Question

    var arr = ["10/27/2017","11/5/2017","11/5/2017","11/10/2017","11/10/2017","12/12/2017"];
    var userDate = "11/10/2017";

如果我想在用户日期之前找到这个（假定排序的）数组中的日期，我该怎么办？所以在这种情况下，我想返回11/5/2017。

基本上，找到数组中指定日期左侧的下一个最近的对象。更进一步，如果我想再次检查前一个元素是什么，但想避免重复，那将是有帮助的。

Answer 1

您可以使用Array#indexOf。

不要忘记检查元素是否存在且不是arr.indexOf(userDate) > 0

的第一个元素

let arr = ["10/27/2017","11/5/2017","11/10/2017","11/10/2017","12/12/2017"];
let userDate = "11/10/2017";

let result = arr.indexOf(userDate) > 0 ? arr[arr.indexOf(userDate) - 1] : null;
console.log(result);

Answer 2

使用Array.indexOf

非常简单

var arr = ["10/27/2017", "11/5/2017", "11/10/2017", "11/10/2017", "12/12/2017"];
var userDate = "11/10/2017";
var prev = arr.indexOf(userDate); // One call to indexOf - DRY
console.log(
   prev > 0 ? // neither the first (==0) nor not found (==-1)
   arr[prev-1] : "no earlier date"
);

Answer 3

您可以使用indexOf获取日期索引。然后很容易得到以前的值。

哟可以尝试以下内容：

＆＃13;

var arr = ["10/27/2017","11/5/2017","11/10/2017","11/10/2017","12/12/2017"];
var userDate = "11/10/2017";
    
var index = arr.indexOf(userDate); // Get the index
var previous = index > 0 ? arr[index - 1] : null; // Check if the date exists. If it does exists get the previous value or set it to null
console.log(previous)

＆＃13;

Answer 4

我认为indexOf最适合这个问题。

var arr = ["10/27/2017", "11/5/2017", "11/10/2017", "11/10/2017", "12/12/2017"];
var userDate = "11/10/2017";
var index = arr.indexOf(userDate);
var result = "";

if (index < 0) { // no match case
  result = 'no match';
} else if (index == 0) { // userdate is first element
  result = 'userDate is first element';
} else { // get the result
  result = arr[index - 1];
}
console.log('result=', result);

Answer 5

arr[arr.indexOf(userData) - 1] ? arr[arr.indexOf(userData) - 1] : throw new Error('...');

如果你坚持。我更愿意处理在我实际使用它时未定义值的情况。

Answer 6

var arr = ["10/27/2017","11/5/2017","11/10/2017","11/10/2017","12/12/2017"];

function findLeft(argum) {
  let firstIndex = arr.indexOf(argum);
  let leftside = firstIndex - 1;
  //you need to figure out error handling for < 0;

  return arr[leftside]; 
}

在数组中查找上一个对象

6 个答案: