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Problematic Isset($_post['submit'])...	View Content

I'm not sure why this code isn't working... I've done this many times before and this time it isn't working at all...

Here's the code I'm using:
<?php

if(isset($_POST['submit'])){
if(filter_var($_POST['value1'], FILTER_VALIDATE_EMAIL)){
echo "Value is a valid email address.";
} else {
echo "Value is NOT a valid email address.";
}
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>

<body>
<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="POST">
<input name="value1" type="text" id="value1" size="40" />
<input id="submit" name="submit" value="Validate Value" type="button" />
</form>
</body>
</html>

I'm wondering whether anything has changed, but I don't see how this couldn't work.

Thanks for any additional information!

Full Excel VBA Course - Beginner to Expert

Why (isset($_post['submit'])) Function Not Working?

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I have a simple log in form. By debugging I found that isset post submit function is not responding. It remains on the same page after submitting form. I have googled all the possible solutions to this problem but none worked. Any help is greatly appreciated. The form is in multiplelogin.php file and the code is in multi.php file. After login redirection is to teachers.php file.

<?php

session_start();

$conn = mysqli_connect("localhost", "root", "", "signup");

if(isset($_SESSION['username'])) {

header('location: teachers.php');

exit();

}

<!DOCTYPE html>

<head>

<title> Teachers Students Login </title>

</head>

<?php

if(isset($_GET['error'])) {

if ($_GET['error'] == 'invalidemail') {

echo "<p class='alert'> Invalid Email Id! </p>";

} else if ($_GET['error'] == 'wrongpassword') {

echo "<p class='alert'> Incorrect Password! </p>";

}

}

?>

</div>

</div>

</div>

</form>

</div>

</body>

</html>

<?php

ini_set( 'display_errors', 1 );
error_reporting( E_ALL );

if (isset($_POST['submit'])) {

$conn = mysqli_connect("localhost", "root", "", "signup");

$email = $_POST['email'];

$password = $_POST['password'];

$sql = "SELECT * FROM teachersRegist WHERE email=?";

$stmt = mysqli_stmt_init($conn);

if (!mysqli_stmt_prepare($stmt, $sql)) {

header("Location: multilogin.php?error=sqlerror");

exit();

} else {

mysqli_stmt_bind_param($stmt, "s", $email);

mysqli_stmt_execute($stmt);

$result = mysqli_stmt_get_result($stmt);

if ($row = mysqli_fetch_assoc($result)) {

$pwdCheck = password_verify($password, $row['password']);

if ($pwdCheck == false) {

header("Location: multilogin.php?error=wrongpassword");

exit();

} else if ($pwdCheck == true) {

session_start();

$_SESSION['username'] = $row['username'];

$_SESSION['email'] = $row['email'];

header("Location: teachers.php?login=success");

exit ();

} else {

header("Location: multilogin.php?error=wrongpassword");

exit();

}

}

else {

header("Location: multilogin.php?error=invalidemail");

exit();

}

}

}

else {

header("Location: multilogin.php");

exit();

}

?>

Edited September 9, 2020 by tarunrathore

Can Not Use Isset($_post)

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I want to link this two file but not working, first file to make drop down option form

Isset($_post)

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Running into some problems with my code. I am having the user search the Database with either a UAID or the user's full name. It works fine when you input both the UAID and name; however, I want it to work if only one is give (i.e. just the name or just the UAID).

I decided to use the isset($_POST) function of php but I am not getting it to work correctly. PHP believe it is set even though it is empty.

Here is my html form.

