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你好编码员我正在尝试在 CODE IGNITER 模型中编写这个 mysql 脚本但没有得到正确的脚本如何将它编码到活动记录中你能帮我吗

SELECT bus_reg_number
FROM t_bus_detail
WHERE bus_reg_number NOT
IN (

SELECT bus_reg_number
FROM t_bus_alot
)

我的模特

 public function get_bus_reg_number()
    {


    $this->db->select('bus_reg_number')->from('t_bus_detail');
 $result=$this->db->where_not_in('bus_reg_number' NOT IN (SELECT 'bus_reg_number' FROM 't_bus_alot')');



    $dropdown_bus_number = array();
    foreach($result as $r) 
    {
    $dropdown_bus_number[$r['bus_reg_number']] = $r['bus_reg_number'];

    }
    return $dropdown_bus_number;
    }
4

1 回答 1

1
$this->db->select('bus_reg_number');
$this->db->from('t_bus_detail');
$this->db->where('`bus_reg_number` NOT IN (SELECT `bus_reg_number` FROM `t_bus_alot`)', NULL, FALSE);

试试这个.. null 和 false 会告诉 CI 不要逃避你的 where 查询..

编辑:这是简单的查询:

$query = $this->db->query("SELECT bus_reg_number FROM t_bus_detail WHERE bus_reg_number NOT IN (SELECT bus_reg_number FROM t_bus_alot)");
$dropdown_bus_number = array();
foreach ($query->result_array() as $row){
   $dropdown_bus_number[$row['bus_reg_number']] = $row['bus_reg_number'];
}
于 2013-04-07T16:16:44.610 回答