arr.push（arr.splice（n，1））“吞下”被移动对象的属性

Question

例如，我想把中间元素放到最后：

let students=[
    {"name":"a","uid":"001"},
    {"name":"b","uid":"002"},
    {"name":"c","uid":"003"},
    {"name":"d","uid":"004"},
    {"name":"e","uid":"005"},
];
students.push(students.splice(students.length/2,1));
console.log(students.length);
for(let s of students){
  console.log(s.name+':'+s.uid+',');
}

但是最后一个元素的属性变得不确定，尽管元素的数量不变，为什么会发生呢？

Answer 1

splice总是返回一个数组，即使只删除了一个元素。如果你想push删除student，你必须先从数组中提取它，否则你将数组推送到students（而不是学生对象）：

let students=[
    {"name":"a","uid":"001"},
    {"name":"b","uid":"002"},
    {"name":"c","uid":"003"},
    {"name":"d","uid":"004"},
    {"name":"e","uid":"005"},
];
const [student] = students.splice(students.length/2,1);
students.push(student);
// or
// students.push(students.splice(students.length/2,1)[0]);
console.log(students.length);
for(let s of students){
  console.log(s.name+':'+s.uid+',');
}

Answer 2

splice方法总是返回数组而不是对象。所以在你的代码中，第5个元素是数组，所以它给出了未定义的值。，只需替换这一行就可以正常工作了

students.push(...students.splice(students.length/2,1));

知道关于...的母马（休息运算符）click here

    let students=[
        {"name":"a","uid":"001"},
        {"name":"b","uid":"002"},
        {"name":"c","uid":"003"},
        {"name":"d","uid":"004"},
        {"name":"e","uid":"005"},
    ];
    students.push(...students.splice(students.length/2,1));
    console.log(students.length);
    for(let s of students){
      console.log(s.name+':'+s.uid+',');
    }

Answer 3

Splice总是返回数组，使用spread operator来获取对象。

let students=[
  {"name":"a","uid":"001"},
  {"name":"b","uid":"002"},
  {"name":"c","uid":"003"},
  {"name":"d","uid":"004"},
  {"name":"e","uid":"005"},
];

students.push(...students.splice(students.length/2,1));

for(let s of students){
  console.log(s.name+':'+s.uid+',');
}