the above revised code, you get:

 

Parse error: syntax error, unexpected T_VARIABLE in /home/collegh9/public_html/one-stride/winterfest_enter.php on line 11

[Maq]

Oops here you go (supposed to be $num_rows = mysql_num_rows($result); ) 

 

include 'header.php'; 
  
$horseid = $_POST['horseid'];
$userid = $_POST['userid']; 
$classtype = $_POST['classtype'];
$classname = $_POST['classname'];

$check_query = "SELECT * FROM winterfest WHERE id = $userid";
$result = mysql_query($check_query) or die(mysql_error())
$num_rows = mysql_num_rows($result);
if($num_rows >= 1)
{
   echo "You have enter 2 or more times.";
}
else {
   $sql="INSERT INTO winterfest (horseid, userid, classname, classtype) VALUES ('$horseid','$userid','$classname','$classtype')";
   mysql_query($sql) or die(mysql_query());
   echo "Success! You have entered Winterfest!"; 
}
?>

[kendallkamikaze]

lol its all good...im new to this whole php thing and my programmer is unavilable...

 

Now im getting:

Parse error: syntax error, unexpected T_VARIABLE in /home/collegh9/public_html/one-stride/winterfest_enter.php on line 11

[premiso]

Look on/near line 11 and solve it. That is a basic syntax issue and easy fix.

 

The issue lies with the $result row there is no ending ; after the die statement.

[Maq]

$result = mysql_query($check_query) or die(mysql_error())

 

needs to be:

 

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

[kendallkamikaze]

the code is fully functional now, but its still allowing double entries...AHAH!

 

I've got it. I realized that user id is not actually posted onto the winterfest table, so by switching user id to horseid its now functional!

 

<?php
include 'header.php'; 
  
$horseid = $_POST['horseid'];
$userid = $_POST['userid']; 
$classtype = $_POST['classtype'];
$classname = $_POST['classname'];

$check_query = "SELECT * FROM winterfest WHERE horseid = $horseid";
$result = mysql_query($check_query) or die(mysql_error());
$num_rows = mysql_num_rows($result);
if($num_rows >= 1)
{
   echo "You have enter 2 or more times.";
}
else {
   $sql="INSERT INTO winterfest (horseid, userid, classname, classtype) VALUES ('$horseid','$userid','$classname','$classtype')";
   mysql_query($sql) or die(mysql_query());
   echo "Success! You have entered Winterfest!"; 
}
?>

 

This website has been amazing. You guys are GREAT help to my lack of PHP knowledge!

I wish I could pay you guys for all your help. If you ever need any work or have time for any work I'm willing to pay to get things fixed/made for my game!

 

<3

[Maq]

I wish I could pay you guys for all your help.

 

Donate!