Date 如何选择显示在特定值之前的行？

我有一张包含以下信息的表格：

CREATE TABLE A (
    ID int,
    start_date varchar(255),
    effdate varchar(255),
    enddate varchar(255),
    lead_effdate varchar(255),
    value varchar(255),
    source varchar(255),
    lead_source varchar(255)
);
INSERT INTO A
VALUES (1,"12/01/2013","12/01/2013","30/07/2014","31/07/2014",4,"REM","REM"),
(1,"31/07/2014","31/07/2014","04/09/2014","12/09/2014",4,"REM","REM"),
(1,"12/09/2014","12/09/2014","12/04/2015","13/04/2015",4,"REM","REM"),
(1,"13/04/2015","13/04/2015","01/01/2016","02/01/2016",4,"REM","REM"),
(1,"02/01/2016","02/01/2016","11/03/2016","13/03/2016",4,"REM","REM"),
(1,"13/03/2016","13/03/2016","22/04/2017","10/07/2016",4,"REM","GS"),
(1,"10/07/2016","10/07/2016",null,"23/04/2017",4,"GS","REM"),
(1,"23/04/2017","23/04/2017","05/05/2018","06/05/2018",4,"REM","REM"),
(1,"06/05/2018","06/05/2018","09/11/2019","10/11/2019",4,"REM","REM"),
(1,"10/11/2019","10/11/2019",null,null,4,"REM",null);

select * from A
order by start_date

我想删除从source=GS到表中最低行的所有行。如您所见，订单由开始日期给出

我试图在（按…划分）上使用一个行数（）来创建GS周围的负值和正值，但我无法计算此索引。您将如何解决它？

如果“到底”是指“更大的id”，那么您可以在

删除
中使用
连接
：

delete a
    from a join
         (select max(id) as gs_id
          from a
          where source = 'GS'
         ) aa
         on a.id > aa.gs_id;

您希望选择日期等于或高于source='GS'的行的所有行。因为可以有多行source='GS'
您必须选择一行。可能是第一行，也可能是最后一行

delete from a where start_date >= (select min(start_date) from a where source = 'GS');

数据没有意义，varchar必须用引号括起来，还必须始终将日期保存为date而不是varchar，并在您仅选择它时格式化它
从source=GS到底部的所有行
？您在说什么顺序？所有
source>='GS'
的行？这不是有效的SQL。提示：使用类型作为日期值，并将其编码为，如中所示<代码> yyy-MM-DD 。提示：“代码> 12”/“01”/“代码”是什么？考虑“代码”>选择“12”/“01”/“2013”/“代码”。对不起，表。我纠正了这些值，现在应该工作。DelabRek不支持删除。您知道另一种方法吗？我不知道数据库。您的问题是标记<代码> mysql < /代码>。我的答案是标准。符合rd的
DELETE
语句，该语句在MySQL和任何其他RDBMS中都能运行……我刚才在Google上搜索了
databricks
和
DELETE
：看起来它是受支持的。SQL语句中的错误：AnalysisException:DELETE destination只支持增量源。正如您所见，除非表是空的，否则databricks不支持DELETE存储在Delta源代码中，我不能这样做。好吧。那我就无能为力了。对不起。我知道SQL和一些RDBMS。我不知道Databricks。