$ query = mysqli_query（$ conn，$ sql）; 在服务器上返回false [重复]

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• How to display errors for my MySQLi query? [duplicate] 3 answers
• Reference - What does this error mean in PHP? 34 answers

    $count = count($down_length);

        for ($i = 0; $i < $count; $i++) {
            $sql = "insert into orders (invoice_id,up_length,up_shoulder,up_sleeve1,up_sleeve2,up_sleeve3,up_chest1,up_chest2,
up_stomach1,up_stomach2,up_hips1,up_hips2,up_neck,down_length,down_fork1,down_fork2,down_bottom1,down_bottom2,down_knee1,down_knee2,
down_thigh1,down_thigh2,down_waist1,down_waist2,down_hips,product,any_other)
values('$last_invoiceid','$top_length[$i]','$top_shoulder[$i]','$top_sleeve1[$i]','$top_sleeve2[$i]','$top_sleeve3[$i]','$top_chest1[$i]','$top_chest2[$i]',
'$top_stomach1[$i]','$top_stomach2[$i]','$top_hips1[$i]','$top_hips2[$i]','$top_neck[$i]','$down_length[$i]','$down_fork1[$i]','$down_fork2[$i]','$down_bottom1[$i]',
'$down_bottom2[$i]','$down_knee1[$i]','$down_knee2[$i]','$down_thigh1[$i]','$down_thigh2[$i]','$down_waist1[$i]','$down_waist2[$i]','$down_hips[$i]','$product[$i]',
'$any_others[$i]')";

            $query = mysqli_query($conn, $sql);
var_dump($query);
echo '<br>';}
            if ($query) {
                header('location:../product_print.php');
            }

its work perfectly on xamp server localhost.but while in a live server with PHP 7 running $query return false. what am I doing wrong here?

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