<?php
@include 'accesscontrol.php';

echo "<!DOCTYPE html PUBLIC \"-//W3C//DTD XHTML 1.0 Transitional//EN\"
      \"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd\">"
     ."<html xmlns=\"http://www.w3.org/1999/xhtml\">"
     ."<title> Hello </title>"
     ."<meta http-equiv=\"Content-Type\"
      content=\"text/html; charset=iso-8859-1\" />"
     ."<head>"
     ."<style type=\"text/css\">"
     ."h1{"
     ."border-bottom-style: double;"
     ."border-bottom-color: black;"
     ."border-bottom-width: 5px;"
     ."}"
     ."p{"
     ."background-color:#4682B4;"
     ."}"
     ."</style>"
     ."</head>"
     ."<body style=\"background-color:#B8B8B8\">"
     ."<table align=\"center\" style=\"background-color:#B8B8B8\" WIDTH=\"100%\" HEIGHT=\"100%\">"
     ."<tr valign=\"middle\">"
     ."<td align=\"center\">"
     ."<form method=\"post\" action=\"user.php\">"
     ."<table align=\"center\" cellpadding=\"10\" style=\"background-color:#5CADFF\" width=\"500\" height=\"400\">"
     ."<tr>"
     ."<th colspan=6 align=\"center\"><h1>Welcome: ".$uid."</h1></th>"
     ."</tr>"
     ."<tr>"
     ."<td colspan=3 width=40% align=center>"
     ."UAID:<br>  <input type=\"text\" size=\"8\" name=\"UAID\" /></td>"
     ."<td colspan=3 width=60% align=center>"
     ."Name (Last, First):<br><input type=\"text\" size=\"8\" name=\"last\"/>, <input type=\"text\" size=\"8\" name=\"first\" />"
     ."</tr>"
     ."<tr>"
     ."<td colspan=6 align=\"center\"><input type=\"submit\" value=\"submit\" />"
     ."</td>"
     ."</tr>"
     ."</tr>"
     ."</table>"
     ."<tr>"
     ."</tr>"
     ."</td>"
     ."</tr>"
     ."</table>"
     ."</body>"
     ."</html>";

?>

Here is the PHP:

<?php
@include 'accesscontrol.php';
//include 'common.php';
include 'db.php';

$fname = $_POST['first'];
$lname = $_POST['last'];

$sql = "SELECT * FROM `user` WHERE UAID='$_POST[UAID]'";
$result = mysql_query($sql);
$rows = mysql_fetch_array($result);

if(isset($fname) && isset($lname))
{

echo "Post first $fname <br>"
     ."Post last $lname";
$firstname = $rows['firstname'];
$lastname = $rows['lastname'];

if($fname != $firstname OR  $lname != $lastname)
{
?>
<html>
<head>
<script language="JavaScript">
    alert(" Hello this is crap" );
    history.back();
</script>";
</head>
</html>
<?
exit;

}

else
{
echo "<!DOCTYPE html PUBLIC \"-//W3C//DTD XHTML 1.0 Transitional//EN\"
      \"URL\">"
     ."<html xmlns=URL>"
     ."<title> Hi </title>"
     ."<meta http-equiv=\"Content-Type\"
      content=\"text/html; charset=iso-8859-1\" />"
     ."<head>"
     ."<style type=\"text/css\">"
     ."h1{"
     ."border-bottom-style: double;"
     ."border-bottom-color: black;"
     ."border-bottom-width: 5px;"
     ."}"
     ."p{"
     ."background-color:#4682B4;"
     ."}"
     ."</style>"
     ."</head>"
     ."<body style=\"background-color:#B8B8B8\">"
     ."<table align=\"center\" style=\"background-color:#B8B8B8\" WIDTH=\"100%\" HEIGHT=\"100%\">"
     ."<tr valign=\"middle\">"
     ."<td align=\"center\">"
     ."<table align=\"center\" cellpadding=\"10\" style=\"background-color:#5CADFF\" width=\"600\" height=\"400\">"
     ."<tr>"
     ."<th colspan=6 align=\"center\"><h1>Welcome: ".$uid."</h1></th>"
     ."</tr>"
     ."<tr>"
     ."<form method=\"Post\" action=\"userview.php\">"
     ."<td colspan=3 width=40%>"
     ."UAID:  <input type=\"text\" size=\"8\" name=\"UAID\" value= ".$_POST['UAID']." /></td>"
     ."<td colspan=3 width=60%>"
     ."Name (LAST, FIRST):<input type=\"text\" size=\"8\" name=\"last\"  value= ".$_POST['last']." />,
      <input type=\"text\" size=\"8\" name=\"first\" value= ".$_POST['first']." />"
     ."</td>"
     ."</tr>"
     ."<tr>"
     ."<td colspan=3 width=40%>"
     ."Department:<input type=\"text\" size=\"8\" name=\"department\" value=" .$rows['BOG'].">"
     ."</td>"
     ."<td colspan=3 width=60%>"
     ."Employee ID:<input type=\"text\" size=\"8\" name=\"empID\" value=" .$rows['employeeID'].">"
     ."</td>"
     ."</tr>"
     ."<tr>"
     ."<td colspan=3 align=\"center\"><input type=\"submit\" value=\"View\">"
     ."</form>"
     ."</td>"
     ."<form method=\"Post\" action=\"homepage.php\">"
     ."<td colspan=3 align=\"center\"><input type=\"submit\" value=\"Back\">"
     ."</td>"
     ."</tr>"
     ."</tr>"
     ."</table>"
     ."<tr></tr></td></tr></table></body></html>";

}}
mysql_free_result($result);
?>

Unset Isset $_post - Is It Possible?

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Is it possible to unset a submit button?

<?PHP

if(isset($_POST['Submit'])) {

$date = $_POST['date'];

if ($date = '1966-10-03'){

     //Do Something

} else {

     //Possible to unset?
    unset($date);
    unset($_POST['Submit']);

}

} else { ?>
<form action="" method="post">
<input type="text" name="date" />
<input type="Submit" name="Submit" value="Submit" />
</form>

<?PHP
}
?>

What I'm wondering is if it's possible to unset after submitting in the event of condition x. Should this not in turn echo out the form as it has been unset?

If Isset Argument From Submit Form Button...

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Hi,

I have a form on my page.

<form id="advanced_search" name="advanced_search" method="POST" action="<?php echo $PHP_SELF;?>">

</form>

Form submits with a regular button
<input type="submit" name="advSearch" id="advSearch" value="Search" />

And i have a detector to see where submit button has been clicked on my next page:
if(isset($_POST['advSearch'])) {
echo "Button was clicked";
}

However, I wanted a custom button for better styling and control.
So i replaced my regular button with this now which uses javascript to submit the form.

<a class="button" href="javascript:document.advanced_search.submit()" name="advSearch" id="advSearch"><span>Search</span></a>

Now the isset($_POST) is not working anymore.
Can anyone help please??

Thanks!

Problematic Headers

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OK you fantastic PHPfreakers I'm not sure how to describe this so I hope you will understand.. I have a predicament where I set the header info to: header('Content-Type: text/html; charset=utf-8'); but later on I have a part of my site that requests an image from somewhere else using getImage() here I need to set header("Content-type: image/jpeg"); in order for the image to show correctly. but this conflicts with the previous page of headers already sent. so I was looking for a way around this OR maybe can I load them in after the page has loaded? I have tried using Jquery to post with ajax but I don't seem to get any response the alert does not trigger for any conversation with an image. I have tested all of this and it works as long as I don't put any OUTPUT before triggering the getImage()

<script type='text/javascript'>
	$(document).ready(function(){
		function loadImages(){										
			$.ajax({
			type: "POST",
			url: "/admin/getImage.php",
			data: {token: '<?php echo $token; ?>', chatId: '<?php echo $chat; ?>', uId: '<?php echo $uId; ?>'},
			success: function(data){		
				alert(data);
				$('#img').html(data);
			}
		});
		return false;
		}
	});
</script>

here is part of the messages page that would show the image. (just imagine there was more content etc..)

<?php 
	header('Content-Type: text/html; charset=utf-8'); 
	## OTHER GENERIC STUFF		
	$parser = new Parser($v1, $v2);			
	$convo = $parser->getConvo($token, $id);

	foreach($convo['messageGroups'] as $con)
	{	
		$chat = $con['messageGroupId'];		
			
		$uId = $con['latestMessage']['messageUid'];
		$key = $con['latestMessage']['messageKind'];						
		if($key == '3'){
										
			$image = $parser->getImage($token, $chat, $uId);
			echo $image;
		}
		
	}
?>

here is the function for getImages()

public function getImage($token, $chat, $uId){
	$this->initCURL();
	$chatter = str_replace("{{chat}}", $chat, $this->url['image']);
	$chatter = str_replace("{{uId}}", $uId, $chatter);
			
	$headerData = $this->defaultGETHeader($token);
			
	curl_setopt($this->curl, CURLOPT_URL, $chatter);
	curl_setopt($this->curl, CURLOPT_HTTPHEADER, $headerData);
			
	$output = curl_exec($this->curl);
	$this->closeCURL();
			
	header("Content-type: image/jpeg");
			
	return $output;
}

Edited by xXREDXIIIXx, 01 June 2014 - 12:18 PM.

Is There A Way To Submit Same Entry Multiple Times On A Single Form Submit?

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Say I have an "Entries" table. I want to submit same multiple entries using a form submission. And If I have other queries submitted in the same form, I want those quarries to be submitted only once. Is that possible to do?

Here's my code.

if(isset($_POST['submit'])) {

  $entries = 10;

  $id   = 55;
  $name = 'Smith';

  $insert = $db->prepare("INSERT INTO entries(id, name) VALUES(:id, :name)");
  $insert->bindParam(':id', $id);
  $insert->bindParam(':name', $name);
  $result_insert = $insert->execute();
  if($result_insert == false) {
    echo 'Fail';
  } else {
    echo 'Success';
  }

}
?>
<form action="" method="post">
  <input type="submit" name="submit" value="SUBMIT" />
</form>

Edited January 13, 2019 by imgrooot

Isset

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I believe that my problem is with the isset on line 7 and 39, but I do not know of anyway around it. Can someone enlighten me.

<?php
session_start();
if (isset($_GET['logout'])){
    unset($_SESSION['logged_in']);
}
?>
<html>
    <head>
        <title>Add a New Car!</title>
        <?php
        if (isset($_POST['submit'])) {
            //print_r($_POST);
            $password = md5($_POST['password']);
            //echo'<p>'.$password.'</p>';
            $username = ($_POST['username']);
            //echo'<p>'.$username.'</p>';
        }
        $db = new PDO('sqlite:login.sqlite');
        $login = 'SELECT * FROM login
        WHERE userid="'.$username.'"
        AND password="'.$password.'";';
        //echo'<p>'.$login.'</p>';
        $attempt = $db->query($login);
        foreach ($attempt as $row) {
        //print_r($row);
        $_SESSION['logged_in'] = $username;
        }
        ?>
    </head>
    <body>
                <?php
        if (!isset($_SESSION['logged_in'])){
        ?>
            <form method="post" action="Db_login.php">
            <p>Username:<br /><input type="text" name="username" /> </p>
            <p>Password:<br /><input type="password" name="password" /></p>
            <p><input type="submit" name="submit" value="Log In!" /></p>
            </form>
        <?php
        } else {
        ?>
        <?php
        if(isset($_POST['submit'])) {
            print_r($_POST);
            $make = ($_POST['make']);
            $model = ($_POST['model']);
            $year = ($_POST['year']);
            $fuel = ($_POST['fuel_type']);
            $vin = ($_POST['vin_number']);
            $reg = ($_POST['reg_number']);
        }
        $db = new PDO('sqlite:cars.sqlite');
        $sqlCars = 'INSERT INTO cars ( )
        VALUES("'.$make.'","'.$model.'","'.$year.'","'.$fuel.'","'.$vin.'","'.$reg.'");';
        echo'<p>'.$sqlCars.'</p.>';
        ?>
        <form method="post" action="Db_new_Cars.php" enctype="multipart/form-data">
            <p>Make:<br /><input type="text" name="make" /> </p>
            <p>Model:<br /><input type="text" name="model" /></p>
            <p>Year:<br /><input type="text" name="year" /></p>
            <p>Fuel Type:<br /><input type="text" name="fuel_type" /></p>
            <p>Vin Number:<br /><input type="text" name="vin_number" /></p>
            <p>Reg Number:<br /><input type="text" name="reg_number" /></p>
            <p>Choose a image to upload:<br />
            <input type="file" name="myfile"></input></p>
            <p><input type="submit" name="submit" value="Submit!" /></p>
        </form>
        <h1> Protected Content!</h1>
        <p>You are logged in as <em><?php echo $_SESSION['logged_in']; ?> </em></p>
        <p><a href="Db_login.php?logout=true">Log out!</a></p>
        <p><a href="New_Cars.php">Add a new car</a></p>
        <p><a href="Book_Out_Cars.php">Book out a car</a></p>
        <?php
        }
        ?>

    </body>
</html>

Need Isset()?

