简体   繁体   English

连接具有多对多关系字段的两个表

[英]Joining two tables with many to many relationship fields

I'm working on google data studio and trying to fetch the data from a third-party system.我正在研究谷歌数据工作室并试图从第三方系统中获取数据。 In that case, there is a problem in the third-party system, because it doesn't allow for SQL join.在这种情况下,第三方系统存在问题,因为它不允许 SQL 加入。 Therefore, I got the data by using 2 URLs separately and joined the two tables(customer_order table & Customer_Order_Demand table).因此,我分别使用 2 个 URL 获取数据并将两个表(customer_order 表和 Customer_Order_Demand 表)连接起来。 But the problem is there is a field(Order_NO) that has many to many relationships in one table.但问题是有一个字段(Order_NO)在一个表中有多对多的关系。 According to my code, I'm getting only first value in Order_No column after joining the two tables.根据我的代码,在加入两个表后,我只在 Order_No 列中获得第一个值。 I want to display the number of orders for each Item by corresponding date.我想按相应日期显示每个项目的订单数量。


  Customer Order Demand Table                                        
+-----------------------------+ 
| id  | item     |    Date    |         
|-----|----------|------------| 
|  1  |  BLU     | 7/10/2020  |  
|  2  |  PM      | 7/20/2020  |   
|  3  |  BLU     | 7/23/2020  |          
+-----------------------------+ 


    Customer Order Table                                        
+---------------------------------------------------------+
| id  |  COD_id         |  Order_NO      |   Date         |
|-----|----------------------------------|----------------|
|  1  |      1          |  #100          |    7/10/2020   |
|  2  |      1          |  #130          |    7/10/2020   |
|  3  |      2          |  #200          |    7/20/2020   |
|  4  |      3          |  #500          |    7/23/2020   | 
+---------------------------------------------------------+

                   Expected  Table                                        
+-----------------------------------------------+
|     |  Item    |    Date    |  Order_NO       |
|-----|----------|------------|-----------------|
|  1  |  BLU     | 7/10/2020  |  #100 /#130     |
|  2  |  PM      | 7/20/2020  |  #200           |
|  3  |  BLU     | 7/23/2020  |  #500           |        
+-----------------------------------------------+



 
 var url=[
    'http://test.smartapps.lk/rest/api/selectQuery?sessionId=' + SessionId + '&maxRows=10&query=select id, Item, Date from Customer_Order_demand &output=json'
  ].join('');
  
  var url2=[
    'http://test.smartapps.lk/rest/api/selectQuery?sessionId=' + SessionId + '&maxRows=10&query=select id,Order_NO from Customer_Order &output=json'
  ].join(''); 
  
  
  var responseID2 = UrlFetchApp.fetch(url);
  var responseID3 = UrlFetchApp.fetch(url2);
   
  var table1 = JSON.parse(responseID2.getContentText());     
  var table2 = JSON.parse(responseID3.getContentText());
   
  var finalArray = [];
  
  for(var i in table1){
      for(var j in table2){

        //Joining two tables 
        if(table1[i][0]==table2[j][0]){

            // converting to json object
             var newObj = {      
             "Item": table1[i][1],
             "Date":table1[i][2],
             "Order_NO": table2[j][1] 
             };
          
         finalArray.push(newObj);
        }
      }
   
  }
  
   return finalArray;
  
  
 

  • Reduce the second table to a map :将第二个表减少map

     const table2map = table2.reduce( (mp,[codId,ordNo]) => mp.set(codId, mp.has(codId)? [...mp.get(codId),ordNo]: [ordNo] ), new Map )
  • Map table1 to a array of objects: Map table1 到对象数组:

     const finalArr = table1.map( ([id,Item,Date]) => ({Item,Date, "Order_No": table2map.get(id).join()}) )

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM