如何使用 javascript 删除 json 数组中的特定字段？ [复制]答案

【问题标题】：How can i delete particular field in json array using javascript? [duplicate]如何使用 javascript 删除 json 数组中的特定字段？ [复制]
【发布时间】：2017-04-14 08:37:28
【问题描述】：

[
  {

    "user_pass": "$PBa5$.cxL91nBU5cG4gqhNp8mWZoJgFY/",
    "user_mobile":"1234567890",
    "user_nicename": "abc",
    "user_email": "abc@gmail.com",
    "user_status": 0

  },
  {

    "user_pass": "$P$BwfQRzajR6R9eLeZbPnTgfQfsfPDhK0",
    "user_mobile":"0987654321",
    "user_nicename": "cdv",
    "user_email": "cdv@gmail.com",
    "user_status": 0
  },
  {

    "user_pass": "$P$Be15Zwqze.9OxoYfTOMd0WjlgXO7xe.",
    "user_mobile":"5432167890",
    "user_nicename": "zxc",
    "user_email": "zxc@gmail.com",
    "user_status": 0
  }
 ]

以上是我的数据。

1 .我想删除每个对象中的 user_pass 字段。

2 。我想使用 user_mobile 作为 user_pass

我的预期结果：

下面我已经提到了我的最终结果

[
  {

    "user_pass": "1234567890",
    "user_mobile":"1234567890",
    "user_nicename": "abc",
    "user_email": "abc@gmail.com",
    "user_status": 0

  },
  {

    "user_pass": "0987654321",
    "user_mobile":"0987654321",
    "user_nicename": "cdv",
    "user_email": "cdv@gmail.com",
    "user_status": 0
  },
  {

    "user_pass": "5432167890",
    "user_mobile":"5432167890",
    "user_nicename": "zxc",
    "user_email": "zxc@gmail.com",
    "user_status": 0
  }
 ]

任何人都可以帮助我，我是这项技术的新手。

【问题讨论】：

使用简单朴素的for循环，迭代数组并更新user_pass。
你能帮我吗@Satpal
jsfiddle.net/as2kg67ufor (var i in users) { users[i].user_pass= users[i].user_mobile; }

标签： javascript json node.js

【解决方案1】：

for (var i in users) { users[i].user_pass= users[i].user_mobile; }

【讨论】：

【解决方案2】：

如果您需要以不可变的方式执行此操作，您可以映射数组并将所有对象属性分配给新对象。

const data = [
  {

    "user_pass": "$PBa5$.cxL91nBU5cG4gqhNp8mWZoJgFY/",
    "user_mobile":"1234567890",
    "user_nicename": "abc",
    "user_email": "abc@gmail.com",
    "user_status": 0

  },
  {

    "user_pass": "$P$BwfQRzajR6R9eLeZbPnTgfQfsfPDhK0",
    "user_mobile":"0987654321",
    "user_nicename": "cdv",
    "user_email": "cdv@gmail.com",
    "user_status": 0
  },
  {

    "user_pass": "$P$Be15Zwqze.9OxoYfTOMd0WjlgXO7xe.",
    "user_mobile":"5432167890",
    "user_nicename": "zxc",
    "user_email": "zxc@gmail.com",
    "user_status": 0
  }
 ]

const newData = data.map(user => 
  Object.assign({}, user, {
    user_pass: user.user_mobile
  }))

console.log(data)
console.log(newData)
console.log(data === newData)
console.log(data[0] === newData[0])

&lt;script src="http://codepen.io/synthet1c/pen/WrQapG.js"&gt;&lt;/script&gt;

【讨论】：

【解决方案3】：

var data = [{

  "user_pass": "$PBa5$.cxL91nBU5cG4gqhNp8mWZoJgFY/",
  "user_mobile": "1234567890",
  "user_nicename": "abc",
  "user_email": "abc@gmail.com",
  "user_status": 0

}, {

  "user_pass": "$P$BwfQRzajR6R9eLeZbPnTgfQfsfPDhK0",
  "user_mobile": "0987654321",
  "user_nicename": "cdv",
  "user_email": "cdv@gmail.com",
  "user_status": 0
}, {

  "user_pass": "$P$Be15Zwqze.9OxoYfTOMd0WjlgXO7xe.",
  "user_mobile": "5432167890",
  "user_nicename": "zxc",
  "user_email": "zxc@gmail.com",
  "user_status": 0
}];

var newData = data.map(function(item) {
  return {
    user_pass: item.user_mobile,
    user_mobile: item.user_mobile,
    user_nicename: item.user_nicename,
    user_email: item.user_email,
    user_status: item.user_status
  };
});

console.log(newData);

【讨论】：

stackoverflow.com/questions/41568946/…

【解决方案4】：

给你

var obj = [
  {

    "user_pass": "$PBa5$.cxL91nBU5cG4gqhNp8mWZoJgFY/",
    "user_mobile":"1234567890",
    "user_nicename": "abc",
    "user_email": "abc@gmail.com",
    "user_status": 0

  },
  {

    "user_pass": "$P$BwfQRzajR6R9eLeZbPnTgfQfsfPDhK0",
    "user_mobile":"0987654321",
    "user_nicename": "cdv",
    "user_email": "cdv@gmail.com",
    "user_status": 0
  },
  {

    "user_pass": "$P$Be15Zwqze.9OxoYfTOMd0WjlgXO7xe.",
    "user_mobile":"5432167890",
    "user_nicename": "zxc",
    "user_email": "zxc@gmail.com",
    "user_status": 0
  }
 ]

for(var i =0 ; i < obj.length; i++){
     obj[i].user_pass= obj[i].user_mobile
}

console.log(obj);

【讨论】：

当我们在 Array.prototype 中实现大量高阶函数时，我仍然不明白为什么我们现在在数组/集合上使用 for 循环
坚守本源
这就是我们不进化的原因，也是我们无法转向更好的编程的原因。当人们这样想的时候。我敢让你在未来十年坚持你的根，我保证你会成为恐龙
Lol 放松... For 循环为您提供更多灵活性...
stackoverflow.com/questions/41568946/…

【解决方案5】：

forEach

var arr=[ { "user_pass": "$PBa5$.cxL91nBU5cG4gqhNp8mWZoJgFY/", "user_mobile":"1234567890", "user_nicename": "abc", "user_email": "abc@gmail.com", "user_status": 0 }, { "user_pass": "$P$BwfQRzajR6R9eLeZbPnTgfQfsfPDhK0", "user_mobile":"0987654321", "user_nicename": "cdv", "user_email": "cdv@gmail.com", "user_status": 0 }, { "user_pass": "$P$Be15Zwqze.9OxoYfTOMd0WjlgXO7xe.", "user_mobile":"5432167890", "user_nicename": "zxc", "user_email": "zxc@gmail.com", "user_status": 0 } ];
arr.forEach(function(a){
a.user_pass=a.user_mobile;
})
console.log(arr);

【讨论】：

【解决方案6】：

您必须遍历对象数组并将user_mobile 引用到user_pass。

var arr = [...data...];

arr.forEach(function(obj) {
    obj.user_pass = obj.user_mobile;
});

【讨论】：