Trying to access array offset on value of type null

最近把十幾年前的php code重構，原本在實現以下代碼時是正常運行的。

$file_show = mysqli_query($link,$file_show) or die('Query failed:'. mysqli_error());
$file_row=mysqli_fetch_array($file_show);
echo $file_row[0];
$print_name = $target_row[2];
if($file_row[0] == "ntof")
{
  $treechange = 1;
  $upnodename = $file_row[2];
}
else
{
  $treechange = 0;
  $upnodename = $target_row[2];
}

後來換到PHP 7.4.16出現這個錯誤。

上stackoverflow查了一下，

ArSeN says:

his happens because $cOTLdata is not null but the index 'char_data' does not exist. Previous versions of PHP may have been less strict on such mistakes and silently swallowed the error / notice while 7.4 does not do this anymore.

To check whether the index exists or not you can use isset():

isset($cOTLdata['char_data'])

Which means the line should look something like this:

$len = isset($cOTLdata['char_data']) ? count($cOTLdata['char_data']) : 0;

Note I switched the then and else cases of the ternary operator since === null is essentially what isset already does (but in the positive case).

換句話說就是原本就是個問題，以前沒有顯示而已，那麼我們只要在使用該變數前使用 isset()方法來判斷是否為null即可。

if(isset($file_row[0])){
if($file_row[0] == "ntof")
{
$treechange = 1;
$upnodename = $file_row[2];
}
else
{
$treechange = 0;
$upnodename = $target_row[2];
}
}

改完後再次運行就正常顯示資料囉!

Refer from https://stackoverflow.com/questions/59336951/message-trying-to-access-array-offset-on-value-of-type-null

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