Passing variable from file to file via select form

Question

Variables are set and everything is displayed correctly in start.php.I can choose between both displayed variables in the form.

Problem is: The variable $projektcheck or $projektcheckk are not passed to nxtfile.php when selected and the form is submitted. When I check it with the print($_POST['prjkttype']); the variable is undefined and nothing gets printed. It instanly switches to my echo"Data is wrong";

Why does it not get passed to the other file although it is set and selected?

start.php:

<?PHP 
    session_start();
    $projektcheck = $_SESSION['projektname'];
    $projektcheckk = $_SESSION['projektnamee'];                
?>
 [other codestuff]
<script>
function validateForm() {
    var fields = ['prjkttype']

    var i, l = fields.length;
    var fieldname;
    for (i = 0; i < l; i++) {
        fieldname = fields[i];
    }   
    f1.method="POST";
    f1.submit();
}
</script>


[other codestuff]

<form action="nxtfile.php" method="POST" name="f1">
    <select class="selectstyle" name="prjkttype" form="submit">
<?PHP 
    if (isset($projektcheck)) {
        echo"<option value='1'>$projektcheck </option>";    
    } else {
    }  

    if (isset($projektcheckk)) {
        echo"<option value='2'>$projektcheckk </option>";   
    } else {
    }  
?>
    </select>

    <input type=button name=s1 value="Insert" onClick="validateForm()">

</form>
[other codestuff]

nxtfile.php:

<?PHP
session_start();
include ("../connection.php");
$prjkttype=$_POST['prjkttype'];


print($_POST['prjkttype']); 

$checkindb="SELECT * FROM projekte WHERE (projektname = '$prjkttype' )";
$dbcheckk=mysqli_query($con ,$checkindb);
$dbcheckedd=mysqli_fetch_array($dbcheckk);

if($dbcheckedd > 0){
    header ("Location: xy.php");
} else {
    echo "Data is wrong.";
}
?>

Some sensible code indentation would be a good idea. It helps us read the code and more importantly it will help **you debug your code** [Take a quick look at a coding standard](http://www.php-fig.org/psr/psr-2/) for your own benefit. You may be asked to amend this code in a few weeks/months and you will thank me in the end. — RiggsFolly, Aug 23 '17 at 14:03
_small note:_ If you dont have anything to do in an `ELSE` then dont code an `ELSE` — RiggsFolly, Aug 23 '17 at 14:05
Your script is at risk of [SQL Injection Attack](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) Even [if you are escaping inputs, its not safe!](http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string) Use [prepared parameterized statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) — RiggsFolly, Aug 23 '17 at 14:06
Its a good idea to code an `exit;` after a `header('Location .......');` as the `header` command does not terminate the script when executed, but all output following it will of course get lost — RiggsFolly, Aug 23 '17 at 14:07
thanks a lot for these helpful informations! I will take a look at it! — CodeNewb, Aug 23 '17 at 14:10
Have you checked the browsers debugger. I am pretty sure that javascript wont do what you think it will — RiggsFolly, Aug 23 '17 at 14:13
Be careful. javascrit errors can be a little Shy unless you know what you are looking for — RiggsFolly, Aug 23 '17 at 14:19
@RiggsFolly even if I delete everything of the script except the `f1.submit();` it still doesn't work. It's kind of mysterious. — CodeNewb, Aug 23 '17 at 14:23

score 1 · Accepted Answer · answered Aug 23 '17 at 14:38

You need to set the values to your variables like below. It will pass the value not the variable. This will work:

 if (isset($projektcheck)) {
        echo"<option value='$projektcheck'>$projektcheck </option>";    
    } 

    if (isset($projektcheckk)) {
        echo"<option value='$projektcheckk'>$projektcheckk </option>";   
    }

Passing variable from file to file via select form

1 Answers1