justlukeyou

March 3, 2013

Hi,

I have a code which inserts a value into a cell when there is a standard button. However I cannot make it work when I use an image as the button. Can anyone advise/quote how I can use an image as a button?

<form method="post" enctype="multipart/form-data">
<?php
// make sure session is started before anything else
if (!isset($_SESSION)) session_start();
?>




<?php
if (isset($_REQUEST['action']) && $_REQUEST['action'] == 'follow') {

    $profileid  = (int)$_GET['ID'];
    $followerid = $_SESSION['userID'];

    if ( !($profileid > 0 && $followerid > 0) ) {
        $errors[] = 'Bad call. You might need to login.';
    } else if ($profileid == $followerid) {
        $errors['sameprofile'] = "You cannot follow your own profile.";
    } else {
        // check if already following
        $query = mysql_query("SELECT * FROM follow WHERE user_id = $profileid AND follow_user_id = $followerid");
        $duplicatefollow = null;
        if (mysql_num_rows($query) > 0) {
            $duplicatefollow = 'You Already Follow This Profile.';
            $errors['duplicatefollow'] = "You Already Follow This Profile.";
        } else {
            $query = "INSERT INTO `follow` (`user_id`, `follow_user_id`) VALUES ('{$profileid}', '{$followerid}')";
            $result = mysql_query($query);
            if (!$result){
                $errors[] = "Query: {$query}<br>Error: " . mysql_error();
            }
        }
    }
}





$loginprofile = intval($_SESSION['userID']);

$query_rsSearch = "SELECT * FROM follow WHERE `follow_user_id` = '$loginprofile'";
$rsSearch = mysql_query($query_rsSearch) or die(mysql_error());
$row_rsSearch = mysql_fetch_assoc($rsSearch);
$totalRows_rsSearch = mysql_num_rows($rsSearch);

// no need to have php end tags if it's the last thing in a file.
// you risk getting the "headers already sent" error


?>


<div class="followbuttonbox">   
<input name="followbutton" type="image"  />
</div>
</form>

<?php
 
$profileid = (int)$_GET['ID'];
$followerid = $_SESSION['userID'];	


$query = mysql_query("SELECT * FROM follow WHERE user_id = $profileid AND follow_user_id = $followerid"); 

$duplicatefollow = null;
if (mysql_num_rows($query) > 0) 
{ 
$duplicatefollow = 'You Already Follow This Profile.'; 
}   
else
{
  
if(isset($_POST['followbutton'])) {


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


	if (!$result){
	$errors[] = "Query: {$query}<br>Error: " . mysql_error();
	}
}  
}     
  
         if(($profileid) == ($followerid)) {
$errors['sameprofile'] = "You cannot follow your own profile.";
}
  
  $loginprofile = intval($_SESSION['userID']); 
  
  $query_rsSearch = "SELECT * FROM follow WHERE `follow_user_id` = '$loginprofile'";
$rsSearch = mysql_query($query_rsSearch) or die(mysql_error());
$row_rsSearch = mysql_fetch_assoc($rsSearch);
$totalRows_rsSearch = mysql_num_rows($rsSearch);
 

        
?>




<form method="post" enctype="multipart/form-data">
<div class="followbuttonbox">   
<input name="followbutton" type="image" type="submit" src="/images/follow.png"   />
<input name="followbutton" type="submit" value="Follow Button" />
</div>
</form>

February 27, 2013

Hi,

I have added this which returns the error message however I cant work out how to stop the duplicate input. Can anyone who knows PHP able to advise how to stop the PHP insert once the error message runs?

    $query = mysql_query("SELECT * FROM follow WHERE follow_user_id  = '". $followerid ."'");
$duplicatefollow = null;
if (mysql_num_rows($query) > 0)
{
$duplicatefollow = 'You Already Follow This Profile.</a>.';
}   

 
if(isset($_POST['followbutton'])) {


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


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

February 26, 2013

Hi,

I am trying to come up with a way of stopping duplicate inserts from being into a table. I have got this point which creates an error message for duplicate follows but I am stuck on what to try next. Any suggestions please?

