Question

我有一个观点，这是它的结果：

ID  Type  Time                                RowNum
18  2     2015-12-14 12:15:00.9330000 +00:00  214
18  6     2015-12-14 12:21:22.4330000 +00:00  215
18  2     2015-12-15 09:04:41.2870000 +00:00  216
18  6     2015-12-15 09:06:22.9400000 +00:00  217
18  5     2015-12-15 09:07:28.0130000 +00:00  218

我试图通过使用下面的查询来比较行和组之间的时差。我的问题是最后一行，我希望得到当前时间的差异，并计算差异的总和。这是我正在使用的查询。如何仅比较最后一行与当前时间并计算总和：

WITH rows AS (SELECT * FROM [dbo].[View])
SELECT  p.ID ,
        p.Type ,
        SUM(DATEDIFF(SECOND, p.Time, q.Time))
FROM    rows p
        JOIN rows q ON p.RowNum = q.RowNum - 1
GROUP BY p.ID ,
         p.Type

Answer 1

如果您使用的是SQL Server 2012或更高版本，则可以使用

                @Override
                public void process(Exchange exchange) throws Exception {
                    XMLSerializer serializer = new XMLSerializer();
                    JSON json=serializer.read(exchange.getIn().getBody(String.class));
                    StringWriter writer=new StringWriter();

                    JsonConfig jsonConfig=new JsonConfig(); 
                    jsonConfig.registerJsonValueProcessor("entero1", new JsonValueProcessor() {
                        @Override
                        public Object processArrayValue(Object value, JsonConfig jsonConfig) {
                            return new BigDecimal(value.toString());
                        }
                        @Override
                        public Object processObjectValue(String key, Object value, JsonConfig jsonConfig) {
                            return new BigDecimal(value.toString());
                        }
                    });
                    JSONSerializer.toJSON(json, jsonConfig).write(writer);
                    writer.close();
                    exchange.getIn().setBody(writer.toString());
                }

参考LEAD（）

https://msdn.microsoft.com/en-IN/library/hh213125.aspx

Answer 2

使用LEFT JOIN和ISNULL功能：

DECLARE @t TABLE
    (
      ID INT ,
      Type INT ,
      Time DATETIMEOFFSET ,
      RowNum INT
    )

INSERT  INTO @t
VALUES  ( 18, 2, '2015-12-14 12:15:00.9330000 +00:00', 214 ),
        ( 18, 6, '2015-12-14 12:21:22.4330000 +00:00', 215 ),
        ( 18, 2, '2015-12-15 09:04:41.2870000 +00:00', 216 ),
        ( 18, 6, '2015-12-15 09:06:22.9400000 +00:00', 217 ),
        ( 18, 5, '2015-12-15 09:07:28.0130000 +00:00', 218 );

SELECT  p.ID ,
        p.Type ,
        SUM(DATEDIFF(SECOND, p.Time, ISNULL(q.Time, GETDATE())))
FROM    @t p
        LEFT JOIN @t q ON p.RowNum = q.RowNum - 1
GROUP BY p.ID ,
         p.Type

输出：

18  2   483
18  5   21186
18  6   74665

如何比较SQL

2 个答案: