我正在尝试在数据库中播种用户,但出现错误提示
Symfony\Component\Debug\Exception\FatalThrowableError : Call to a member function random() on bool
我有用户表和性别表,用户表中的 sex_id 指向性别表中具有 hasMany 关系的 Man 或 Woman 列。当我播种数据库并创建新用户时,我希望能够在用户表中自动写入性别 ID。目前,使用此代码,我从上面得到了该错误,并且在 sex_id 列中得到了 NULL,但其余的它会正确插入到用户和性别表中。当我删除 random() 函数时,它总是在性别 ID 中插入 1,但我希望能够随机写入 1 或 2。另外,当我转储 $genders 时,它会返回 TRUE。有什么办法可以解决这个问题,我们将不胜感激。这是我的代码。
UserSeeder.php
<?php
use Carbon\Carbon;
use Illuminate\Database\Seeder;
use Illuminate\Support\Facades\DB;
use Illuminate\Support\Facades\Hash;
class UsersTableSeeder extends Seeder
{
/**
* Run the database seeds.
*
* @return void
*/
public function run()
{
$genders = DB::table('genders')->insert([
[
'genders' => 'Woman',
],
[
'genders' => 'Woman Looking For Woman',
],
[
'genders' => 'Man',
]
]);
//dd($genders);
DB::table('users')->insert([
'gender_id' => $genders->random(),
'name' => 'authuser',
'email' => '<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="e8899d9c809d9b8d9aa8899d9c80c68b8785" rel="noreferrer noopener nofollow">[email protected]</a>',
'email_verified_at' => now(),
'password' => Hash::make('auth123456'),
'age' => 18,
'remember_token' => Str::random(10),
'created_at' => Carbon::now(),
'updated_at' => Carbon::now(),
]);
}
}
用户表
<?php
use Illuminate\Support\Facades\Schema;
use Illuminate\Database\Schema\Blueprint;
use Illuminate\Database\Migrations\Migration;
class CreateUsersTable extends Migration
{
/**
* Run the migrations.
*
* @return void
*/
public function up()
{
Schema::create('users', function (Blueprint $table) {
$table->bigIncrements('id');
$table->unsignedBigInteger('gender_id')->nullable();
$table->string('name');
$table->string('email')->unique();
$table->timestamp('email_verified_at')->nullable();
$table->string('password')->default();
$table->integer('age')->default()->nullable();
$table->rememberToken();
$table->timestamps();
});
}
/**
* Reverse the migrations.
*
* @return void
*/
public function down()
{
Schema::dropIfExists('users');
}
}
性别表
<?php
use Illuminate\Database\Migrations\Migration;
use Illuminate\Database\Schema\Blueprint;
use Illuminate\Support\Facades\Schema;
class CreateGendersTable extends Migration
{
/**
* Run the migrations.
*
* @return void
*/
public function up()
{
Schema::create('genders', function (Blueprint $table) {
$table->bigIncrements('id');
$table->string('genders');
});
}
/**
* Reverse the migrations.
*
* @return void
*/
public function down()
{
Schema::dropIfExists('genders');
}
}
用户.php
public function gender()
{
return $this->belongsTo(Gender::class, 'gender_id', 'id');
}
性别.php
public function users()
{
return $this->hasMany(User::class, 'gender_id', 'id');
}
最佳答案
您可以从 Gendre
中提取您的 id 值,并对其进行随机操作,如下所示:
$genders = DB::table('genders')->insert([
['genders' => 'Woman'],
['genders' => 'Woman Looking For Woman'],
['genders' => 'Man']
]);
$gendreIds = Genders::pluck('id');
DB::table('users')->insert([
'gender_id' => $gendreIds->random(),
...
]);
这将为您提供数据库中存在的性别。
有时种子不会给你 1 到 3 的 id。
所以我认为使用rand(1,3)
并不是最好的解决方案。
祝你好运!
关于php - 在 Laravel 中将用户播种到数据库时出现问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59391682/