如何在CI4中显示另一个表(外键)的值

问题描述 投票:0回答:0

在“表 A”中,我有 id_costumer,在“表 B”中,我也有 id_costumer 作为外键。 要为“table_b”创建新数据,我需要从“table a”中获取值并使其在视图中可选择。 我的问题是,如何获取该值以便在视图中显示它并为“表 b”创建新数据? 我知道我放在这里的代码仍然是基本的,但也许可以让任何人更容易解决或提供解决方案。

$validation = \Config\Services::validation();
        $validation->setRules(['id_costumer'=>'required']);
        $isDataValid = $validation->withRequest($this->request)->run();
        
        if($isDataValid){
            $repair = new RepairModel();
            $repair->insert([
                "id_costumer" => $this->request->getPost('id_costumer'),
                "serial_number" => $this->request->getPost('serial_number'),
                "tgl_terima" => $this->request->getPost('tgl_terima'),
                "user_terima" => $this->request->getPost('user_terima'),
                "kerusakan" => $this->request->getPost('kerusakan'),
                "solusi" => $this->request->getPost('solusi'),
                "user_teknisi" => $this->request->getPost('solusi'),
                "status" => $this->request->getPost('status'),
                "garansi_batal" => $this->request->getPost('garansi_batal'),
                "tgl_selesai" => $this->request->getPost('tgl_selesai'),

            ]);
            return redirect('admin/repair');
        }

        $header['title']='Detail Repair';
        echo view('partial/header', $header);
        echo view('partial/top_menu');
        echo view('partial/side_menu');
        echo view('admin_repair/admin_create_repair');
        echo view('partial/footer');
foreign-keys add codeigniter-4
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