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left outer join 会从左边的集合名(collection1_name)中返回所有的记录,即使在右边的集合名(collection2_name)中没有匹配的记录。
<collection1_name | (select_set1)> as <alias1_name> left outer join <collection2_name | (select_set2)> as <alias2_name> [on condition]
参数名 | 参数类型 | 描述 | 是否必填 |
---|---|---|---|
collection1_name/collection2_name | string | 集合名 | 是 |
select_set1/select_set2 | set | 结果集 | 是 |
alias1_name/alias2_name | string | 别名 | 是 |
condition | expression | 集合之间关联条件 | 否 |
left outer join 会从左边的集合(collection1_name)那里返回所有的记录,即使在右边的集合(collection2_name)中没有匹配的记录。
集合 sample.persons 中记录如下:
{ "Id_P": 1, "LastName": "Adams", "FirstName": "John", "Address": "Oxford Street", "City": "London" } { "Id_P": 2, "LastName": "Bush", "FirstName": "George", "Address": "Fifth Avenue", "City": "New York" } { "Id_P": 3, "LastName": "Carter", "FirstName": "Thomas", "Address": "Changan Street", "City": "Beijing" }
集合 sample.orders 中记录如下:
{ "Id_O": 1, "OrderNo": 77895, "Id_P": 3 } { "Id_O": 2, "OrderNo": 44678, "Id_P": 3 } { "Id_O": 3, "OrderNo": 22456, "Id_P": 1 } { "Id_O": 4, "OrderNo": 24562, "Id_P": 1 } { "Id_O": 5, "OrderNo": 34764, "Id_P": 65 }
列出所有客户的订单, 如果该用户没有订单信息,则用 null 替代其订单信息
> db.exec("select t1.LastName, t1.FirstName, t2.OrderNo from sample.persons as t1 left outer join sample.orders as t2 on t1.Id_P=t2.Id_P") { "LastName": "Adams", "FirstName": "John", "OrderNo": 22456 } { "LastName": "Adams", "FirstName": "John", "OrderNo": 24562 } { "LastName": "Bush", "FirstName": "George", "OrderNo": null } { "LastName": "Carter", "FirstName": "Thomas", "OrderNo": 77895 } { "LastName": "Carter", "FirstName": "Thomas", "OrderNo": 44678 } Return 5 row(s).