Question

查询

SELECT strftime('%W', [DatumEintrag], 'weekday 1') Woche, 
       SUM([FOO]), Count(Distinct [DatumEintrag]) as Tage 
FROM MainData
Where [BAR] LIKE 'BC%'
Group By Woche

导致出现类似的表

Woche | FOO | Tage  
41    | 4218| 3  
42    | 5624| 4  
43    | 7030| 5  
44    | 2812| 2

我如何仅获取Tage> = 4的数据。

SELECT strftime('%W', [DatumEintrag], 'weekday 1') Woche, SUM([FOO]), Count (Distinct [DatumEintrag]) as Tage 
FROM MainData
Where [BAR] LIKE 'BC%' AND Tage >= 4
Group By Woche

导致“滥用总计：Count（）”

Answer 1

使用having子句

 public void renameFile(ActionEvent actionEvent) {
    TextInputDialog dialog = new TextInputDialog("Rename");
    dialog.setTitle("Rename");
    dialog.setHeaderText("Rename File");
    dialog.setContentText("New name");
    Optional<String> res = dialog.showAndWait();
    if (res.isPresent()) {
        try {
            Path paths = Paths.get("client_storage/" + ClientListView.getSelectionModel().getSelectedItem());
            Files.move(paths, paths.resolveSibling(res.get()));
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

Answer 2

一种选择是使用嵌套查询：

SELECT *
FROM(
SELECT strftime('%W', [DatumEintrag], 'weekday 1') Woche, 
       SUM([FOO]), 
       Count (Distinct [DatumEintrag]) as Tage 
FROM MainData 
Where [BAR] LIKE 'BC%' Group By Woche)t
WHERE Tage>=4

Answer 3

SQLite允许您在having子句中使用列别名。所以你可以这样写：

select strftime('%W', DatumEintrag, 'weekday 1') as Woche, 
       sum([FOO]), 
       count(distinct DatumEintrag) as Tage 
from MainData
where [BAR] like 'BC%' 
group by Woche
having tage > 4

计为列的条件

3 个答案: