在laravel中使用外鍵遷移時出錯

Question

我試圖在函數下面遷移，但它每次都給我同樣的錯誤，我沒能找出原因，即使我嘗試了文檔中的其他語法

//向上函數

public function up()
    {
        Schema::create('categories', function (Blueprint $table) {
            $table->id('cat_id');
            $table->string('cat_name');
            $table->string('cat_desc');
            $table->string('cat_image');
            $table->timestamps();
        });



Schema::create('sellers', function (Blueprint $table) {
        $table->id('seller_id');
        $table->string('seller_name');
        $table->string('seller_desc');
        $table->string('seller_image');
        $table->timestamps();
    });

    Schema::create('products', function (Blueprint $table) {
        $table->id('prd_id');
        $table->string('prd_name');
        $table->string('prd_price');
        $table->string('prd_image');
        $table->string('prd_desc');
        $table->foreignId('product_cat_id')->constrained('categories');
        $table->foreignId('product_seller_id')->constrained('sellers');
        $table->boolean('prd_status');
        $table->timestamps();
    });

}

//錯誤

SQLSTATE[HY000]: General error: 1005 Can't create table ecommerce.products (errno: 150 "Foreign key constraint is wronglyformed") (SQL: alter table products add constraint products_product_cat_id_foreign外鍵(product_cat_id)引用類別(id))

我還從數據庫中刪除了遷移文件並進行了新的遷移，但沒有幫助，請指導我，在此先感謝

Answer 1

public function up()
    {
        Schema::create('categories', function (Blueprint $table) {
            $table->increments('id');
            $table->integer('cat_id')->unsigned();
            $table->string('cat_name');
            $table->string('cat_desc');
            $table->string('cat_image');
            $table->timestamps();
        });



Schema::create('sellers', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('seller_id')->unsigned();
        $table->string('seller_name');
        $table->string('seller_desc');
        $table->string('seller_image');
        $table->timestamps();
    });

    Schema::create('products', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('prd_id')->unsigned();
        $table->string('prd_name');
        $table->string('prd_price');
        $table->string('prd_image');
        $table->string('prd_desc');
        $table->foreignId('product_cat_id')->constrained('categories');
        $table->foreignId('product_seller_id')->constrained('sellers');
        $table->boolean('prd_status');
        $table->timestamps();
    });

}

Answer 2

這是因為，在您的categories和sellers ，表主鍵都不是id ，默認情況下受約束的方法使用id作為主鍵作為外鍵引用，因此更改：

$table->foreignId('product_cat_id')->constrained('categories');
$table->foreignId('product_seller_id')->constrained('sellers');

到，

$table->unsignedBigInteger('product_cat_id');
$table->foreign('product_cat_id')->references('cat_id')->on('categories');

$table->unsignedBigInteger('product_seller_id');
$table->foreign('product_seller_id')->references('seller_id')->on('sellers');

如果您使用的是模型，那么還要在模型上定義主鍵，例如：

protected $primaryKey = 'cat_id';