Mysql 使用字符串列选择特定输出

我正在做一份关于一家酒店的报告，我需要总结一些关于他们预订的董事会的信息

现在，我的输出如下所示：

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

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

我的输出如下所示：

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

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

我想要的输出是这样的，并将不同的值放在一起，比如所有的早餐（早餐、早餐盒、美式早餐和欧式早餐）或所有的全包（从下午3点开始全包和全包）。不幸的是，ID不是顺序的

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餐名

我希望你能帮助我，请对我放松点，我对MySQL还很陌生

提前谢谢你

一个“快速而肮脏”的解决方案将利用字符串函数：

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

适当的解决方案是添加一列，其中包含膳食类别：

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

并通过

deal.category

而不是

deal.title

加入特定的膳食和分组。一个“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

适当的解决方案是添加一列，其中包含膳食类别：

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

---

并通过

膳食加入特定的膳食和分组。类别

而不是

膳食。标题

请在所需输出中发布您的餐桌数据是野餐7或47？应该是47，抱歉！请在所需输出中发布您的表的数据是野餐7还是47？应该是47，对不起！非常感谢，它工作得很好！！我想我在那里坐得太久了，太想念森林了！谢谢：）万分感谢，它工作得很好！！我想我在那里坐得太久了，太想念森林了！谢谢：）