在更改表迁移中将现有外键列设置为可为空

Question

我首先创建了这样的迁移：

Schema::create('table1',function(Blueprint $table){
        $table->bigIncrements('id');
        $table->string('name')->unique();
        $table->integer("user_id")->unsigned();
        $table->foreign("user_id)->references("id")->on("users");
});

然后我想向 user_id 列添加可空属性，我写了这个迁移：

Schema::table('f_subjects', function (Blueprint $table) {
        $table->integer('user_id')->nullable()->change();
        $table->foreign('original_law_id')->references('id')->on('f_original_law');
    });

但我收到了这个错误：

Cannot change column 'user_id': used in a foreign key constraint 'table1_user_id_foreign'

Answer 1

1.需要先删除约束：

$table->dropForeign(['user_id']);

2.或者您可以暂时禁用 FK 约束：

Schema::disableForeignKeyConstraints();

然后启用约束：

Schema::enableForeignKeyConstraints();

https://laravel.com/docs/5.5/migrations#foreign-key-constraints

Answer 2

1-删除您的外键

$table->dropForeign('table1_user_id_foreign');

2- 更改 user_id 列定义：

//If user_id is not unsigned remove unsigned function
$table->integer('user_id')->nullable()->unsigned()->change();   

3- 创建索引

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

完整迁移：

Schema::table('table1',function(Blueprint $table){
    //Or disable foreign check with: 
    //Schema::disableForeignKeyConstraints();
    $table->dropForeign('table1_user_id_foreign');
    $table->integer('user_id')->nullable()->unsigned()->change();
    //Remove the following line if disable foreign key
    $table->foreign('user_id')->references('id')->on('users');
});