Question

我很困惑在SQL中使用AS查看三种情况，这是我的表

id | Fruit
___________
1  | Apel
2  | Orange
3  | Apel
4  | Manggo
___________

这是我的第一个查询：

SELECT
COUNT(fruit) AS apelfruit
FROM fruit_table
WHERE fruit = 'Apel'

输出：

apelfruit
_________
    2

但是当我这样做时：

SELECT
COUNT(fruit) AS apelfruit,
COUNT(fruit) AS orangefruit,
COUNT(fruit) AS manggofruit
FROM fruit_table
WHERE fruit = 'Apel'
AND fruit = 'Orange'
AND fruit = 'Manggo'

输出为零（0）：

 apelfruit | orangefruit | manggofruit
 ____________________________________
       0            0            0

我有点困惑什么错。任何人都可以帮助我，我将不胜感激。谢谢。

Answer 1

尝试以下方式

SELECT
sum(case when fruit = 'Apel' then 1 else 0 end) AS apelfruit,
sum(case when fruit = 'Orange' then 1 else 0 end) AS orangfruit,
sum(case when fruit = 'Manggo' then 1 else 0 end) AS manggofruit
FROM fruit_table

Answer 2

为简洁起见，我更喜欢在这里使用COUNT：

SELECT
    COUNT(CASE WHEN fruit = 'Apel'   THEN 1 END) AS apelfruit,
    COUNT(CASE WHEN fruit = 'Orange' THEN 1 END) AS orangfruit,
    COUNT(CASE WHEN fruit = 'Manggo' THEN 1 END) AS manggofruit
FROM fruit_table;

正如@nicomp指出的那样，以这种方式进行数据透视查询的局限性在于，输出中出现的列是硬编码的，并且必须添加更多代码以进行新的条件计数。但是，解决此问题的唯一真正方法是使用动态SQL。

Answer 3

条件是您遇到的问题是水果不能等于“ apel”，“ ”和它应该是“ apel” 或“ apel”

由于只计算水果，将始终计算所有水果，因此，您还需要使用一个条件进行计数，在此解决方案中，根据条件求和1或0：

  this.selectedFilters$ = merge(
      this.filter$.pipe(
        switchMap((allFilters) => {
          return of({
            itemId: allFilters[0].itemId,
            filterOptions: this.getDefaultFilterOptions(allFilters)
          });
        }),
      ),
      this.newFilter$.pipe(
        switchMap((newFilter: FilterOption) => {
          return of(this.getNewFilterOptions(newFilter, this.currentFilters));
        }),
      )
    ).pipe(tap((filters) => this.currentFilters = filters));
  }

  // LESS IMPORTANT CODE BELOW...

  private getNewFilterOptions(newFilter: FilterOption, currentFilters: ItemFilter): ItemFilter {
    const changedFilter = currentFilters.filterOptions.findIndex(f => f.filterId === newFilter.filterId);
    currentFilters.filterOptions[changedFilter] = newFilter;

    return currentFilters;
  }

  private getDefaultFilterOptions(filters: ItemFilter[]): FilterOption[] {
    return filters.reduce((acc, item) => acc.concat(item.filterOptions[0]), []);
  }

SQL中的三个别名

3 个答案: