Laravel 4 不能删除外键
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Laravel 4 cannot drop foreign key
提问by Joshua
I am trying to drop Foreign key on a table, but I got this message:
我正在尝试将外键放在表上，但收到以下消息：
[Illuminate\Database\QueryException]
SQLSTATE[42000]: Syntax error or access violation: 1091 Can't DROP 'user_id
'; check that column/key exists (SQL: alter table `posts` drop foreign key
user_id)
And I am using migration to do this:
我正在使用迁移来做到这一点：
Schema::table('posts', function($table) {
$table->dropForeign('user_id');
$table->foreign('user_id')
->references('id')->on('users')
->onDelete('cascade');
});
I am sure that the 'user_id' exists in the 'posts' table:
我确定 'user_id' 存在于 'posts' 表中：
Field | Type | Null | Key | Default | Extra |
+------------+------------------+------+-----+---------------------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| user_id | int(10) unsigned | NO | MUL | NULL | |
How can I fix this? Is it a bug or something?
我怎样才能解决这个问题？是bug还是什么？
------------------
update:
I found that the reason might be that "you are trying to delete a key which is being used by another table."
Does that means I should drop those tables that uses 'posts' table first?
------------------
更新：
我发现原因可能是“您正在尝试删除另一个表正在使用的键”。
这是否意味着我应该先删除那些使用“posts”表的表？
回答by Marwelln
When creating a foreign key, the name will be table_fields_foreignunless you set a name in the second parameter (see createIndexNamemethod).
创建外键时，table_fields_foreign除非您在第二个参数中设置名称（请参阅createIndexName方法），否则名称将为。
So if you didn't specify a name your foreign key should be posts_user_id_foreignand not user_id.
因此，如果您没有指定名称，您的外键应该是posts_user_id_foreign而不是user_id.
