Can some one help me with this code with a error on scalar value as an array?
$getal = $_POST['getal'];
For($teller=1; $teller<=11; $teller=$teller+1)
{
$uitkomst[$teller]=$teller*$getal;
}
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Shikiryu
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user2991218
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1What is `$uitkomst` variable ? – OlivierH Nov 14 '13 at 08:46
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And what is the exact error you are getting? – qrazi Nov 14 '13 at 08:48
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What does `var_dump($uitkomst);` output? It's probably a number. You need to use an array instead. – Amal Murali Nov 14 '13 at 08:49
1 Answers
1
$getal = $_POST['getal'];
$uitkomst = array();
For($teller=1; $teller<=11; $teller=$teller+1)
{
$uitkomst[$teller]=$teller*$getal;
}
Dmitry Seleznev
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Still curious about your error, since I could run your code (assuming $_POST['getal'] = 5) without error. PHP will deduce from this code that $uitkomst should be an array, although your code doesnt adhere to strict coding standards. – qrazi Nov 14 '13 at 08:52
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Maybe because I allready got a other $_post $getal code in my script?"; } $getal = $_POST['getal']; $uitkomst = array(); For($teller=1; $teller<=11; $teller=$teller+1) { $uitkomst[$teller]=$teller*$getal; } – user2991218 Nov 14 '13 at 09:43