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Okay, first things first. I have never done anything to URL's when coding n PHP, but...

 

I want to make a redirect page, as in, if the user has met a certain condition, they are redirected to a different page, and they are given a specific error message depending on what the url is.

 

(sorry if I have not explained it well.)

 

But I do not know how to go about doing this.

 

Any help at all please?

 

Thanks

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From what you say, I guess you will be sending them to a page, lets say error.php that you made, that will then redirect them to your index.php page.

 

Now on the index page you can use $_SERVER['HTTP_REFERER'] to find out where they just came from, if is from page error.php , then display a message.

http://php.net/manual/en/reserved.variables.server.php

 

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But instead of looking for where everyone came from all the time, it may be better to just delay the redirect, show the message and send them back to index.php

 

<?php
  header( "refresh:5;url=index.php" );
  echo "You'll be redirected in about 5 secs. If not, click <a href='index.php'>here</a>.";
?>

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From what you say, I guess you will be sending them to a page, lets say error.php that you made, that will then redirect them to your index.php page.

 

Now on the index page you can use $_SERVER['HTTP_REFERER'] to find out where they just came from, if is from page error.php , then display a message.

http://php.net/manual/en/reserved.variables.server.php

 

 

I want't going to have an error page. it is for a login script, where, if the cookie id does not match the id in the database, then they are logged out. Here is a bit of pseudo code to show what i mean.

 

//page.php

if ($cookieid != database['cookie'])
{
header(Redirect: index.php/98eyux9dsn29dn;
}

//index.php

while (redirect: 98eyux9dsn29dn)
{
echo "You have been logged out because your cookie has been compromised";
}

 

Something along the lines of that

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if that's the case, why not just this

 

if($logged_in){
if ($cookieid != database['cookie']) {
session_destroy();
header( "refresh:5;url=index.php" );
echo "You have been logged out because your cookie has been compromised";
}
}

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Sorry if this is off topic, but I want to disable php errors (or not display them) for a specific page on my website, is this possible? I am asking because if the cookie has been modified, then lots of undefined constants and variable errors appear. While it does not alter the way the script works, I just don't like the way it looks (I'm a bit OCD in that sense)

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Then you should check if those variables exist before using them.

 

You should use isset

 

I was getting the errors because I was using the cookie to get the username from the correct row in the database, so I could display it on screen. And if the cookie has been modified, then it has been destroyed, so the variable would not have been defined (if there was originally no cookie, then i do not get these errors, as the variable is only defined if the cookie is present)

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But you must understand, that hiding errors, also untreated variables is very very bad !

 

I know. I have only disabled errors for the following code

if($_COOKIE)
{
error_reporting(0);
if (isset($_COOKIE['Key_my_site']) == $cookiedbid)
		{
			echo "Your cookie is okay.";
		}

		elseif (isset($_COOKIE['Key_my_site']) !== $cookiedbid)
		{
			header( "refresh:5;url=index.php" );
			echo "You have been logged out because your cookie has been compromised";
			setcookie(Key_my_site, 0, $past);
		}

		else 
		{
			echo "Your cookie is no longer there.";
		}
	error_reporting(1);
}

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(if there was originally no cookie, then i do not get these errors, as the variable is only defined if the cookie is present)

For this, you must do this

if(!isset($_COOKIE['name']))
    set_cookie('name', 'value');
else
    // Do something with existing COOKIE

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melloorr

Your code is incorrect.

 

Below, my code, it is correct

error_reporting(E_ALL);
if(isset($_COOKIE))
{
    if (isset($_COOKIE['Key_my_site']) && $_COOKIE['Key_my_site'] == $cookiedbid)
        echo "Your cookie is okay.";

    elseif (isset($_COOKIE['Key_my_site']) && $_COOKIE['Key_my_site'] !== $cookiedbid)
    {
        header("refresh:5;url=index.php" );
        echo "You have been logged out because your cookie has been compromised";
        setcookie(Key_my_site, 0, $past);
    }

    else 
        echo "Your cookie is no longer there.";
}

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(if there was originally no cookie, then i do not get these errors, as the variable is only defined if the cookie is present)

For this, you must do this

if(!isset($_COOKIE['name']))
    set_cookie('name', 'value');
else
    // Do something with existing COOKIE

 

I do not understand this. That code is saying, if there is no cookie, then set one?

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Yes.

 

The code was for example

 

I know but the variable is set in a different file, and the whole point of this is to check if the if in the cookie is the same as the one in the database, and if it is not, then the cookie is deleted and they are logged out due to the cookie being deleted. All the code works as it is, they are logged out if the cookie is not correct, and they will have to log in again. It is all secure (i think). I may upload it to be beta tested

 

But thanks for all your help :)

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