获取column_id在id数组中的所有行，然后按id排序
我正在尝试这个，不确定我是否正确执行此操作：
User.where("region_id => ?", region_ids).order("id ASC")
region_ids = [1234,234322,234324,2343,....]
另外，如果region_ids为空（不是空，而是空），这会起作用吗
我看到一个错误：
check the manual that corresponds to your MySQL server version for the right syntax to use near '=> NULL) ORDER BY id ASC' at line 1:
当我处于调试器模式时，我输出region_ids及其曾是 []。
I am trying this, not sure if I am doing this correctly:
User.where("region_id => ?", region_ids).order("id ASC")
region_ids = [1234,234322,234324,2343,....]
Also, will this work if the region_ids is empty (not null, but empty)
I am seeing an error:
check the manual that corresponds to your MySQL server version for the right syntax to use near '=> NULL) ORDER BY id ASC' at line 1:
When I was in debugger mode, I output the region_ids and it was [].
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您在字符串中混合了 Ruby 和 SQL 语法。您想要的更像是
("region_id in (?)", Region_ids)或({ "region_id" => Region_ids })。You're mixing up Ruby and SQL syntax within the string. What you want is more like
("region_id in (?)", region_ids)or({ "region_id" => region_ids }).您需要颠倒两行的顺序，region_ids 应设置在 User.where 之前
因为现在当你调用 where() 时，region_ids 是 NULL。
You need to reverse the order of the two lines, region_ids should be set before the User.where
becuase the way it is now region_ids is NULL when you call where().