Question

我有一个mysqli_result对象执行此操作：

    $mysqli->real_query("SELECT * FROM users WHERE `level` > 2");
    $user_list = $mysqli->use_result();
    print_r($user_list); die();

这样做，print_r给我：

mysqli_result Object ( [current_field] => 0 [field_count] => 6 [lengths] => [num_rows] => 0 [type] => 1 )

稍后我只是使用while ($row = $user_list->fetch_assoc())来解析所有内容。想想在做之前我想从mysqli_result对象中检索num_rows但只做$ user_list ['num_rows']不起作用：

致命错误：无法使用mysqli_result类型的对象作为数组

Answer 1

$user_list->num_rows是否有效？

Answer 2

您必须使用num_rows

获取$result->num_rows

if ($result = $mysqli->query("")) {

    /* determine number of rows result set */
    $row_cnt = $result->num_rows;
}

获得num_rows后，您可以获取它。

Answer 3

试试这个，在我的情况下它可以工作:)

$query  = "SELECT * FROM users WHERE `level` > 2";
$result = mysqli_query($connection, $query);
while ($rows = mysqli_fetch_array($result, MYSQLI_NUM)) {
     var_dump($rows);
}

Answer 4

$mysqli->real_query("SELECT * FROM users WHERE `level` > 2");
$user_list = $mysqli->use_result();
echo $user_list->num_rows;

从mysqli_result对象到字符串

4 个答案: