Laravel 4迁移问题中的外键

2022-01-08 00:00:00 mysql laravel laravel-4

我刚刚创建了一个新的 Laravel 4 项目，发现架构构建器的外键方面发生了奇怪的事情.如果我在任何迁移中使用 ->foreign() 方法，我会抛出 MySQL 错误 150 和一般错误 1005.根据 laravel.com/docs 上的文档，这两个场景位于底部应该管用?有人知道他们为什么不这样做吗?

I've just created a new Laravel 4 project and am finding strange things happening with the foreign key aspect of the schema builder. If I use the ->foreign() method in any of my migrations I get thrown MySQL errors 150 and general error 1005. According to the documentation at laravel.com/docs the two scenario's at the bottom should work? Anyone know why they don't?

以下确实有效:

    Schema::create('areas', function($table)
    {
        $table->engine ='InnoDB';
        $table->increments('id');

        $table->integer('region_id')->references('id')->on('regions');

        $table->string('name', 160);
        $table->timestamps();
    });

但这两个不起作用:

    Schema::create('areas', function($table)
    {
        $table->engine ='InnoDB';
        $table->increments('id');

        $table->foreign('region_id')->references('id')->on('regions');

        $table->string('name', 160);
        $table->timestamps();
    });

    Schema::create('areas', function($table)
    {
        $table->engine ='InnoDB';
        $table->increments('id');

        $table->integer('region_id');
        $table->foreign('region_id')->references('id')->on('regions');

        $table->string('name', 160);
        $table->timestamps();
    });

推荐答案

检查您的 id 类型.Laravel 4 创建一个带有 int(10) 无符号的增量 id.如果您创建一个基本整数并尝试在其上放置一个外键，它将失败.

Check your id type. Laravel 4 creates an incremental id with a int(10) unsigned. If you create a basic integer and try to put a foreign key on it, it will fail.

如 此链接，您应该使用 $table->unsignedInteger(YOUR_ID_NAME); 创建外部 id 以使其工作.

As suggested in the documentation at this link, you should create the foreign id with $table->unsignedInteger(YOUR_ID_NAME); to make it work.

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