IF statement in button or new php??

[jsk1gcc]

hi again, having trouble getting the result from the drop down box to direct to another page

 

the if statement works when i don't try and do anything with it.

 

should the IF statement go on another page and the submit button POST the $result, i'm really not sure. any help would be great..

 

code so far

 

index.php

<?php
$query = "SELECT * FROM `ride` ORDER BY name";  
$result = mysql_query($query);  

        echo"<select name='name'><option value=''>Select Ride</option>";
                while( $row = mysql_fetch_array($result) ) 
                { 
                echo '<option>'.$row['name'].'</option>'; 
                }  
                        echo '</select>'; 
        mysql_free_result( $result ); 

//// do rest of form when done/////

echo "<form method=post name=f1 action='dd-check.php'>";
echo "<input type=submit value=Submit>";
echo "</form>";

?>

 

dd-check.php

<?php
$result =$_POST['WHAT GOES HERE?']; <-----------???? is this even needed?
echo  "Ride Choosen = $result";     <-----------???? how do i get the $result from index.php?

if ($result == "Swinging Ship") {
                 header('Location: Swing.php');
                } elseif ($result == "Roller Coaster") {
                 header('Location: Roller.php');
                } elseif ($result == "Ice Blast") {
                 header('Location: Ice.php');
                } else {
        echo "choose a ride";
                }
?>

[kenrbnsn]

Your form is not correct, since the drop down list is outside the form definition.

Try something like this:

<?php
echo '<form method="post" name="f1" action="dd-check.php">';
$query = "SELECT * FROM `ride` ORDER BY name";  
$result = mysql_query($query);  

        echo "<select name='name'><option value=''>Select Ride</option>";
                while( $row = mysql_fetch_array($result) ) 
                { 
                echo '<option>'.$row['name'].'</option>'; 
                }  
                        echo '</select>'; 

//// do rest of form when done/////


echo "<input type='submit' value='Submit'>";
echo "</form>";
?>

 

In the processing script:

<?php
$result =$_POST['name']; 
echo  "Ride Choosen = $result";  
//
// etc...
//
?>

 

Ken

[jsk1gcc]

Thankyou, 

 

It's messed up how strict and loose php can be, order is everything in this language. Once I get used to it, i'll be fine untill then i'll keep asking for help and learning

 

Thanks again for a quick reply