Given two tables:
给定两个表:
- t1: a table with elements, and
- t1:包含元素的表
- t2: another table with list of IDs processed from t1
- t2:从t1处理的id列表的另一个表
How do you query for all the elements in t1, which were not yet processed, that is, does not have a corresponding pid in t2?
如何查询t1中尚未处理的所有元素，即t2中没有相应的pid ?
For clarity, I'm looking for the functional equalent of
为了清晰，我在寻找函数等式
Select * from t1 where t1.id not in (select pid from t2)
从t1中选择*。id不在(从t2中选择pid)
Alternatively, a way to implement a simple processing queue to determine which new elements in t1 have not been processed yet.
另一种方法是实现一个简单的处理队列，以确定t1中的哪些新元素尚未被处理。
Update:
更新:
Currently I'm at the following:
目前我的观点如下:
r.db("mydb").table("t1").filter( function(u) { return r.db("mydb").table("t2")("pid").contains( u("id") ).not(); })
r.db(mydb).table(t1)。过滤器(函数(u){返回r.db(mydb).table(t2)(“pid”)。包含(u(" id "))自身之外();})
However, this is horribly inefficient: specifically, it requires full tablescans for every element in t2 against t1. Are there any ways to return this more efficiently?
然而，这是非常低效的:具体来说，它要求t2中的每个元素都要用大汤匙来对付t1。有什么方法可以更有效地回报这个问题吗?
Many thanks!
很多谢谢!
1 个解决方案
#1
3
You can do this by creating a secondary index on t2 for pid:
你可以通过在t2上创建一个二级索引来实现这一点:
r.table('t2').indexCreate('pid')
r.table('t1').filter(function(u) {
return r.table('t2').getAll(u('id'), {index: 'pid'}).isEmpty();
})
#1
3
You can do this by creating a secondary index on t2 for pid:
你可以通过在t2上创建一个二级索引来实现这一点:
r.table('t2').indexCreate('pid')
r.table('t1').filter(function(u) {
return r.table('t2').getAll(u('id'), {index: 'pid'}).isEmpty();
})