Php error with an array
I have this code:
The last part of the output looks like the attached. Something is wrong with my array definition (I think)
Any help?
Array.JPG
$ckd_sels = array();
$nc = 0;
for ($i = 0; $i < $ns; $i++) {
$s = mysqli_fetch_array($ress,MYSQLI_ASSOC);
$cbnam = $s['pid'] . "-" . $s['ruid'] . "-" . $s['selno'];
$ck = "ck" . $cbnam;
echo "ck + cbnam = " . $ck . "<br>";
echo "checked = " . $_POST[$ck] . "<br>";
if ($_POST[$ck] == "on") {
echo "post-ck = " . $_POST[$ck] . "<br>";
$ckd_sels[$i] = $ck;
echo "ckdcells = " . $ckd_sels[$i] . "<br>";
$nc++;
}
}
echo "ckd_sels array = " . $ckd_sels . "<br>";
$_SESSION['ckdsels'] = $ckd_sels;
The last part of the output looks like the attached. Something is wrong with my array definition (I think)
Any help?
Array.JPG
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PHP
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You can't show an array using echo with concatenated string, you need to use :
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If you want to show the content of an array inline you may need to convert it to string first using implode or json_encode.