Laravel 4迁移中的外键问题 [英] Foreign keys in Laravel 4 migrations issue

问题描述

我刚刚创建了一个新的Laravel 4项目，并且发现架构生成器的外键方面发生了奇怪的事情.如果在任何迁移中都使用->foreign()方法，则会抛出MySQL错误150和常规错误1005.根据laravel.com/docs上的文档，底部的两种情况应该起作用吗?有人知道为什么他们不这样做吗?

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?

以下方法有效:

    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，则应使用$table->unsignedInteger(YOUR_ID_NAME);创建外部ID，以使其正常工作.

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

这篇关于Laravel 4迁移中的外键问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！