mysqli_fetch_array错误未定义的偏移量：1

Question

我有5个sp_id，$ count也给了5个。但是当我通过mysqli_fetch_array检查大小时，它给出了我的大小2并且只打印1并且给出了错误未定义的偏移：1进一步。我想获得id的值，我的意思是通过for循环打印ID。

     $selectId="Select sp_id from stitchedproduct";
$selectIdR = mysqli_query($connection,$selectId) or die(mysqli_error($connection));
$count = mysqli_num_rows($selectIdR);
$fetchedResult= mysqli_fetch_array($selectIdR);

for ($i=0;$i < $count; $i++) {
print $fetchedResult[$i];
     }

Answer 1

替换

$fetchedResult= mysqli_fetch_array($selectIdR);

for ($i=0;$i < $count; $i++) {
    print $fetchedResult[$i];
}

带

while( $fetchedResult= mysqli_fetch_array($selectIdR)) {
    print $fetchedResult['sp_id'];
}