As far as I can tell I need something in the insert code that says if the value is already in the table it cannot be re-inserted and then echo the message.

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

$result = mysql_query($query);



    if (!$result){

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

    }

}



if(($profileid) == ($followerid)) {

$errors['sameprofile'] = "You cannot follow your own profile.";

}



$loginprofile = intval($_SESSION['userID']);



$query_rsSearch = "SELECT * FROM follow WHERE `follow_user_id` = '$loginprofile'";

$rsSearch = mysql_query($query_rsSearch) or die(mysql_error());

$row_rsSearch = mysql_fetch_assoc($rsSearch);

$totalRows_rsSearch = mysql_num_rows($rsSearch);



$duplicate = $profileid;





if(($loginprofile) == $row_rsSearch['follow_user_id']) {

$errors['duplicatefollow'] = "You already follow this profile.";

}

I have this across a number of pages so I should be able to re-use it quite often.

February 25, 2013

I have it now so that it only runs when the button is pressed however I want to remove the button and only use the image. Any suggestions please? (Almost there with it)

<code>

</div>

</code>

February 25, 2013

I've spent THREE months on it. I need to get it sorted. Once its sorted its done then.

February 25, 2013

Hi,

I have been trying to make this work for 3 months now. Firstly I got it working only when the page is loaded. Now it works when both the page is loaded and the button is pressed.

But I just cant seem to make it work only when the button is pressed.

I somehow need to match these two parts together but I cant seem to be able to do it.

<code>

if($ID != "") {

</code>

<code>

<?php

$ID = mysql_real_escape_string($_GET['ID']);

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

$profileid = $ID;

if($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();

}

?>

</div>

</form>

</code>

February 23, 2013

Sorted.

I tried do this earlier but put the $accounty = Y; in the wrong place.

$accounty = Y;	
$loginPassword = md5($loginPassword);
$query = 'SELECT * FROM users WHERE email = "' . mysql_real_escape_string($loginEmail) . '" AND password = "' . $loginPassword . '" AND accountconfirmed = "' . $accounty . '"';
           $result = mysql_query($query);

February 23, 2013

Oh right, so I need something like this? ....

But this still allows me to login into unconfirmed accounts.

       $query_rsSearch = "SELECT * FROM users WHERE `email` = '$emailAddress'";
$rsSearch = mysql_query($query_rsSearch) or die(mysql_error());
$row_rsSearch = mysql_fetch_assoc($rsSearch);
$totalRows_rsSearch = mysql_num_rows($rsSearch);

   $accounty = ('Y');

   // Login attempt
   if(isset($_POST['loginSubmit']) && $_POST['loginSubmit'] == 'true')
   {
       $loginEmail = filter_var(trim($_POST['email']), FILTER_VALIDATE_EMAIL);
       $loginPassword    = trim($_POST['password']);
       $accounty = $row_rsSearch['accountconfirmed'];

February 23, 2013

Hi,

I dont understand what you mean. If their account is not confirmed I dont want a user to log in. Are you suggesting that I let someone logging in even if they have not confirmed their account?

I do not know how to block a user from logging in if they have not confirmed their account.

February 23, 2013

But I dont know how to stop them from logging in.

$emailAddress = $_POST['email'];

$query_rsSearch = "SELECT * FROM users WHERE `email` = '$emailAddress'";
$rsSearch = mysql_query($query_rsSearch) or die(mysql_error());
$row_rsSearch = mysql_fetch_assoc($rsSearch);
$totalRows_rsSearch = mysql_num_rows($rsSearch);

$activeStatus = $row_rsSearch['accountconfirmed']; 

if($totalRows_rsSearch == 0) {
$errors['loginEmail'] = "Your email address is not registered";
} else {
if($activeStatus != "Y") {

$errors['confirmedaccount'] = "Your account has not yet been confirmed. Please request a confirmation email.";
}
}

February 23, 2013

Thanks,

I have now got it like this. It returns the correct error messages when someone has not confirmed their account but only their enters their email address. However it does block someone from logging in when they enter both their email address and password.

