Question

我正在尝试使用表格数据制作一个对象

 var cond = [
        { 
    "locality": "Dwarka",
    "Created ->OFP": "4.73",
    "OFP -> Picked": "2.16",
    "Picked - > Delivery": "14.91",
    "Over_All_TAT": "22.65",
    "Total_Order": "159" 
  },
  {
    "locality": "IIT & AIIMS",
    "Created ->OFP": "2.56",
    "OFP -> Picked": "9.74",
    "Picked - > Delivery": "19.41",
    "Over_All_TAT": "32.61",
    "Total_Order": "54"
  },
  {
    "locality": "Gk1 & Nehru Place",
    "Created ->OFP": "2.39",
    "OFP -> Picked": "6.87",
    "Picked - > Delivery": "14.06",
    "Over_All_TAT": "24.13",
    "Total_Order": "31"
  },
  {
    "locality": "Mayur Vihar",
    "Created ->OFP": "3.52",
    "OFP -> Picked": "1.24",
    "Picked - > Delivery": "15.3",
    "Over_All_TAT": "21.03",
    "Total_Order": "33"
  },
  {
    "locality": "Snapdeal_Mundhka",
    "Created ->OFP": "17.32",
    "OFP -> Picked": "4.65",
    "Picked - > Delivery": "100.32",
    "Over_All_TAT": "123.16",
    "Total_Order": "31"
  }
  ]

我知道密钥名称不符合规则，但这些是列名。我可以访问第二列说创建 - ＆gt; OFP 使用索引（例如cond[0][1]）。

我只能访问cond[i]如何访问内部块元素。

当我使用R时，json对象是使用jsonlite包创建的，所以，我有局限性。

Answer 1

您可以像这样访问内部块：

cond[i]['Created ->OFP']

循环示例：

＆＃13;

var cond = [{
  "locality": "Dwarka",
  "Created ->OFP": "4.73",
  "OFP -> Picked": "2.16",
  "Picked - > Delivery": "14.91",
  "Over_All_TAT": "22.65",
  "Total_Order": "159"
}, {
  "locality": "IIT & AIIMS",
  "Created ->OFP": "2.56",
  "OFP -> Picked": "9.74",
  "Picked - > Delivery": "19.41",
  "Over_All_TAT": "32.61",
  "Total_Order": "54"
}, {
  "locality": "Gk1 & Nehru Place",
  "Created ->OFP": "2.39",
  "OFP -> Picked": "6.87",
  "Picked - > Delivery": "14.06",
  "Over_All_TAT": "24.13",
  "Total_Order": "31"
}, {
  "locality": "Mayur Vihar",
  "Created ->OFP": "3.52",
  "OFP -> Picked": "1.24",
  "Picked - > Delivery": "15.3",
  "Over_All_TAT": "21.03",
  "Total_Order": "33"
}, {
  "locality": "Snapdeal_Mundhka",
  "Created ->OFP": "17.32",
  "OFP -> Picked": "4.65",
  "Picked - > Delivery": "100.32",
  "Over_All_TAT": "123.16",
  "Total_Order": "31"
}]
var val;
cond.forEach(function(el) {
  val = el['Created ->OFP'];
  console.log(val);
});

＆＃13;

Answer 2

据我了解，您希望能够做的是：

cond[0][1];   // "4.73"

你不能这样做，因为cond[0]是和对象，并且不保证对象中键的顺序。因为Created ->OFP是您输入的第二个键并不意味着它实际上是对象中的第二个键。您可以查看Object.keys(cond[0])，可能的密钥将以相同的顺序返回，但不能依赖此。

你需要做的是将数组中的对象（不是其他人指出的JSON）转换为数组本身。所以你最终得到的结果是：

[
    [
        "Dwarka",
        "4.73",
        "2.16",
        "14.91",
        "22.65",
        "159"
    ],...
]

然后你可以这样做：

cond[0][1];   //and get the first row and second column.

要将原始数据转换为数组数组，您可以执行以下操作：

// Note: this is the order **you** want the columns to be iterated in
var columns = ["locality",
    "Created ->OFP",
    "OFP -> Picked",
    "Picked - > Delivery",
    "Over_All_TAT",
    "Total_Order"];

var arrayOfArrays = cond.map(function (i) {
    return columns.map(function (prop) {
        return i[prop];
    });
});

console.log(arrayOfArrays[0][1]);   // logs 4.73

Fiddle

Answer 3

您可以像这样访问您的元素：

cond[i]["Created ->OFP"]; // "4.73"

如果你想使用像1,2这样的索引，你有两个选择：使用数组而不是对象，它将如下所示：

 var cond = [
        ["Dwarka", "4.73", "2.16", "14.91", "22.65", "159" ],
        [ ... ]
  }

或者您可以在对象上使用索引1,2 ...，如下所示：

var cond = [
        { 
    "0": "Dwarka",
    "1": "4.73",
    "2": "2.16",
    "3": "14.91",
    "4": "22.65",
    "5": "159" 
  },
  {...}
]

但除非您将其作为元素添加到对象中，否则您将无法找到属性length。

我认为最适合您的解决方案是保持您的对象不变，并使用循环for(key in cond)

for(var i=0; i<cond.length; i++) {
    for(var key in cond[i]) {
       console.log(cond[i][key]);
    }
}

如何使用索引表示法访问对象

3 个答案: