Php 如何在迁移中删除表的外键和主键?

Php 如何在迁移中删除表的外键和主键?,php,laravel,laravel-5,laravel-5.5,Php,Laravel,Laravel 5,Laravel 5.5,如何在迁移文件中删除多个外键和主键 下面是我的迁移文件代码 迁移文件 public function up() { Schema::create('role_user', function(Blueprint $table){ $table->integer('role_id')->unsigned(); $table->integer('user_id')->unsigned(); $table->fore

如何在迁移文件中删除多个外键和主键

下面是我的迁移文件代码

迁移文件

public function up()
{
    Schema::create('role_user', function(Blueprint $table){
        $table->integer('role_id')->unsigned();
        $table->integer('user_id')->unsigned();

        $table->foreign('role_id')
                ->references('id')
                ->on('roles')
                ->onDelete('cascade');

        $table->foreign('user_id')
                ->references('id')
                ->on('users')
                ->onDelete('cascade');

        $table->primary(['role_id', 'user_id']);
    });
}
public function down()
{
    Schema::drop('role_user');
    //how drop foreign and primary key here ?
}

Blueprint类提供了dropForeigndropPrimary方法,允许您删除外键约束和主键

下面应该可以做到这一点:

public function down()
{
    Schema::table('role_user', function (Blueprint $table) {
        $table->dropForeign('role_user_role_id_foreign');
        $table->dropForeign('role_user_user_id_foreign');
        $table->dropPrimary();
    });
}

hi@jedzej我可以在dropForeign(['foreign_one','foreign_two])这样的单个dropForeign函数中删除两个外键吗;你不能。你为什么要这么做?只是为了在一个小的迁移文件中保存一行?