How would I block someone from logging in if their account is not confirmed?

$emailAddress = $_POST['email'];

$query_rsSearch = "SELECT * FROM users WHERE `email` = '$emailAddress'";
$rsSearch = mysql_query($query_rsSearch) or die(mysql_error());
$row_rsSearch = mysql_fetch_assoc($rsSearch);
$totalRows_rsSearch = mysql_num_rows($rsSearch);

$activeStatus = $row_rsSearch['accountconfirmed'];

if($totalRows_rsSearch == 0) {
$errors['loginEmail'] = "Your email address is not registered";
} else {
if($activeStatus == "Y") {
//'Y' MEANS ACCOUNT HAS BEEN ACTIVATED - SHOW WEBSITE

} else {
//ACCOUNT HAS NOT BEEN ACTIVATED YET .
$errors['confirmedaccount'] = "Your account has not yet been confirmed. Please request a confirmation email.";
}
}

February 23, 2013

Hi,

It is 21:45 and I have spent 8 hours today trying to finish off my login page. The last stage is stop someone from logging in if their account is not confirmed. This is the closest I have got to it.

If the user has confirmed their account it should have a Y in the 'accountconfirmed' row.

Can anyone please advise how I return a message and block someone from logging into an unconfirmed account?

$accounty = ('Y'); 

$query_rsSearch = "SELECT * FROM users WHERE `accountconfirmed` = '$accounty'";
$rsSearch = mysql_query($query_rsSearch) or die(mysql_error());
$row_rsSearch = mysql_fetch_assoc($rsSearch);
$totalRows_rsSearch = mysql_num_rows($rsSearch);

if($totalRows_rsSearch != 0) {
$errors['confirmedaccount'] = "Your account has not yet been confirmed.  Please request a confirmation email.";
}

February 19, 2013

Hi,

This is the all the code. It is not returning any error messages. It logs in and does everything fine. If you wanted a user to be able to check if their email address is in the table how would you do this? I cant wee why what I am trying to do doesn't work.


<?php
     if ($_SESSION['userLoggedIn'])

        session_start();
       $_SESSION['userLoggedIn'] = 0;
       $_SESSION['userEmail'] = '';
       $_SESSION['userID'] = '';
$_SESSION['userfirstname'] = '';
$_SESSION['usersurname'] = '';

       // Reset errors and success messages
       $errors = array();
       $success = array();


   // Login attempt
   if(isset($_POST['loginSubmit']) && $_POST['loginSubmit'] == 'true')
   {
       $loginEmail = filter_var(trim($_POST['email']), FILTER_VALIDATE_EMAIL);
       $loginPassword    = trim($_POST['password']);

		$accounty = ('y'); 


       if(count($errors) === 0)
       {
$loginPassword = md5($loginPassword);
$query = 'SELECT * FROM users WHERE email = "' . mysql_real_escape_string($loginEmail) . '" AND password = "' . $loginPassword . '" AND accountconfirmed = "' . $accounty . '"LIMIT 1';
           $result = mysql_query($query);
           if (!$result)
           {
               die('Invalid query: ' . mysql_error());
           }

           if(mysql_num_rows($result) === 1)
           {
               $row = mysql_fetch_assoc($result);
               $_SESSION['userLoggedIn'] = 1;
               $_SESSION['userEmail'] = $loginEmail;
               $_SESSION['userID'] = $row['id'];
               $_SESSION['userfirstname'] = $row['firstname'];
			$_SESSION['usersurname'] = $row['surname'];

               header('Location: /test/board.php');
               exit;
           } else {
               $errors['login'] = 'No user was found with the details provided.1.';
           }
       }
   }
   /*
     The rest of your login page code
   */ 

