迁移中的外键错误（Laravel 6.0）

Question

我试图弄清楚 Laravel 中的迁移是如何工作的。在这里你可以看到我正在创建 2 个表。在第二个表（RealNationalLeague）中，我在第一个表（RealNationalLeagueLevel）中有两列（real_nation_id，level）的外键。

public function up()
    {

        ...

        Schema::create(Model\RealNationalLeagueLevel::TABLE, function (Blueprint $table) {
            $table->unsignedInteger("real_nation_id");
            $table->unsignedTinyInteger("level");

            $table->foreign("real_nation_id")->references("id")->on(Model\RealNation::TABLE);
            $table->primary(["real_nation_id", "level"]);
        });


        Schema::create(Model\RealNationalLeague::TABLE, function (Blueprint $table) {
            $table->increments("id");
            $table->unsignedInteger("real_nation_id");
            $table->unsignedTinyInteger("level");
            $table->string("name", 32);

            $table->foreign("real_nation_id")->references("real_nation_id")->on(Model\RealNationalLeagueLevel::TABLE); // this works
            $table->foreign("level")->references("level")->on(Model\RealNationalLeagueLevel::TABLE); // this does not
        });
    }

运行迁移不起作用。它抛出 QueryException：

SQLSTATE[HY000]: General error: 1005 Can't create table `testdb`.`#sql-17a00_448a37` (errno: 150 "Foreign key constraint is incorrectly formed") (SQL: alter table `real_national_league` add constraint `real_national_league_level_foreign` foreign key (`level`) references `real_national_league_level` (`level`))

知道为什么它不起作用吗？感谢您的帮助。

Answer 1

您要引用的表，您必须先创建它，然后在另一个迁移中，创建从该表到该引用表的列。