谷歌应用脚本数组答案

【问题标题】：Google App Script arrays谷歌应用脚本数组
【发布时间】：2015-12-06 12:15:16
【问题描述】：

我正在研究 Google Script，并且正在测试创建二维数组的不同方法。我创建了一个这样的数组：

  var codes = new Array(6);
  for (var i = 0; i < 6; i++) {
  codes[i] = new Array(4);
  }
  codes[0][0]="x"; 
  codes[0][1]="x"; 
  codes[0][2]="x"; 
  codes[0][3]="x";
  codes[1][0]="x";
  codes[1][1]="x";
  codes[1][2]="x"; 
  codes[1][3]="x";
  codes[2][0]="x";
  codes[2][1]="x";
  codes[2][2]="x";
  codes[2][3]="x";
  codes[3][0]="x";
  codes[3][1]="x";
  codes[3][2]="x";
  codes[3][3]="x";
  codes[4][0]="x";
  codes[4][1]="x";
  codes[4][2]="x";
  codes[4][3]="x";
  codes[5][0]="x";
  codes[5][1]="x";
  codes[5][2]="x";
  codes[5][3]="x";

而且它工作正常。

我阅读了以下链接 here、here 和 here。

但是当我这样做时：

var codes = new Array(6);
  for (var i = 0; i < 6; i++) {
  codes[i] = new Array(4);
  }
codes[0]=["x","x","x","x"];
codes[1]=["x","x","x","x"];
codes[2]=["x","x","x","x"];
codes[3]=["x","x","x","x"];
codes[4]=["x","x","x","x"];
codes[5]=["x","x","x","x"];

没用，所以我这样尝试：

var codes = new Array([["x","x","x","x"],["x","x","x","x"],["x","x","x","x"],["x","x","x","x"],["x","x","x","x"],["x","x","x","x"]]);

它也没有用。

当代码不起作用时，我没有收到任何错误，只是没有显示值。

我做错了什么？它看起来是相同的代码，并且在许多文档中都推荐了这两种不工作的方式。

【问题讨论】：

标签： javascript arrays multidimensional-array google-apps-script

【解决方案1】：

W3schools 表示不需要使用new Array()。为了简单、可读性和执行速度，请使用文字方法 ex:

var animals = ["cat", "rabbit"];

您的代码无法正常工作的原因是您在循环内等于codes，并且在循环范围“代码”结束后仅获取最后一组数组。相反，您应该将这些数组推送到codes。

var codes = [];
for (var i = 0; i < 6; i++) {
  codes.push([i]);
}
console.log(codes)
codes[0]=["x","x","x","x"];
codes[1]=["x","x","x","x"];
codes[2]=["x","x","x","x"];
codes[3]=["x","x","x","x"];
codes[4]=["x","x","x","x"];
codes[5]=["x","x","x","x"];

【讨论】：

我没有看到 new Array() 是一个糟糕的解决方案。谢谢。

【解决方案2】：

更好的是，两个 for 循环来创建双数组：

var codes = [], // Initiate as array, in Javascript this is actually fastre than using new (I don't know any cases you should use new)
    rows = 6,
    columns = 6;

for (var i = 0; i < rows; i++){
    codes.push([]); // Initiate 
    for (var j = 0; j < columns; j++){
        codes[i][j] = 'x';
    }
}

其他想法，使用正确的列预先启动一个数组，然后复制：

var arrTemp = [],
    codes = [],
    rows = 6,
    columns = 6;

for (var j = 0; j < columns; j++)
    arrTemp[i] = 'x';

for (var i = 0; i < rows; i++)
    codes.push( arrTemp.slice(0) ); // If you just push the array without slice it will make a reference to it, not copy

使用 'x's 预初始化数组的其他方式：

arrTemp = Array.apply(null, Array(columns)).map(function () {return 'x'});

【讨论】：