提交表單后，Laravel 5.2一對一關系給出錯誤消息“ SQLSTATE [23000]：”

Question

我在laravel 5.2中創建了兩個表，一個表稱為“用戶”，另一個表稱為“ artists_details”，它們具有一對一的關系。 用戶表的架構如下

Schema::create('users', function (Blueprint $table) {
        $table->engine = 'InnoDB';
        $table->increments('id');
        $table->boolean('admin')->nullable();
        $table->boolean('manager')->nullable();
        $table->string('name');
        $table->string('email')->unique();
        $table->string('password', 60);
        $table->rememberToken();
        $table->timestamps();
    });

並且Artists表的架構如下

Schema::create('artists_details', function (Blueprint $table) {
        $table->engine = 'InnoDB';
        $table->integer('user_id')->unsigned();
        $table->string('artists_image_path');
        $table->string('name');
        $table->integer('phone_no');
        $table->integer('passport');
        $table->string('city');
        $table->string('county');
        $table->string('facebook_name');
        $table->string('twitter_handle');
        $table->string('email');
        $table->string('alternative_email');
        $table->string('website');
        $table->text('biography');
        $table->timestamps();

        $table->foreign('user_id')
              ->references('id')
              ->on('users')
              ->onDelete('CASCADE');

    });

我已經在模型中指出了以下關系用戶模型

public function artists_relation()
{
    return $this->hasOne('App\artists_details_model');
}

並在artist_details_model上如下

public function user_artist_details()
{
    return $this->belongsTo('App\User');
}

在控制器中處理表單提交的php代碼如下

public function artists_details_store(Request $request)
{
  $input = \Request::all();
  $file = $request->file('file');

   $name = time(). $file->getClientOriginalName();

   $file->move('artists_image/photo', $name);



  $artists_input = new artists_details_model;

  $artists_input->artists_image_path  = 'artists_image/photo/'. $name;
  $artists_input->name                = $input['name'];
  $artists_input->phone_no            = $input['phone_no'];
  $artists_input->passport            = $input['passport'];
  $artists_input->city                = $input['city'];
  $artists_input->county              = $input['county'];
  $artists_input->facebook_name       = $input['facebook_name'];
  $artists_input->twitter_handle      = $input['twitter_handle'];
  $artists_input->email               = $input['email'];
  $artists_input->alternative_email   = $input['alternative_email'];
  $artists_input->website             = $input['website'];
  $artists_input->biography           = $input['biography'];
  $artists_input->save();


    return redirect('create');
}

當我單擊提交按鈕時，出現以下錯誤信息

Connection.php第669行的QueryException：

SQLSTATE [23000]：違反完整性約束：1452無法添加或更新子行：外鍵約束失敗

我似乎看不到我要去哪里或問題出在哪里

Answer 1

對於表“ artists_details”，您提到的是

$table->foreign('user_id') ->references('id') ->on('users')

因此，每當您嘗試將詳細信息保存在“ artists_details”中時，都需要提供“ user_id”，否則將無法保存信息。

您需要傳遞“ UserID”作為隱藏參數，或者如果UserID保存在會話中，則需要從會話中檢索它。

Answer 2

您似乎沒有在表單中包含外鍵{user_id}