我是laravel的初学者.从来没有使用过框架.我创建了一个名为’tabletest’的数据库.它有两张桌子.一个是用户表,另一个是电话表.用户表有两列(id和name).电话表有3列(id phone和user_id).我正在尝试的是,我将使用表单输入并将输入发送到数据库表.尽管名称和电话已正确保存在不同的表中,但user_id(外键列)未更新.它总是0.我现在该怎么办?
迁移文件是:
用户表:
public function up()
{
Schema::create('user',function (Blueprint $table) {
$table->increments('id');
$table->string('name');
$table->timestamps();
});
}
电话表:
public function up()
{
Schema::create('phone',function (Blueprint $table) {
$table->increments('id');
$table->string('phone');
$table->unsignedInteger('user_id');
$table->timestamps();
});
}
用户模型:
use App\Model\Phone;
class User extends Model
{
protected $table = "user";
protected $fillable = ['name'];
public function phone(){
return $this->hasOne(Phone::class);
}
}
手机型号:
use App\Model\User;
class Phone extends Model
{
protected $table = "phone";
protected $fillable = ['phone','user_id'];
public function user(){
return $this->belongsTo(User::class);
}
}
PhoneController.PHP
PHP
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use App\Http\Requests;
use App\Http\Controllers\Controller;
use App\model\User;
use App\model\Phone;
class PhoneController extends Controller
{
public function store(Request $request)
{
User::create([
'name' => $request->name
]);
Phone::create([
'phone' => $request->phone
]);
}
}
最佳答案
您永远不会指定要为其创建电话号码的用户.即使您有外键,也只能索引和约束user_id列. MysqL无法“读取”您的PHP代码并假设您正在为哪个用户创建电话号码.
有几个选择.
你可以做到这一点:
$user = User::create([
'name' => $request->input('name')
]);
Phone::create([
'phone' => $request->phone,'user_id' => $user->id
]);
或者另一种方式是:
$user = User::create([
'name' => $request->input('name')
]);
$user->phone()->create([ // this uses the create method on the users phone relationship
'phone' => 999999999,]);
不确定是否也可以进行转换,但为了便于阅读,我不会推荐它(即User :: create([]) – > phone() – > create([]);)