 // Reset errors and success messages  
   $errors = array();  
   $success = array();  
   // Login attempt  
   if(isset($_POST['loginSubmit']) && $_POST['loginSubmit'] == 'true'){  
       $loginEmail = trim($_POST['email']);  
       $loginPassword  = trim($_POST['password']);  

   }  


  	if(!isset($loginEmail) || empty($loginEmail)) {   
       $errors['loginEmail'] = "Please enter your email address.";
   }	  

if(!isset($loginPassword) || empty($loginPassword)) { 
       $errors['loginPassword'] = "Please enter your password.";
   }



$accounty = ('Y'); 


		  if(($accounty) != ($row['accountconfirmed'])) {
       $errors['confirmedaccount'] = "Your account has not yet been confirmed.  Please request a confirmation email.";
   } 

			  if(($loginEmail) != ('email@yahoo.com')) {
       $errors['incorrectemail'] = "Your email address is incorrect.";
   } 

	  	 if(($loginPassword) != ($row['password'])) {
       $errors['incorrectpassword'] = "Your password is incorrect.";
   } 



   ?>

February 19, 2013

Apologies,

I will try and get more information. I just the row command would search the row.

 if(($loginEmail) != ($row['email'])) {
	$errors['incorrectemail'] = "Your email address is incorrect.";
}

February 19, 2013

Hi,

Does anyone have any suggestions please. I can log in and create a session, I just cant seem to be able to read the table.

February 18, 2013

You've got to be fucking kidding me.

Hi Jessica,

How do I scan the table to check if the email address is in the table?

February 18, 2013

Hi,

I am trying to check if an email address is in the table.

I can log in and create a session but I cant seem to return a message which says the email address is not in the table.

February 18, 2013

Hi,

I am trying scan my table on the login page to detect an email address but I dont know what code I should use to read a row.

If I enter "emailaddress@yahoo.com" in the form it does return the error message. However if enter "hello@yahoo.com" it returns the error message.

			  if(($loginEmail) != ('emailaddress@yahoo.com')) {
       $errors['incorrectemail'] = "Your email address is incorrect.";
   }

However when I try do to the following it is not scanning the table. Can anyone advise how I scan the table to determine if an email address in table?

if(($loginEmail) ($row['email'])) {

$errors['incorrectemail'] = "Your email address is incorrect.";

}

{/code]

February 18, 2013

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;

February 17, 2013

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.

February 17, 2013

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>

February 14, 2013

Hi,

I have launched a project on KickStarter and would be grateful for any feedback please. I have added a simple web page. Is it to basic?

http://www.onemillionmonsters.com/

February 13, 2013

Thanks,

What process would I follow. I cant see how I can use /test/searchoutput.php?search=green-widgets

February 13, 2013

Hi,

Do I need a form to the link. I cant add a query to the end of the link such as /test/searchoutput.php?search=green-widgets.

So Im a bit confused what the next step is. I have tried Googling but nothing suitable seems to come up.

I have tried "search database with link" etc

February 12, 2013

Thanks, how would I do this please?

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justlukeyou

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Posts posted by justlukeyou

Use Image As Button

Stopping Duplicate Inserts

Stopping Duplicate Inserts

Only Insert On Image Click

Only Insert On Image Click

Only Insert On Image Click

Block Unconfirmed Logins

Block Unconfirmed Logins

Block Unconfirmed Logins

Block Unconfirmed Logins

Block Unconfirmed Logins

Block Unconfirmed Logins

Scanning Table On Login Page

Scanning Table On Login Page

Scanning Table On Login Page

Scanning Table On Login Page

Scanning Table On Login Page

Scanning Table On Login Page

Finish Final Stage of Social Networking Button

Finish Final Stage of Social Networking Button

Finish Final Stage of Social Networking Button

KickStarter Project Feedback Please

Using A Link To Perform Search

Using A Link To Perform Search

Using A Link To Perform Search

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