Question

我有一张表，其中第一行是天数。第二和第三行是这些天数内不同公司的日常生产。它是这样的。

我的要求是在2D数组中获取行数和特定公司生产详细信息行。

这是我的代码。

function GetData()
 {
  myFunction("Konabay");
 }

 function myFunction(Company)
 {
   var dSheet = SpreadsheetApp.getActiveSpreadsheet();
   var aSheet= dSheet.getSheetByName("Data");
   var range = aSheet.getDataRange();
   var values = range.getValues();
   var VNeeded=[];

   for (var i = 0; i < values.length; i++) 
   {
   //var row = "";
   var companyCheck="No";

   for (var j = 0; j < values[i].length; j++) 
     {
       if (values[i][j]) 
         {
           if (i==0)//Need the first row(days) in my new array always
           {
             VNeeded[i,j]=values[i][j];
             Logger.log("VNeeded i=0 (0,3) = "+ VNeeded[0,3]);
             Logger.log("VNeeded i=0(1,3)= "+ VNeeded[1,3]);
             //row = row + values[i][j];
           }
           else if (i>0)//From second row onwards check for the company needed
           {
             if(values[i][0]==Company)
             {
               companyCheck="Yes";
             }
             if(companyCheck=="Yes")
             {
               VNeeded[i,j]=values[i][j];
               Logger.log("VNeeded i>0 (0,3) = "+ VNeeded[0,3]);
               Logger.log("VNeeded i>0 (1,3)= "+ VNeeded[1,3]);
               //row = row + values[i][j];
             }
           }

         }
      //row = row + ",";
     }
      //Logger.log(row);

 }
}

我正在寻找像这样的东西

VNeeded [] = [10,2.5]，[30,4]，[40,5]，[30,5]

但我得到的是

VNeeded [] = [2.5,2.5]，[4,4]，[5,5]，[5,5]

检查发生了什么我使用了Logger.log，它给出了这个结果。

[18-01-28 22:36:30:121 IST] VNeeded i=0 (0,3) = undefined
[18-01-28 22:36:30:122 IST] VNeeded i=0(1,3)= undefined
[18-01-28 22:36:30:123 IST] VNeeded i=0 (0,3) = undefined
[18-01-28 22:36:30:123 IST] VNeeded i=0(1,3)= undefined
[18-01-28 22:36:30:124 IST] VNeeded i=0 (0,3) = undefined
[18-01-28 22:36:30:124 IST] VNeeded i=0(1,3)= undefined
[18-01-28 22:36:30:125 IST] VNeeded i=0 (0,3) = 30
[18-01-28 22:36:30:126 IST] VNeeded i=0(1,3)= 30
[18-01-28 22:36:30:126 IST] VNeeded i=0 (0,3) = 30
[18-01-28 22:36:30:127 IST] VNeeded i=0(1,3)= 30
[18-01-28 22:36:30:127 IST] VNeeded i>0 (0,3) = 30
[18-01-28 22:36:30:128 IST] VNeeded i>0 (1,3)= 30
[18-01-28 22:36:30:128 IST] VNeeded i>0 (0,3) = 30
[18-01-28 22:36:30:129 IST] VNeeded i>0 (1,3)= 30
[18-01-28 22:36:30:130 IST] VNeeded i>0 (0,3) = 30
[18-01-28 22:36:30:130 IST] VNeeded i>0 (1,3)= 30
[18-01-28 22:36:30:131 IST] VNeeded i>0 (0,3) = 5
[18-01-28 22:36:30:132 IST] VNeeded i>0 (1,3)= 5
[18-01-28 22:36:30:132 IST] VNeeded i>0 (0,3) = 5
[18-01-28 22:36:30:133 IST] VNeeded i>0 (1,3)= 5

那我在哪里弄错了？当我将值设置为VNeeded [1,3]时，相同的值也被设置为另一个维度（VNeeded [0,3]）。怎么可能？

请帮我解决这个问题。

我如何获得理想的结果。我做错了什么

我的测试表的链接是：https://docs.google.com/spreadsheets/d/15hOjIr11Czf1KtrvEgo00pvE0kc7fMfSvRhmrn1tPBM/edit?usp=sharing

请帮忙

Logger.log(JSON.stringify(VNeeded)) gives this result :  ["Konabay",2.5,4,5,5]

Answer 1

由于comma operator，JavaScript VNeeded[i,j]评估为VNeeded[j]。

请改用VNeeded[i][j]。

我必须将代码粘贴到https://www.typescriptlang.org/play/以实现此问题。

Answer 2

您对数组位置VNeeded[i,j]的引用并未提供您期望的结果。实际上，您将位置整数数组传递给VNeeded，这将解析为最后一个数字。即VNeeded[j]

请改用VNeeded[i][j]。

VNeeded[i][j]=values[i][j];

<强>无论其！

这将在您的代码中引发错误，因为＆＃34;数组＆＃34;在位置i，你试图把价值放在位置＆＃39; j＆＃39;不存在。因此，如果需要，您需要使用数组预填充该位置：

VNeeded[i] = VNeeded[i] || [];
VNeeded[i][j] = values[i][j];

<强>无论其！

获得您需要的结果：

[10,2.5],[30,4],[40,5],[30,5];

您需要在行条件检查的两个部分中将VNeeded[i][j]切换为VNeeded[j][i]。

VNeeded[j] = VNeeded[i] || [];
VNeeded[j][i] = values[i][j];

因为您正在尝试实际上是对数据透视图的过滤器，所以通过执行数据透视可以大大简化您的代码，首先，检查列表中公司的位置并以这种方式返回结果对：

function getData(company) {

  var dSheet = SpreadsheetApp.getActiveSpreadsheet();
  var aSheet= dSheet.getSheetByName("Data");
  var range = aSheet.getDataRange();
  var values = range.getValues();
  var vNeeded=[];
  var pivotedData = []; // Used to carry the transposed data

  for (var i = 0, l = data.length; i < l; i += 1) {
    for (var j = 0, k = data[i].length; j < k; j += 1) {    
      pivotedData[j] = pivotedData[j] || []; // allows for entry into preexisting array.
      pivotedData[j][i] = data[i][j];
    }
  }

  var companyIndex = pivotedData[0].indexOf(company); // find company in the list

  if (companyIndex < 0) return; // There is no matching company, returns undefined

  for (var i = 1, l = pivotedData.length; i < l; i += 1) { 
    vNeeded[i-1] = [pivotedData[i][0], pivotedData[i][companyIndex]]; // [expected,found]
  }
  return vNeeded;
}

Answer 3

您可以使用此函数展平数组：

function flattenArray(array){
var flatArray = []
  for (var i = 0; i < array.length; i++) {
    flatArray.push(array[i][0])
  }
  return flatArray
}

在Google Script中使用多维数组

3 个答案: