Question

<>proBLEM>:

我想提出一个问题,这将引发类似的东西。

select * from users where code in (1,2,4);

缩略语

www.un.org/Depts/DGACM/index_spanish.htm WHAT I TRIED:

这是一部单一的法典:

named_scope :of_code, lambda {|code| {:conditions => ["code = ?", code]}}

我尝试过这样的事情。

named_scope :of_codes, lambda {|codes| {:conditions => ["code in ?", codes]}}

送交

user.of_codes( (1,2,4) )

it triggers select * from users where code in (1,2,4) which raises a MySQL error because of the extra quotes.

<PS:Ideally I 我想送<>code>user.of_codes ([1,2,4]

Answer 1

这将只是找到工作,而且不会使你遭受到卡片注射袭击:

named_scope :of_codes, lambda { |codes|
  { :conditions => [ code in (?) , codes] }
}

User.of_codes([1, 2, 3])
# executes "select * from users where code in (1,2,3)"

如果你想稍微增加一点点,你可以这样做:

named_scope :of_codes, lambda { |*codes|
  { :conditions => [ code in (?) , [*codes]] }
}

然后,你可以打上<代码>。 www.un.org/chinese/ga/president

Answer 2

最简单的办法是在条件而不是阵列中使用 has:

named_scope :of_codes, lambda { |*codes| { :conditions => { :code => codes } } }

这将按预期运作。

User.of_codes(1, 2, 3) # => SELECT ... code IN (1,2,3)
User.of_codes(1) # => SELECT ... code IN (1)

Answer 3

页: 1

named_scope :of_codes, lambda {|codes| {:conditions => ["code in "+codes]}}

以及

user.of_codes( (1,2,4) )

参考文件

named_scope :of_codes, lambda {|codes| {:conditions => ["code in (?) ", codes]}}

以及

user.of_codes([1,2,4])

友情链接