sql小提琴可以处理什么？sql小提琴不编译任何东西,并且不返回任何错误消息

Question

我最近创建了一个成功编译并返回所需结果的查询.当我使用那段代码作为另一段代码中的子查询时,stackoverflow上的用户想出了我,我遇到了一些问题,最终解决了.我尝试将此查询用作给我的那段代码中的子查询.但是,sql小提琴不会返回任何内容.没有错误或编译的消息.当我尝试输入语法错误时 - 就像一个随机+符号,没有任何反应.是因为查询太长了？

模式

CREATE TABLE sampleData 
    (
     id MEDIUMINT NOT NULL AUTO_INCREMENT,
      timecode int, 
     count int,
      PRIMARY KEY (id)
    )
#ENGINE=MyISAM
;

INSERT INTO sampleData
(timecode, count)
VALUES
(1344893440, 1), ( 1346014720, 1),( 1344898688,1),( 1345654784,1),( 1345978368,1),
( 1345959296,1), (1345064704,1), ( 1345156352,1),( 1345225600,1),
(1345017984,1),( 1345640960,1),( 1346019968,1),( 1345834752,1),
( 1345438464,1),( 1344986880,1),( 1345045632,1),( 1345557888,1),( 1344973056,1),( 1345087232,1),( 1345433216,1),( 1345691008,1),
( 1344917760,1),( 1345253248,1),( 1344934912,1),( 1345890048,1),( 1345272448,1), (1345829504,1),( 1345798400,1),( 1345203200,1),( 1344741120,1),
( 1345175552,1),( 1344824192,1),( 1344926336,1),( 1345571712,1),( 1344931584,1),( 1345211776,1),( 1345059456,1),( 1345516288,1),( 1345441920,1),( 1346009472,1)

询问

select t_0.*,
           (coalesce(t_3.average_number_of_votes_per_previous_period_days, 0) - coalesce(t_4.average_number_of_votes_per_previous_period_days, 0)) * 100.0
    from 
        (select t.*,
           (coalesce(t_1.count, 0) - coalesce(t_2.count, 0)) * 100.0 as "percentage increase in count in %"
    from  
    (
      SELECT sum(1) AS ordr,
      t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"

      FROM 
        (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
        FROM sampleData
        GROUP BY DAY) t1
        INNER JOIN 
        (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
        FROM sampleData
        GROUP BY DAY) t2 
        on t1.day >= t2.day                                                                                                                                
        GROUP BY t1.day, t1.count
        ORDER BY t1.day
        )t 
      left outer join
         (
          SELECT sum(1) AS ordr,
            t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"
          FROM 
            (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
            FROM sampleData
            GROUP BY DAY) t1
          INNER JOIN 
            (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
            FROM sampleData
            GROUP BY DAY) t2 
          on t1.day >= t2.day                                                                                                                                
          GROUP BY t1.day, t1.count
          ORDER BY t1.day
          )t_1
       on t.ordr = t_1.ordr + 1 left outer join
         (
          SELECT sum(1) AS ordr,
            t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"
          FROM 
              (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
              FROM sampleData
              GROUP BY DAY) t1
            INNER JOIN 
              (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
              FROM sampleData
              GROUP BY DAY) t2 
            on t1.day >= t2.day                                                                                                                                
            GROUP BY t1.day, t1.count
            ORDER BY t1.day
            ) t_2
        on t.ordr = t_2.ordr + 2)t_0 
    left outer join
         (select t.*,
           (coalesce(t_1.count, 0) - coalesce(t_2.count, 0)) * 100.0 as "percentage increase in count in %"
    from  
    (
      SELECT sum(1) AS ordr,
      t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"

      FROM 
        (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
        FROM sampleData
        GROUP BY DAY) t1
        INNER JOIN 
        (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
        FROM sampleData
        GROUP BY DAY) t2 
        on t1.day >= t2.day                                                                                                                                
        GROUP BY t1.day, t1.count
        ORDER BY t1.day
        )t 
      left outer join
         (
          SELECT sum(1) AS ordr,
            t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"
          FROM 
            (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
            FROM sampleData
            GROUP BY DAY) t1
          INNER JOIN 
            (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
            FROM sampleData
            GROUP BY DAY) t2 
          on t1.day >= t2.day                                                                                                                                
          GROUP BY t1.day, t1.count
          ORDER BY t1.day
          )t_1
       on t.ordr = t_1.ordr + 1 left outer join
         (
          SELECT sum(1) AS ordr,
            t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"
          FROM 
              (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
              FROM sampleData
              GROUP BY DAY) t1
            INNER JOIN 
              (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
              FROM sampleData
              GROUP BY DAY) t2 
            on t1.day >= t2.day                                                                                                                                
            GROUP BY t1.day, t1.count
            ORDER BY t1.day
            ) t_2
        on t.ordr = t_2.ordr + 2) t_3
         on t.ordr = t_3.ordr + 1 
    left outer join
         (select t.*,
           (coalesce(t_1.count, 0) - coalesce(t_2.count, 0)) * 100.0 as "percentage increase in count in %"
    from  
    (
      SELECT sum(1) AS ordr,
      t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"

      FROM 
        (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
        FROM sampleData
        GROUP BY DAY) t1
        INNER JOIN 
        (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
        FROM sampleData
        GROUP BY DAY) t2 
        on t1.day >= t2.day                                                                                                                                
        GROUP BY t1.day, t1.count
        ORDER BY t1.day
        )t 
      left outer join
         (
          SELECT sum(1) AS ordr,
            t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"
          FROM 
            (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
            FROM sampleData
            GROUP BY DAY) t1
          INNER JOIN 
            (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
            FROM sampleData
            GROUP BY DAY) t2 
          on t1.day >= t2.day                                                                                                                                
          GROUP BY t1.day, t1.count
          ORDER BY t1.day
          )t_1
       on t.ordr = t_1.ordr + 1 left outer join
         (
          SELECT sum(1) AS ordr,
            t1.id,t1.day, t1.count, SUM(t2.count) as aggregate, (SUM(t2.count)-t1.count)/(sum(1)-1) as "average_number_of_votes_per_previous_period_days"
          FROM 
              (SELECT  id, date(FROM_UNIXTIME( timecode))  AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
              FROM sampleData
              GROUP BY DAY) t1
            INNER JOIN 
              (SELECT date(FROM_UNIXTIME( timecode) ) AS day,(FROM_UNIXTIME( timecode)) AS original, COUNT(1)  as 'count'
              FROM sampleData
              GROUP BY DAY) t2 
            on t1.day >= t2.day                                                                                                                                
            GROUP BY t1.day, t1.count
            ORDER BY t1.day
            ) t_2
        on t.ordr = t_2.ordr + 2) t_4
        on t_0.ordr = t_4.ordr + 2

Answer 1

我已经在这个小提琴中插入了你的查询,我现在看到了问题.您的查询超过8000个字符(确切地说是8423),我没有在结果面板上显示该消息.基本上,这是我在之前没有注意到的SQL Fiddle中的显示错误(所以,感谢您引起我的注意!).

同时,您可以尝试删除一些字符,使其符合8000个字符的限制.