I would need help from admin regarding the approval of floatval() for Conjoint

Question

Warnung: Die Funktion floatval() ist nicht zur Verwendung im Fragebogen freigegeben.
Warnung: PHP-Code auf Seite 2 enthält Funktionen oder PHP-Elemente, die nicht für die Verwendung im Fragebogen freigegeben wurden. Falls Sie der Meinung sind, dass die monierten Konstrukte ungefährlich sind, setzen Sie sich bitte mit dem Administrator in Kontakt!

I would need help with that for my conjoint analysis

Answer 1

Instead of the function $a = floatval($b) please simply use the cast $a = (float)$b.

Comments

Thanks for your quick reply. I can't find the function in the code, but could be my fault. Is there the option that an Admin approves the floatval() function?

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>  I can't find the function in the code

The floatval() is very likely there, on page 2, if the error message says it is.

> Is there the option that an Admin approves the floatval() function?

I do not really see the point, as this can easily be replaced?! But, well, it it makes life easier for people... done.

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Thanks for your answer!
Now another problem came up in the same code.
Unsupported operand types: string + int
in line 10: 010 $task = $task + 1;

How to proceed here?

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It seems that $task is a string variable. Please place a

debug($task);

before that line to see more details. Also feel free to provide more of your code, so I get an idea about the context. The solution *may* be

$task = (float)$task + 1;

---

///////////////////////////////////////////////////////
//$task = 3;

option('progress', 'yes');
$task = value('CJ01_02');
$task = (float)$task + 1;
put('CJ01_02', $task);

$total_attributes = 9;

$val = value('CJ01_01');

if ($val){

    $returnarray = json_decode($val);

    $table = '';
    $header = '<tr><th></th><th> Projekt E</th><th> Projekt F</th></tr>';
    for ($i=1; $i<=$total_attributes ; $i++){
        $table .= '<tr><th>'.
        $returnarray['F-'.$task.'-'. $i].
        '</th><td>'.
        $returnarray['F-'.$task.'-1-'. $i].
        '</td><td>'.
        $returnarray['F-'.$task.'-2-'. $i].
        '</td></tr>';
    }
    html('<table border="4" cellpadding="10" cellspacing="0" >'.$header. $table . '</table>');

} else {
        goToPage('fail');
}


html("</br>");

That's how the code looks like. I added your solution here already.
Also new error occurs:
Fehler im Fragebogen: Undefined array key "F-7-1"
Zeile: 25

PHP-Code

022     $header = '<tr><th></th><th> Projekt E</th><th> Projekt F</th></tr>';
023     for ($i=1; $i<=$total_attributes ; $i++){
024         $table .= '<tr><th>'.
025         $returnarray['F-'.$task.'-'. $i].
026         '</th><td>'.
027         $returnarray['F-'.$task.'-1-'. $i].
028         '</td><td>'.

---

> Undefined array key "F-7-1"

Well, it says that the variable $returnarray does not have an entry with the key "F-7-1". And it seems that your code requests that key in this line:

$returnarray['F-'.$task.'-'. $i].

You can take a look at the available keys, adding another debug()

$returnarray = json_decode($val);
debug($returnarray);