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Isn't this code redundant...

if (isset($_SESSION['loggedIn']) && $_SESSION['loggedIn'] == TRUE){

After all, if $SESSION['logged_in'] == TRUE then it is *obviously* set...

Debbie

Isset()

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I want to get the COUNT of the fields (ArrayM11 - ArrayM19), which has the word "YES". The code isn't working. Appreciate any help.

$Array1 = array();
for($j=0;$j<count($ArrayM1);++$j) {
$result = array_count_values($ArrayM1[$j]);
if (isset($result['YES'])) {
$points1 = $result['YES'];
}
}

Need Help With Isset()

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guys this thing called isset() is bugging me.What is the difference in when you say
Code: [Select]
if(isset($_POST['name']))
and
Code: [Select]
if($_POST['name'])
and
Code: [Select]
if(!$_POST['name'])

and why would someone require a code such as Code: [Select]
if($_POST['submit']=='Submit')
And what is that code which tells you the user has entered something into the form
Thanks People

Isset Help

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why dose this not work

<?php

if(isset($_SESSION['SESS_ADMIN']))
$loc = ../admin/admin.php;
else
$loc= ../head/head.php";

?>
<form><INPUT TYPE="button" value="Main Menu" onClick="parent.location='<?php echo $loc; ?>'"></form>
</body>

</HTML>

Isset

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i Have two page login.php and verify.php

In login.php user is entering Username and Password and in the action of the form i am writing verify.php.

If verification is a success then i will start a session()

Code: [Select]
session_start();

$_SESSION['ls']="loggedin"; // ls stands for login status
$_SESSION['id']=$row['id'];
$_SESSION['user_name']=$row['user_name'];
$_SESSION['email']=$row['email'];
header('location : members.php');
die (" ");
and will redirect them to members.php
....

Everything is working fine...

But in login.php i have written a code on the top...

Code: [Select]
<?php
if(isset($_SESSION['ls']) && isset($_SESSION['id']) && isset($_SESSION['user_name']) && isset($_SESSION['email']) )
{
header('location:members.php');
die(" ");

}
?>
I want if a session is started and if user will access login.php they will automatically go to members.php

But this script is not working...

Iam not getting whats the problem...
also i m not getting any error login.php or anything

CODE WRITTEN ON THE TOP OF members.php IS :

Code: [Select]
<?php
session_start();
if(@$_SESSION['ls']!=="loggedin" ) // Checks if the person has done stage one (Registration.php)
{
header('Location: login.php'); //If they haven't done it, it sends them back.
die (" ");
}
?>

Using If Isset More Than Once

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Hi all

I need to do a multiple if isset to get three variables from the URL. Here is my code for just one:

if(isset($_GET['category'])) {
$fetchproducts = mysql_query(" SELECT * FROM `products` WHERE category = '".$_GET['category']."'");
while($returnedProduct = mysql_fetch_array($fetchproducts)) {
include('product-cell.php');
}

How do I combine this to check what is the 'type' and 'colour' from the URL and fetch from the SQL accordingly?

Many thanks for you help

Pete

Php 404 Isset Help

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Hello all, i am hoping someone can help me.

i have some code i have been working on and i am trying to add a 404 page if it cant find the page, i know it will be some sort of if statment, but with me been pretty much a noob at php i am stuck.

heres the code that the page calls.

Code: [Select]
function get_content() {

//Get page name from browser if none call it home
$page = (isset($_GET['page'])) ? $_GET['page'] : 'home';

//Get The Page Content
$sql = "SELECT * FROM content WHERE page = '$page'";
$res = mysql_query($sql) or die (mysql_error());
while($row = mysql_fetch_assoc($res)){

//Show The Page Content
echo '<h1>' . $row['title'] . '</h1>';
echo $row['body'];
}
}

thanks all.

Isset With || And &&

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This is blowing my mind. I have this function and all I want it to do is check to make sure that the var $race is set and = '0'. The default value from the form is '0'. I also want it to check the value of race and if it is anything but the two listed , submit error. Yes I've tried to replace the || with &&.. Nothing works(Meaning that it will error when its not suppose to or it wont error when its suppose to).. Any ideas?

Code: [Select]
function check_Race ($race,$display_name,$member_id,$description,$ip) {

//echo "This is the race". $race;
//exit;

if (isset($race) && $race ='0' && $race !== 'White' && $race !== 'Black'){

echo "This is an errror.!";
$t_error="19";
$member_id = $member_id;
notify_Admin($t_error,$member_id,$ip);
// logOut ($t_error);
exit;

}

Isset Help?

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Hi there,

I'm inserting data into a mysql database. My submissions works fine, however when a part of the form is not filled out the php scripit crashes. How can I submit all data that is set and skip pieces that aren't filled out?

For example I have a plant database which has an option to insert different types of common names for the plant. It's ok if they aren't all filled out.

My code is as follows: Code: [Select]
<?php

//insert general plant data into db
$queryplant="INSERT INTO actualplanttable (id, scientific, common1, common2, common3, family, genus, description, planting, growing, care, duration, reference)
VALUES ('NULL', '$_POST[scientific]', '$_POST[common1]', '$_POST[common2]', '$_POST[common3]', '$_POST[family]', '$_POST[genus]', '$_POST[description]', '$_POST[planting]', '$_POST[growing]', '$_POST[care]','$_POST[duration]','$_POST[reference]')";
mysql_query($queryplant)
or die ('Error updating actualplanttable');

//insert maintenance data into db
$querymaintenance="INSERT INTO plantmaintenancelinktable (plant_id, maintenance_id)
VALUES ('$inserted_id', '$_POST[maintenance]')";
mysql_query($querymaintenance)
or die ('Error updating plantdurationtable');

?>

Should I be trying to work an isset function into this code? Any help would be greatly appreciated.

Thanks!

Bill...

If [form Validated] Submit Form1 Else Submit Form2?

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Hi. Pretty straight forward I guess but as the name suggests am a newbie. I have a form that requires the user to enter certain parameters. If the values are blank it submits to itself and loads the error messages. What I want to do is create PHP code that submits the form to a different url. What I thought was create two forms (the second with hidden fields replicating the first form), each form having a different url in the action"" code. What I cant work out is the PHP IF ELSE code to submit form 2 if Form1 is is validated correctly. This is the PHP code relevant to the form validation. Help?

<?php
      //If form was submitted
      if ($_POST['submitted']==1) {
        $errormsg = ""; //Initialize errors
        if ($_POST[width]){
           $title = $_POST[width]; //If title was entered
        }
        else{
           $errormsg = "Please enter width";
        }
        if ($_POST[drop]){
           $textentry = $_POST[drop]; //If comment was entered
        }
        else{
           if ($errormsg){ //If there is already an error, add next error
              $errormsg = $errormsg . " & content";
           }else{
              $errormsg = "Please enter drop";
           }
        }
      }

      if ($errormsg){ //If any errors display them
      echo "<div class=\"box red\">$errormsg</div>";
      }

      //If all fields present
      if ($title && $textentry){
       //Do something

echo 'THIS IS WHERE I WANT THE CODE TO SUBMIT FORM 2 or SUBMIT FORM 1 TO A DIFFERENT URL';
      }
      ?>

Isset Still Not Working

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Code: [Select]
if (isset($_COOKIE['hide_div']['0']) && isset($_COOKIE['hide_div']['2'])){
$cookie1 = $_COOKIE['hide_div']['0'];
$cookie2 = $_COOKIE['hide_div']['2'];
}
if ($cookie1 == $cat_id OR $cookie2 == $cat_id) {
foreach ($this->forums as $forum_id => $forum_data)
{
if ($forum_data['category'] == $cat_id)
{
//-----------------------------------
// We store the HTML in a temp var so
// we can make sure we have cats for
// this forum, or hidden forums with a
// cat will show the cat strip - we don't
// want that, no - we don't.
//-----------------------------------

$temp_html .= $this->process_forum($forum_id, $forum_data);
}
}

if ($temp_html != "")
{
$this->output .= $this->html->CatHeader_Expandedhidden($cat_data);
$this->output .= $this->html->end_this_cat();
}

unset($temp_html);
}else{
$cookie1 = '';
$cookie2 = '';
why in the world is it displaying Undefined variable: cookie1 and cookie2 as errors?

