Laravel外鍵未輸出預期結果

Question

我正在開發一個Laravel 9 Web 應用程序，其中有兩個表（users 和feedbacks），它們使用名為username 的外鍵進行連接。一個用戶可以有很多回饋。據我所知，如果我獲得用戶的詳細信息，這些數據也包含相關回饋。我的問題是，用戶資料已正確獲取，但它附帶所有回饋，而不是連接到該特定用戶的回饋。 Laravel 執行這樣的查詢。

select * from `feedback` where `feedback`.`username` = 0 and `feedback`.`username` is not null

據我了解，0 應該替換為使用者的使用者名稱。這裡有什麼問題嗎？

回饋模型-

#

class Feedback extends Model
{
    use HasFactory;

    //One single user can have many feedbacks.
    public function user() {
        return $this->belongsTo(User::class);
    }
}

User模型-

class User extends Authenticatable
{
    use HasApiTokens, HasFactory, Notifiable;

    /**
     * The attributes that are mass assignable.
     *
     * @var array<int, string>
     */
    protected $fillable = [
        'name',
        'username',
        'gender',
        'email',
        'password',
        'is_admin',
    ];

    /**
     * The attributes that should be hidden for serialization.
     *
     * @var array<int, string>
     */
    protected $hidden = [
        'password',
        'remember_token',
        'is_admin',
    ];

    protected $primaryKey = 'username';

    public function feedbacks() {
        return $this->hasMany(Feedback::class, 'username');
    }

    /**
     * The attributes that should be cast.
     *
     * @var array<string, string>
     */
    protected $casts = [
        'email_verified_at' => 'datetime',
    ];
}

create_users_table遷移-

public function up()
    {
        Schema::create('users', function (Blueprint $table) {
            $table->increments('userId');
            $table->string('name');
            $table->string('username')->unique();
            $table->string('gender');
            $table->string('email')->unique();
            $table->timestamp('email_verified_at')->nullable();
            $table->string('password');
            $table->boolean('is_admin')->default(0);
            $table->rememberToken();
            $table->timestamps();
        });
    }

create_feedback_table遷移-

public function up()
    {
        Schema::create('feedback', function (Blueprint $table) {
            $table->increments('feedbackId');
            $table->text('feedback');
            $table->string('username');
            $table->timestamps();
            $table->foreign('username')
                ->references('username')
                ->on('users')
                ->onDelete('cascade');
        });
    }

FeedbackController取得數據，

class FeedbackController extends Controller
{
    public function giveFeedback($username)
    {
        $userData = User::find($username);

        dd($userData->feedbacks);

        return view('feedback.givefeedback', compact('userData'));
    }
}

users 表-

feedback 表-

這是刀片上的輸出，正如您所看到的，它輸出了所有回饋，即使我只使用路由請求了 nerigupex 的回饋。

如果您需要更多程式碼來解決此問題，請提出請求，我將相應更新問題。 TIA。

Answer 1

這樣做（僅解決資料載入問題）

1。重構遷移

用戶遷移

Schema::create('users', function (Blueprint $table) {
    $table->bigIncrements('id'); # change your current primary key to this

    .... rest of the code
}

回饋遷移

Schema::create('feedback', function (Blueprint $table) {
    $table->bigIncrements('id'); # change your current primary key to this
    $table->unsignedBigInteger('user_id');
    $table->foreign('user_id')->references('id')->on('users');

    .... rest of the code
}

2。重構模型

//protected $primaryKey = 'username'; --> remove this

public function feedbacks() {
    return $this->hasMany(Feedback::class);
}

3。在回饋控制器中

class FeedbackController extends Controller
{
    public function giveFeedback($username)
    {
        $userData = User::with('feedbacks')->where('username', $username)->get();

        dd($userData->feedbacks);

        return view('feedback.givefeedback', compact('userData'));
    }
}