Noo*_*der 5 mysql database foreign-key-relationship eloquent laravel-5.1
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();
});
}
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电话表:
public function up()
{
Schema::create('phone', function (Blueprint $table) {
$table->increments('id');
$table->string('phone');
$table->unsignedInteger('user_id');
$table->timestamps();
});
}
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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);
}
}
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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);
}
}
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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
]);
}
}
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您永远不会指定要为哪个用户创建电话号码。即使您有外键,它也只会索引和约束 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。
有几种选择。
你可以这样做:
$user = User::create([
'name' => $request->input('name')
]);
Phone::create([
'phone' => $request->phone,
'user_id' => $user->id
]);
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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,
]);
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不确定改变这是否也可能,但为了可读性我不会推荐它(即User::create([])->phone()->create([]);)
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