Question

使用代码点火器！我有两个表student和student_class与外键student_id，我想选择存在于学生表但在class_student上找不到的数据

这是我的sql

function student_class(){
    $this->db->SELECT ('student.student_id, student.firstname, student.middlename, student.lastname');
    $this->db->FROM('student');
    $this->db->WHERE('student.status',0);
    $this->db->JOIN('student_class', 'student_class.student_id=student_class.student_id', 'left');
    $this->db->where_not_in('student_class.student_id');
    $query =$this->db->get();
    return $query->result_array();
}

它不起作用!! 我可以得到帮助..

Answer 1

试试这样..

首先查找与student_class表匹配的所有学生ID。然后使用$this->db->where_not_in获取所需的结果。

    function student_class(){
        $this->db->select ('student.student_id');
        $this->db->from ('student');
        $this->db->join('student_class', 'student.student_id=student_class.student_id', 'left');
        $this->db->where('student.status',0);
        $data = $this->db->result_array();//array of matched id's

       $this->db->select('student_id,firstname, middlename,lastname');
       $this->db->from('student');
       $this->db->where_not_in($data);
       $query = $this->db->get();

       return $query->result_array();
    }

希望它能奏效。

从一个表上存在但在另一个表上不存在的数据中选择

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