Question

我有一个数组：

results = [
  {"Created Date ": "20181012", "Created By ": "A", "Job Number ": "001", "Department": "FS"},
  {"Created Date ": "20181012", "Created By ": "B", "Job Number ": "002", "Department": "DS"},
  {"Created Date ": "20181012", "Created By ": "C", "Job Number ": "004", "Department": "FS"}
]

每个值都会在以下某个时间出现。（不更改空白字符和列顺序，但列名是固定的）

results = [
  {"Created By": "A", "Department": "FS", "Created Date": "20181012", "Job Number": "001" },
  {"Created By": "B", "Department": "DS", "Created Date": "20181012", "Job Number": "002"},
  {"Created By": "C", "Department": "FS", "Created Date": "20181012", "Job Number": "004"}
]

现在，我只想获取一个新数组，其中包含3列的值，即“创建者”，“创建日期”和“作业编号”，且具有特定顺序。期望值应为：

results = [
  {"20181012", "A", "001"},
  {"20181012", "B", "002"},
  {"20181012", "C", "004"}
]

有什么好的解决方法吗？

Answer 1

首先请注意，对象没有可靠的排序。因此，第一列或第二列没有任何意义。

但是，由于您知道要获取其值的键，因此可以使用map并同时考虑键的两种变化来计算额外的字符：

var results = [{
    "Created Date ": "20181012",
    "Created By ": "A",
    "Job Number ": "001",
    "Department": "FS"
  },
  {
    "Created Date": "20181012",
    "Created By ": "B",
    "Job Number ": "002",
    "Department": "DS"
  },
  {
    "Created Date": "20181012",
    "Created By ": "C",
    "Job Number ": "004",
    "Department": "FS"
  }
]

var output = results.map(d => [
  d['Created Date'] || d['Created Date '],
  d['Created By'] || d['Created By '],
  d['Job Number'] || d['Job Number ']
]);

console.log(output);

Answer 2

这演示了使用数组对象的map（）方法的另一种方法，如下所示：

var results = [{
"Created Date ": "20181012",
"Created By ": "A",
"Job Number ": "001",
"Department": "FS"
  },
  {
"Created Date": "20181012",
"Created By": "B",
"Job Number": "002",
"Department": "DS"
  },
  {
" Created Date": "20181012",
" Created By": "C",
" Job Number": "004",
"Department": "FS"
  }
];

Array.prototype.remove = function(index){
  this.splice(index,1);
}

var nu_arry = results.map(item => {
  let d = [];
  let keys = Object.keys(item);
  keys.remove(3);  // remove Department key name
  keys.forEach(function(key) {
    d.push(item[key]);
  });
  return d;
});
console.log(nu_arry);

在键名之前或结尾是否存在空格都没有什么区别，因为代码可以利用Object.keys()来精确检测所有键的名称。该解决方案的优势在于，如果编码人员决定修改键的名称，由于没有对变体键进行硬编码，因此该代码仍然可以使用。注意：short arrow function语法消除了创建单独的用户定义函数的需要。

如何针对特定条件缩小数组？

2 个答案: