Question

我正在报道一家酒店，我需要总结一些关于他们预订的董事会的信息。

现在我的输出看起来像这样：

SELECT meal.id, meal.title,
    SUM(CASE WHEN meal.id >= 600 THEN '1' ELSE '0' END) as NoofMeals
    FROM meal
    JOIN itinerary i ON i.id = meal.journeyid
    WHERE i.arrival BETWEEN '2015-01-01' AND '2015-01-01'
    GROUP BY meal.title

我的输出如下：

ID    | title                  | NoofMeals

608   | Dinner                 | 15
621   | Breakfast              | 478
648   | Lunch                  | 74
649   | Breakfast Box          | 5
655   | Lunch-Box              | 7
659   | American breakfast     | 73
671   | Continental breakfast  | 102
674   | All Inclusive          | 312
689   | All Inclusive from 3pm | 7
693   | Picnic                 | 47

我想要的输出看起来像是这样，并且汇集了不同的价值，如所有的早餐（早餐，早餐盒，美式早餐和欧式早餐）或所有全包（全包和全包从下午3点）。不幸的是，ID不是连续的。

title                  | NoofMeals

Dinner                 | 15
Breakfast              | 658
Lunch                  | 81
All Inclusive          | 319
Picnic                 | 7

我不再需要ID了，我只需要能够重命名用餐的标题。

我希望你能帮助我，请放轻松我，我对MySQL很陌生。

提前谢谢！

Answer 1

“快速而肮脏”的解决方案将使用字符串函数：

SELECT
  case
    when meal.title like '%Breakfast%' then 'Breakfast'
    when meal.title like '%All inclusive%' then 'All Inclusive'
    else meal.title
  end as title,
  ...
...
GROUP BY
  case
    when meal.title like '%Breakfast%' then 'Breakfast'
    when meal.title like '%All inclusive%' then 'All Inclusive'
    else meal.title
  end

虽然正确的解决方案是添加包含膳食类别的列：

id | title                  | category
---|------------------------|--------------
1  | Breakfast              | Breakfast
2  | American breakfast     | Breakfast
3  | Continental breakfast  | Breakfast
4  | All Inclusive          | All Inclusive
5  | All Inclusive from 3pm | All Inclusive

并按meal.category代替meal.title加入特定膳食和小组。

使用字符串列选择特定输出

1 个答案: