使用CodeIgniter显示查询，条件是只有相同的ID

Question

我们有2个表tb_kriteria和tb_subkriteria，我们必须从tb_subkriteria获取所有记录，其中id_kriteria的{​​{1}}与tb_kriteria相同在id_kriteria

中

Answer 1

运行codeigniter联接查询以获取其他结果，请尝试以下代码以获取输出

$this->db->select('tb_subkriteria.*');
$this->db->from('tb_subkriteria');
$this->db->join('tb_kriteria', 'tb_kriteria.id_kriteria = tb_subkriteria.id_kriteria'); 
$query = $this->db->get();
return $query->result();

Answer 2

对此查询使用联接

function tb_subkriteria_by_tb_kriteria($id_kriteria){
    $this->db->join('tb_kriteria','tb_kriteria.id_kriteria=
    tb_subkriteria.id_kriteria');
    return $this->db->where('tb_subkriteria.id_kriteria',$id_kriteria)->get('tb_subkriteria)->result_array();
}

Answer 3

在这里，您需要使用join和where来获取数据

function get_subkriteria($id_kriteria){
     $id_kriteria = 18;//suppose 
     $this->db->select('tb_subkriteria.*');
     $this->db->from('tb_subkriteria');
     $this->db->join('tb_kriteria', 'tb_kriteria.id_kriteria = tb_subkriteria.id_kriteria');
     $this->db->where('tb_subkriteria.id_kriteria',$id_kriteria); 
     $query = $this->db->get();
     echo $this->db->last_query();die;//to see your executed query
     if($query->num_rows() > 0){
       return $query->result();
     }else{
       return array();
     }
}

Answer 4

对此查询使用内部联接

function getresult($tid){
       $query = $this->db->query("select tb_subkriteria.* from tb_subkriteria inner 
       join tb_kriteria on tb_subkriteria.id_subkriteria=tb_kriteria.id_kriteria);
       return  $query->result_array();      
}