Question

我的用例就是这样。

我有餐厅。
一个餐厅有很多餐厅。
每个餐厅项目都有一个评分值。

我希望将餐厅以及该餐厅中评分最高的商品带到一起。

我的伪查询将是这样的。

<html ng-app="app">
<head>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
</head>
  <body>
    
    <div ng-controller="createCtrl as main">
      <h1>create Ctrl</h1>
    <input ng-model="title" placeholder="title" /><br/>

      <input ng-model="name" placeholder="name" /><br/>

      <input ng-model="post" placeholder="post" /><br/>

      <input ng-model="tag" placeholder="tag" /><br/>
      <button ng-click="display()">  SAVE DATA </button>     
    </div>

<br/> <hr/>

    <div ng-controller="newsCtrl">
      <h2>news Ctrl </h2>
      <button ng-click="getData()">  GET DATA </button>

      <p> title : {{array.title}} </p>
      <p> name : {{array.name}} </p>
      <p> post : {{array.post}} </p>
      <p> tag : {{array.tag}} </p>
      
    </div>
    
  </body>
</html>

在RESTAURANTS表中。

SELECT * FROM RESTAURANTS 
WHERE RESTAURANT_IDS = 1,2,3,4,5 
LEFT JOIN RESTAURANT_ITEMS 
WHERE ratings = MAX(ratings)

在RESTAURANT_ITEMS表中

id|name
-------------
1 |McDonalds
2 |Pizzahut
3 |Buger king
4 |KFC

所需的输出为

id|name      |rating| FK_RES_ID
----------------------------
1 |Pizza     |5     |2
2 |Lava cake |4     |2
3 |Veg burger|3     |4
4 |Chicken b |5.6   |4
5 |Ice cream |4.3   |1
6 |Burger    |5     |1

Answer 1

我想这就是您要寻找的。尝试并确认。：）

SELECT 
      res.ResName
      ,item.Name as ResItem
      ,(Select max(SubItem.rating) from RESTAURANT_ITEMS SubItem where res.RestaurantID= SubItem.RestaurantID) as Rating
FROM 
      RESTAURANTS res
LEFT JOIN 
      RESTAURANT_ITEMS item on res.RestaurantID = item.RestaurantID
WHERE 
      res.Restaurant_ID in (1,2,3,4,5)

问题后的编辑部分也已更新

SELECT 
      res.Name as [Restaurant Name]
      ,item.Name as [Top Item]
FROM 
      RESTAURANTS res
LEFT JOIN 
      RESTAURANT_ITEMS item on res.RestaurantID = item.RestaurantID 
                            and rating = (select max(subItems.rating) from RESTAURANT_ITEMS subItems where subItems.RestaurantID = res.RestaurantID)
WHERE 
      res.Restaurant_ID in (1,2,3,4,5)

希望这对您有所帮助：）

Answer 2

您可以在餐厅的派生表中获得最高的评分值。
将其重新加入主表以获得所需的输出

尝试以下操作：

SELECT r.id, 
       r.name, 
       ri.name AS top_item 
FROM RESTAURANTS AS r 
LEFT JOIN (SELECT FK_RES_ID, MAX(rating) AS max_rating 
           FROM RESTAURANT_ITEMS 
           GROUP BY FK_RES_ID) AS dt 
  ON dt.FK_RES_ID = r.id 
LEFT JOIN RESTAURANT_ITEMS AS ri ON ri.FK_RES_ID = dt.FK_RES_ID AND 
                                    ri.rating = dt.max_rating 

WHERE r.id IN (1,2,3,4)

SQL如何从内部联接获取最大值的行

2 个答案: