有这个问题,

$acc = $this->db->get_where('accounts', array('username' => $username), 1);

我遇到的问题是,每当我尝试回显它(通过print_r)时,这就是结果,为什么它返回一个对象?
CI_DB_mysqli_result Object (
    [conn_id] => mysqli Object (
        [affected_rows] => 1
        [client_info] => mysqlnd 5.0.11-dev - 20120503 - $Id: 76b08b24596e12d4553bd41fc93cccd5bac2fe7a $
        [client_version] => 50011
        [connect_errno] => 0
        [connect_error] =>
        [errno] => 0
        [error] =>
        [error_list] => Array ( )
        [field_count] => 5
        [host_info] => localhost via TCP/IP
        [info] =>
        [insert_id] => 0
        [server_info] => 5.7.14
        [server_version] => 50714
        [stat] => Uptime: 16045 Threads: 3 Questions: 382 Slow queries: 0 Opens: 121 Flush tables: 1 Open tables: 114 Queries per second avg: 0.023
        [sqlstate] => 00000
        [protocol_version] => 10
        [thread_id] => 3
        [warning_count] => 0
    )
    [result_id] => mysqli_result Object (
        [current_field] => 0
        [field_count] => 5
        [lengths] =>
        [num_rows] => 1
        [type] => 0
    )
    [result_array] => Array ( )
    [result_object] => Array ( )
    [custom_result_object] => Array ( )
    [current_row] => 0
    [num_rows] =>
    [row_data] =>
)

它不应该返回一个值数组吗?
https://www.codeigniter.com/userguide2/database/active_record.html#select
它没有显示在那里,所以我只是假设active records类中的任何查询都会返回值。

最佳答案

$acc = $this->db->get_where('accounts', array('username' => $username), 1)->result();


$acc = $this->db->get_where('accounts', array('username' => $username), 1)->result_array();

您打印了查询生成器对象,而不是mysql result对象。

09-10 09:49
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