這就是你想要的嗎？ Option ExplicitPrivate Declare PtrSafe Function GetSystemMetrics Lib "user32" _(ByVal Index As Long) As LongPrivate Declare PtrSafe Function GetDC Lib "user32" (ByVal hWnd As Long) As LongPrivate Declare PtrSafe Function ReleaseDC Lib "user32" _(ByVal hWnd As Long, ByVal hDC As Long) As LongPrivate Declare PtrSafe Function GetDeviceCaps Lib "gdi32" _(ByVal hDC As Long, ByVal Index As Long) As LongPrivate Sub Workbook_Open() Dim ScreensWidth As Single Dim ScreensHeight As Single ScreensWidth = GetSystemMetrics(78) ScreensHeight = GetSystemMetrics(79) ScreensWidth = ConvertPixelsToPoin

考慮在將數據聚合到日水平后，通過窗口函數計算總運行時間。由于使用單個條件進行聚合，FILTER條件可以轉換為基本WHERE： SELECT daily, SUM(total_balance) OVER (ORDER BY daily) AS total_value_per_dayFROM ( SELECT DATE_TRUNC('DAY', (time)) AS daily, SUM(balance) AS total_balance FROM tbl WHERE is_spent_column IS NULL GROUP BY 1) AS daily_aggORDER BY daily

通常，不可能使用SELECT list at WHERE子句中定義的別名。 另一個限制是不能在WHERE子句中使用窗口函數，因此以下操作也會失敗： SELECT innings_pitched, birth_year, player, NTILE(100) OVER(PARTITION BY innings_pitched) pctl, num_seasonsFROM pitching_cteWHERE season = 2021 AND (NTILE(100) OVER(PARTITION BY innings_pitched)) >= 98; 解決方案是使用另一個級別的cte： WITH pitching_cte AS ( SELECT player, player_id, season, CAST(strftime('%Y', birth_date) AS TEXT) AS birth_year, COUNT(*) AS num_seasons, printf("%i.%i", outs/3, outs % 3) AS innings_pitched FROM pitching INNER JOIN player USING(player_id) GROUP BY player), cte AS ( SELECT innings_pitched, birth_year,

實際上，這里不需要窗口函數，因為您可以按desc順序對數據進行排序，然后使用limit(1)返回第一條記錄。 但為了便于練習，您可以使用如下窗口函數： import org.apache.spark.sql.functions.{col, row_number}import org.apache.spark.sql.expressions.Windowval windowSpec = Window.orderBy(col("insert_dttm").desc)val lastSessionId = df.withColumn("row_number", row_number.over(windowSpec)).filter("row_number=1").first.getString(0)

您需要使用如下子查詢： WITH T AS (SELECT CASE WHEN SUM(ISNULL(epf.emp_contribution, 0)) OVER(PARTITION BY epf.emp_no, epf.FinancialYear ORDER BY(epf.pcm_year * 100 + pcm_month)) + SUM(ISNULL(epf.vpf, 0)) OVER(PARTITION BY epf.emp_no, epf.FinancialYear ORDER BY(epf.pcm_year * 100 + epf.pcm_month)) < 3000 THEN SUM(ISNULL(epf.emp_contribution, 0)) OVER(PARTITION BY epf.emp_no, epf.FinancialYear ORDER BY(epf.pcm_year * 100 + pcm_month)) + SUM(ISNULL(epf.vpf, 0)) OVER(PARTITION BY epf.emp_no, epf.FinancialYear ORDER BY(epf.pcm_year * 100 + epf.pcm_month)) ELSE NULL END AS CALCFROM EmpPFContributionTest epf

正如@Larnu指出的，您需要PARTITION BY和CASE表達式來檢查負數。 Try this: -- To check for negative numbers we need another select in order to be able to use the result calculated by LAGSELECT TOP(200) [value], ( CASE WHEN difference < 0 THEN NULL ELSE difference END ) AS difference, readDate, symbol, description, VINFROM ( SELECT -- Added PARTITION BY A.Value as value,A.Value-LAG(A.Value, 1) OVER (PARTITION BY VIN ORDER BY readDate) as difference, readDate, symbol, description, VIN FROM Table1 as V JOIN Table2 ON Table2.VX_Id=V.VX_Id JOIN Table3 ON Table3.FX_Id =Table2.FX_Id JOIN Table4 on Table3.FX_Id=Table4.FX

當然，排名函數滿足這個定義。 當不使用partition by時，整個表就是“表行集合”。 與當前行的關系是當前行中相對于其他行的值。 它遵循定義。

合并兩個表日期，引導它們，然后使用相關子查詢 with cte as(SELECT employeeid,effectivefrom from EMPLOYEEROLEunion select employeeid,effectivefrom from employeesalary),cte1 as(select employeeid,effectivefrom, coalesce( date_sub(lead(effectivefrom) over (partition by employeeid order by effectivefrom),interval 1 day) , now()) nextefffrom cte)select *, datediff(nexteff,effectivefrom) + 1 diff, (select grade from employeerole e where e.effectivefrom <= cte1.effectivefrom order by e.effectivefrom desc limit 1) grade, (select salary from employeesalary e where e.effectivefrom <= cte1.nexteff order by e.effectivefrom

我可以給你一個簡單的例子，你可以使用rank。 假設您有student_marks表，并且每個類有一個測試。（1級，2級。。為簡單起見，所有數據都在一個表中考慮） 如果你現在想給每個班級的學生一個結果（最高分產生第一個數字，以此類推），那么你可以使用rank，如下所示： Rank() over (partition by class order by marks desc) -- student_rank_in_class Here, Partition by class-意味著每個類應該給出排名。 Order by marks desc-意味著在每一個班級中，得分最高的學生（marks desc）應該得到第一名，然后第二名應該得到第二名，依此類推。 希望這個解釋能給你一些關于rank窗口函數的提示。

不幸的是，現在還不可能。 我們認為沒有足夠廣泛的需求來保證該功能的維護成本。 你可以參考這里和這里

>我通過使用JPA nativeQuery=true找到了解決方案 @Query(value = "select " + "b.q_stop_id as q_stop_id," + "b.q_work_day as q_work_day," + "b.q_work_end as q_work_end," + "b.q_mc_num as q_mc_num," + "b.q_mc_name as q_mc_name," + "b.Hnbn as q_hnbn," + "b.q_koku as q_koku," + "b.q_stop_st_time as q_stop_st_time," + "b.q_stop_en_time as q_stop_en_time," + "b.q_stop_time_s as q_stop_time_s," + "b.q_stop_maj_code as q_stop_maj_code," + "b.q_stop_major as q_stop_major," + "b.q_stop_num as q_stop_num," + "b.q_stop_minor as q_stop_minor," + "b.q_pause_flag as q_pause_flag " + "from" + "(" + "select ROW_NUMBER() OV

您確定此處需要窗口功能嗎？您定義的窗口通過id對數據進行分區，我假設它是唯一的，因此該窗口生成的每個組將只有一行。看起來您需要一個覆蓋整個數據幀的窗口，這意味著您實際上不需要一個窗口。如果您只想添加一個最大值的列，可以使用原始數據幀上的聚合獲得最大值，并與之交叉連接： val maxDF = df1.agg(max("salary").as("top"))val res = df1.crossJoin(maxDF)

WITHcte AS ( SELECT release_year, COUNT(film_id) cnt, DENSE_RANK() OVER (ORDER BY COUNT(film_id) DESC) drnk FROM film GROUP BY release_year )SELECT release_year, cntFROM cteWHERE drnk = 1;

只有當他們的第二次登錄連續時，才會要求您計數。第一個查詢通過只計算差為1的第二行來實現這一點。 WHERE t.rn = 2AND t.difference = 1 第二個查詢將采用任何連續登錄。 Demonstration 請注意，這兩個查詢都需要group by tp.all_players。在某些模式下，MySQL有時會為您推斷分組。但不要指望它，其他數據庫也不會。請參閱分組的MySQL處理。在學習SQL時考慮在ANSI模式中運行MySQL以使其更好地遵循標準。

注意，count只統計non-null項，分組只由partitionBy子句定義，而不是由orderBy子句定義。 指定排序列時，默認窗口范圍為（根據文檔） (rangeFrame, unboundedPreceding, currentRow) 所以你的窗口定義實際上是 w = (Window().partitionBy("I_id","p_id") .orderBy(F.col("xyz")) .rangeBetween(Window.unboundedPreceding, Window.currentRow) ) 因此窗口只包含當前行中從xyz = -infinity到值xyz的行。這就是為什么第一行的計數為零，因為它統計從xyz = -infinity到xyz = null的non-null項，即數據幀的前兩行。 對于xyz = 2所在的行，計數包括xyz = -infinity到xyz = 2的non-null項，即前四行。這就是為什么計數為2，因為non-null項是1和2。