Finish Final Stage of Social Networking Button

[justlukeyou]

Hi,

 

I am trying to complete a social networking button but I just cant sort one last peice of out. When I remove the following piece of code the social networking button works just by visiting the page.

 

Does anyone have any suggestions on how I can get this peice of working. To be totally honest I dont even know what its supposed to be doing. The concept is to only run when the button is pressed. Am I making it to complicated?

 

if(isset($_POST['followbutton']) && $_POST['followbutton'] == 'true'){

 

 

<?php



$followerid = intval($_SESSION['userID']);
       $profileid  = intval($row['id']);



if(isset($_POST['followbutton']) && $_POST['followbutton'] == 'true'){


if($profileid =  $followerid) {
       $errors['profileid'] = "This is a test error.";
   }



if(!$errors){
       //Validation of input vars passed, attempt to run query
       //Force vars to be ints to be safe for query statement


	    $followerid = intval($_SESSION['userID']);
       $profileid  = intval($row['id']);

       $query = "INSERT INTO `follow` (`user_id`, `follow_user_id`) VALUES ('{$profileid}', '{$followerid}')";
       $result = mysql_query($query);


       if (!$result)
       {
               $errors[] = "Query: {$query}<br>Error: " . mysql_error();

       }
}	
}



?>




      <?php if($errors['profileid']) print '<div class="invalid">' . $errors['profileid'] . ''; ?>  



</div>
<div class="followbuttonbox"> 
<a href="<?php echo $_SERVER['PHP_SELF']; ?>?ID=<?php echo $profileid; ?>"><img src="/images/follow.png"  id="followbutton"   /></a>
<input type="hidden" id="followbutton" name="followbutton"  value="true">
<submit button>
</form>

[denno020]

I can't see an opening <form> tag? That omission of that will be halting anything coming through $_POST with your current code.

 

As for what the code is doing:

 

isset($_POST['followbutton']) - checks to see if the variable is set (which at the moment it probably isn't getting set without the opening form tag

$_POST['followbutton'] == 'true' - makes sure the value of followbutton is true, but will only check this condition if the first condition is satisfied.

 

Hope that helps,

Denno

[justlukeyou]

Hi,

 

The form starting here. Should I starting it here?

 

   <form action="" method="post">
<a href="<?php echo $_SERVER['PHP_SELF']; ?>?ID=<?php echo $profileid; ?>"><img src="/images/follow.png"   /></a>
<input type="hidden" id="followbutton" name="followbutton" value="true">
</form>

 

When I echo 'profileid' and 'followerid' I get the two different ID number which Im expecting but I just cant seem to insert them into the database when the button is pressed.

 

 

[PaulRyan]

This:

if($profileid = $followerid) {

 

Should be:

if($profileid == $followerid) {

[justlukeyou]

Thanks,

 

I changed that but it has had no affect. I have tried to echo the two ID numbers in different places. The thing is the second set of ID numbers do not echo. I cant understand why though as the third set do echo.

 

Any suggestions please why this may be?

 

echo $profileid;
echo $followerid;


if(isset($_POST['followbutton']) && $_POST['followbutton'] == 'true'){




echo $profileid;   	    THIS DOES NOT ECHO
echo $followerid;           THIS DOES NOT ECHO






       $query = "INSERT INTO `follow` (`user_id`, `follow_user_id`) VALUES ('{$profileid}', '{$followerid}')";
       $result = mysql_query($query);


       if (!$result)
       {
               $errors[] = "Query: {$query}<br>Error: " . mysql_error();

               }
       }       


     		echo $profileid;
echo $followerid;