[SOLVED] What is wrong with my logic?

[izza56]

Hi Folks,

 

I have a bit of php here that lists registered users on a site and you can follow (twitter like) a user.  If a user clicks 'follow' the code does a quick check to see if they are already being followed, if they are then state it, otherwise insert into the database the follow details.  However, when i try to add a user that is already being followed the message appears ok, but the code then continues to add the user to the database.  So, when you want to report on who you are following the same user appears multiple times?  Can you see why this is happening below?

 

<?php

require('dbUtils.php');

$connection = dbConnect();

 

$checkiffollow = "SELECT * FROM follow

WHERE userfollowid IN (

SELECT user.userid FROM user

WHERE username = '$followinguser')

AND userid IN (

SELECT user.userid FROM user

WHERE username = '$username')";

 

$insertfollow = "INSERT INTO follow (userid, userfollowid, followdate)

VALUES ((SELECT userid FROM user WHERE username = '$username'),

(SELECT userid FROM user WHERE username = '$followinguser'),

NOW())";

 

$result = mysql_query($checkiffollow, $connection);

$insertrow = mysql_query($insertfollow, $connection);

 

if(!$result) {

 

die("could not query database");

}

 

$count = mysql_num_rows($result);

 

if ($count != 0 ) {

echo("Sorry you are already subscribed to this user");

}

 

else {

$insertrow;

header("location:http://127.0.0.1/mysubs.php");

}

?>

 

 

 

Many thanks.

[Zhadus]

You have this line:

 

$insertrow = mysql_query($insertfollow, $connection);

 

Your logic is that it doesn't actually execute that line of code until you run the line that states

$insertrow;

 

That logic is false because it isn't a function

 

Where you have the $insertrow by itself on a line where it fails, execute this instead:

mysql_query($insertfollow, $connection) or die(mysql_error());

 

And remove all of the $insertrow variable lines and it should solve it.

[Zhadus]

I apologize, that was a little confusing, here is everything fixed:

 

<?php
require('dbUtils.php');
$connection = dbConnect();

$checkiffollow =    "SELECT * FROM follow
               WHERE userfollowid IN (
                  SELECT user.userid FROM user
                  WHERE username = '$followinguser')
               AND userid IN (
                  SELECT user.userid FROM user
                  WHERE username = '$username')";
                  
$insertfollow =    "INSERT INTO follow (userid, userfollowid, followdate)
               VALUES ((SELECT userid FROM user WHERE username = '$username'),
               (SELECT userid FROM user WHERE username = '$followinguser'),
               NOW())";
                  
$result = mysql_query($checkiffollow, $connection);
   
   if(!$result) {
   
      die("could not query database");
   }
   
   $count = mysql_num_rows($result);
   
   if ($count != 0 ) {
      echo("Sorry you are already subscribed to this user");
   }
   
   else {      
      mysql_query($insertfollow, $connection) or die(mysql_error());
      header("location:http://127.0.0.1/mysubs.php");
   }
?>

[izza56]

Zhadus, you are an absolute star mate.  Thats it sorted and I see from your explaination where the issue was.

 

If you're ever in Dublin, theres a cold guinness with your name on it.