When I declared both of them even AFTER my if statement,

this is bogus dude, lol

this isset stuff is really confusing or am i just not doing it right?

How Do I Isset All My Variables?

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In my index.php I use a global $_GET variable he

Code: [Select]
$ibforums->input = $std->parse_incoming();
parse_incoming function is:

Code: [Select]
function parse_incoming()
{
global $HTTP_GET_VARS, $HTTP_POST_VARS, $HTTP_CLIENT_IP, $REQUEST_METHOD, $REMOTE_ADDR, $HTTP_PROXY_USER, $HTTP_X_FORWARDED_FOR;
$return = array();

if( is_array($HTTP_GET_VARS) )
{
while( list($k, $v) = each($HTTP_GET_VARS) )
{
//-- mod_sec_update_131 begin
if (strpos($k, "amp;") === 0)
$k = substr($k, 4);
//-- mod_sec_update_131 end

if ( $k == 'INFO' )
{
continue;
}

if( is_array($HTTP_GET_VARS[$k]) )
{
while( list($k2, $v2) = each($HTTP_GET_VARS[$k]) )
{
$return[$k][ $this->clean_key($k2) ] = $this->clean_value($v2);
}
}
else
{
$return[$k] = $this->clean_value($v);
}
}
}

// Overwrite GET data with post data

if( is_array($HTTP_POST_VARS) )
{
while( list($k, $v) = each($HTTP_POST_VARS) )
{
if ( is_array($HTTP_POST_VARS[$k]) )
{
while( list($k2, $v2) = each($HTTP_POST_VARS[$k]) )
{
$return[$k][ $this->clean_key($k2) ] = $this->clean_value($v2);
}
}
else
{
$return[$k] = $this->clean_value($v);
}
}
}

//----------------------------------------
// Sort out the accessing IP
// (Thanks to Cosmos and schickb)
//----------------------------------------

$addrs = array();

foreach( array_reverse( explode( ',', $HTTP_X_FORWARDED_FOR ) ) as $x_f )
{
$x_f = trim($x_f);

if ( preg_match( '/^\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}$/', $x_f ) )
{
$addrs[] = $x_f;
}
}

$addrs[] = $_SERVER['REMOTE_ADDR'];
$addrs[] = $HTTP_PROXY_USER;
$addrs[] = $REMOTE_ADDR;

//header("Content-type: text/plain"); print_r($addrs); print $_SERVER['HTTP_X_FORWARDED_FOR']; exit();

$return['IP_ADDRESS'] = $this->select_var( $addrs );

// Make sure we take a valid IP address

$return['IP_ADDRESS'] = preg_replace( "/^([0-9]{1,3})\.([0-9]{1,3})\.([0-9]{1,3})\.([0-9]{1,3})/", "\\1.\\2.\\3.\\4", $return['IP_ADDRESS'] );

$return['request_method'] = ( $_SERVER['REQUEST_METHOD'] != "" ) ? strtolower($_SERVER['REQUEST_METHOD']) : strtolower($REQUEST_METHOD);

return $return;
}
Now, I use my $ibforums->input throughout my forum system, and I use it in EVERY SINGLE file, I have around 50 Files and I get "unidetified variable" errors.

is there a way I can just make sure my global $ibforums->input is ISSET so it it's a GLOBAL ISSET? So I don't have to go through each file that uses $ibforums->input and put a isset( ) around it? It would take over prob 10 days to go through each file and change all 300-500 errors, using the $ibforums->input, there has to be a way I can globally defined my $ibforums->input Variable with a isset..

Thank you