[Resolved] How do I redirect after submit?

[AndrewWest]

This is a form of error handling. I can't seem to get it to work.

<?php
   $path =  $_SERVER['PHP_SELF'];
   $path_parts = pathinfo($path);
   $string = $path_parts["basename"];
   $exclude = ".jpg.php";

   $path = substr("$string",0,strpos($string,$exclude));
   echo "Image: $path ";

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

$ip = $HTTP_SERVER_VARS["REMOTE_ADDR"];

$query = "SELECT last_ip FROM photographs WHERE name=$path";

if ($query = $ip) {
header("Location: error.php");
}


$link = mysql_connect('localhost', 'andrew', '1357908642');
if (!$link) {
   die('Not connected : ' . mysql_error());
}

$db_selected = mysql_select_db('andrew_site', $link);

$message = stripslashes(trim($message));
$message = nl2br($message);
$message = htmlentities($message);

$query = "INSERT INTO photos (ip, comment, photo)
VALUES ('$ip', '$message', '$path')";
mysql_query($query);

mysql_close();

echo "Thanks $name your comment has been added";
}

?>

You can see the part that is wrong

if ($query = $ip) {
header("Location: error.php");
}

Headers can't be sent after submit has been made.. How can I redirect if it's wrong?

[zapa]

change = to ==

if ($query == $ip) {
header("Location: error.php");
}

[AndrewWest]

No errors, but it still won't send it to error.php

[zapa]

it works if you had teh sql right
get your Select right

it should be

$query = "SELECT last_ip FROM photographs WHERE name='".$path."'";

insted of
$query = "SELECT last_ip FROM photographs WHERE name=$path";

[AndrewWest]

If that's incorrect, I'm curious why it works every where else on my site?

I'll try it though.

[zapa]

you can test it out by this..

<?PHP
$ip = $HTTP_SERVER_VARS["REMOTE_ADDR"];

$query = $HTTP_SERVER_VARS["REMOTE_ADDR"];

if ($query = $ip) {
header("Location: error.php");
}

?>

if you fix the Sql part of you script it will work

[tommynanda]

From your code

<?php 
   $path =  $_SERVER['PHP_SELF']; 
   $path_parts = pathinfo($path); 
   $string = $path_parts["basename"]; 
   $exclude = ".jpg.php"; 

   $path = substr("$string",0,strpos($string,$exclude)); 
   echo "Image: $path "; 

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

$ip = $HTTP_SERVER_VARS["REMOTE_ADDR"]; 

$query = "SELECT last_ip FROM photographs WHERE name=$path"; 

if ($query = $ip) { 
header("Location: error.php"); 
} 

I couldn't see a connection to MySQL DB before SELECT Statment and also, you didnt execute $query

change that part to...

<?php 
   $path =  $_SERVER['PHP_SELF']; 
   $path_parts = pathinfo($path); 
   $string = $path_parts["basename"]; 
   $exclude = ".jpg.php"; 

   $path = substr("$string",0,strpos($string,$exclude)); 
   echo "Image: $path "; 

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

$ip = $HTTP_SERVER_VARS["REMOTE_ADDR"]; 
$link = mysql_connect('localhost', 'andrew', '1357908642');
$db_selected = mysql_select_db('andrew_site', $link); 
$query = "SELECT last_ip FROM photographs WHERE name='$path'"; 
$result = mysql_query($query);
$extip = mysql_result($result,0);
if ($extip == $ip) { 
header("Location: error.php"); 
} 
// And attach rest of your code

You were comparing IP address with some string "SELECT..." which is not true any time... you should execute query to execute query u should connect to DB...

Good Luck
TommYNandA
200

[zapa]

$query = "SELECT last_ip FROM photographs WHERE name=$path";

$query = "SELECT last_ip FROM photographs WHERE name=$path";

should be:
name='$path'

[AndrewWest]

Zapa, you're suggesting syntax problems. The syntax of the coding is correct, as it works other places.

Tommynanda, When I replace my code with your code i get

"Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/andrew/public_html/port/modules/fullsize/90242.jpg.php on line 83"

Line 81 $query = "SELECT last_ip FROM photographs WHERE name=$path";

Line 82$result = mysql_query($query);

Line 83 $extip = mysql_result($result2,0);

[AndrewWest]

tommy,

Your code just had a small issue with it.. you called result2 from somewhere when the only result was just result...

Now.. I'm back to the original error.. headers cannot be sent.. already output

There's got to be a different way than headers.. is there?

[tommynanda]

yeah typo... I changed that to $result... (edited 😉 )
If it says headers already sent... you are echoing some data to browser or you have a white space before <?... as I can see you are echoing $image path before header() function you cant do that...

Anyway if you are done with header() then Javascript is alternative

if ($extip == $ip) {

?>
<script language="Javascript">
location.href="error.php"
</script>
<?

}

This works exactly the way u want... but no headaches like headers already sent 😃

Happy Coding
TommYNandA

NOTE: note my editing again 😉

[AndrewWest]

Thanks! Javascript works fine. I also take that it's univcersal (looks simple enough) - I would remove the header, but I want everything to be uniqely generated as I have it so I will use the javascript. 🙂

[tommynanda]

Just comment the echo line before header() and header() works fine too
// echo "Image: $path ";

this makes header() work you can echo this after header() function....
Anyways Javascript does the job w/o this... Mark This thread resolved if you are done 😉 Its at the bottom of this page

Good Luck
TommYNandA

[AndrewWest]

Ok - Everything is working fine, but for my knowledge. I can // an echo.. and it will echo it later?

[tommynanda]

I mean... use
echo "Image: $path ";
after
header("Location: error.php");

something like this

<?php  

   $path =  $_SERVER['PHP_SELF'];  

   $path_parts = pathinfo($path);  

   $string = $path_parts["basename"];  

   $exclude = ".jpg.php";  

   $path = substr("$string",0,strpos($string,$exclude));  


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

$ip = $HTTP_SERVER_VARS["REMOTE_ADDR"];  

$link = mysql_connect('localhost', 'andrew', '1357908642'); 
$db_selected = mysql_select_db('andrew_site', $link);  

$query = "SELECT last_ip FROM photographs WHERE name='$path'";  

$result = mysql_query($query); 
$extip = mysql_result($result,0); 
if ($extip == $ip) {  

header("Location: error.php");  

}
echo "Image: $path "; 
// Rest of your code here

This should do your Job w/o Javascript 😉

Have Fun!
TommYNandA

[AndrewWest]

You know what the overall problem was? Whitespace kills!

[tommynanda]

I Told yaa 😃