Codeigniter MySQL加入复杂

Question

我有几张桌子，希望一次显示一个id。

表A

id | isi a1 | isi a2

表B

id | id_a | isi b1 | isi b2

表C

id | id_a | isi c1 | isi c2

表D

id | id_a | isi d1 | isi d2

表E

id | id_a | isi e1 | isi e2

我想显示isi B1-E2，其中每个表在表A中都有id_a = id。我不知道mysql join，我试过这段代码

$this->db->select('BaseTbl.id, BaseTbl.tanggal, BaseTbl.atas_nama, BaseTbl.kerugian, BaseTbl.keterangan, BaseTbl.admin, BaseTbl.status');
$this->db->from('data_blacklist as BaseTbl');
$this->db->join('bl_rekening as Rekening, bl_telefon as Telefon, bl_bukti as Bukti, bl_pelapor as Pelapor', 'Rekening.id_blacklist = BaseTbl.id, Telefon.id_blacklist = BaseTbl.id, Bukti.id_blacklist = BaseTbl.id, Pelapor.id_blacklist = BaseTbl.id','left');

但它总是给我

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' bl_telefon as Telefon, bl_bukti as Bukti, bl_pelapor as Pelapor ON `Rekening`.`' at line 3

我现在不知道。任何的想法？我使用codeigniter

Answer 1

试试这个。您必须查询每个表的连接。你无法将它们组合起来

$this->db->select('BaseTbl.id, BaseTbl.tanggal, BaseTbl.atas_nama, BaseTbl.kerugian, BaseTbl.keterangan, BaseTbl.admin, BaseTbl.status');
$this->db->from('data_blacklist as BaseTbl',);
$this->db->join('bl_rekening as Rekening','Rekening.id_blacklist = BaseTbl.id','left');
$this->db->join('bl_telefon as Telefon','Telefon.id_blacklist = BaseTbl.id','left');
$this->db->join('bl_bukti as Bukti','Bukti.id_blacklist = BaseTbl.id','left');
$this->db->join('bl_pelapor as Pelapor','Pelapor.id_blacklist = BaseTbl.id','left');

Answer 2

$this->db->select('a.*,b.*,c.*,d.*,e.*');
        $this->db->from( TABLES::$A . ' AS a' );
        $this->db->join(TABLES::$B .  ' AS b', 'a.id = b.id_a');
        $this->db->join(TABLES::$C .  ' AS c', 'a.id = c.id_a');
        $this->db->join(TABLES::$D .  ' AS d', 'a.id = d.id_a');
        $this->db->join(TABLES::$E .  ' AS e', 'a.id = e.id_a');

        $query=$this->db->get();
        //  echo $this->db->last_query();
        $result = $query->result_array();

        return $result;

Answer 3

我可以帮助您在CodeIgniter中进行手动查询。

return $this->db->query("select A.*, B.*, C.*, D.*, E.* from Table A, Table B, Table C, Table D, Table E WHERE A.id = B.id_a AND A.id = C.id_a AND A.id = D.id_a AND A.id = E.id_a")->result_array();

A，B，C，D和E是表的别名。如果值存在，它将为您带来数据。或者您也可以编写手动查询