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left outer join

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。

语法

<collection1_name | (select_set1)> as <alias1_name> left outer join <collection2_name | (select_set2)> as <alias2_name> [on condition]

参数

参数名 参数类型 描述 是否必填
collection1_name/collection2_name string 集合名。
select_set1/select_set2 set 结果集。
alias1_name/alias2_name string 别名。
condition expression 集合之间关联条件。

返回值

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。

示例

  • 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).
    Takes 0.11811s.
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