javascript重组,js重组数组

JS两个数组根据相同的ID进行重新组合数据？

var c = a.concat(b),//合并成一个数组

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temp = {},//用于id判断重复

result = [];//最后的新数组

//遍历c数组，将每个item.id在temp中是否存在值做判断，如不存在则对应的item赋值给新数组，并将temp中item.id对应的key赋值，下次对相同值做判断时便不会走此分支，达到判断重复值的目的；

c.map((item,index)={

if(!temp[item.id]){

result.push(item);

temp[item.id] = true

}

})

console.log(result)

js json解析并重组成二维数组

var arr = [{a:1},{a:2},{a:3},{b:5},{b:7},{b:10}];

var newArr = [];

arr.map(e=Object.keys(e)[0]).filter((e,i,o)=o.indexOf(e)===i).map(e={

arr.filter((e1,i)=Object.keys(e1)==e).forEach((e,i)={

if(!newArr[i])newArr[i]=[];

newArr[i].push(e);

});

console.log(newArr);

js获取json字符串后如何按格式重组？

题目没有怎么看懂啊

用正则啊

function rookitReg(){

var reg = /\(\d+,\d+\)/g;

var str = "某某某(02,000002)某某某(02,000005)某某某(02,000101)某某(02,000111)某某某......";

alert(new RegExp(reg).test(str));

str = str.replace(reg,""); // 替换你要替换的，什么下拉框，单选框啥子哦

}

JS数组对象中有相同值的数据拿出相同的对象重组

const List = [{"end_date":"2022-02-10","end_seconds":1644488295.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-10 18:08:34","id":10319,"last_heart_beat_time":"2022-02-10 18:18:15","scanner_id":167,"start_date":"2022-02-10","start_seconds":1644487714.0,"start_total_seconds":581},{"end_date":"2022-02-11","end_seconds":1644550497.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-11 11:19:08","id":10338,"last_heart_beat_time":"2022-02-11 11:34:57","scanner_id":167,"start_date":"2022-02-11","start_seconds":1644549548.0,"start_total_seconds":949},{"end_date":"2022-02-11","end_seconds":1644558220.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-11 11:48:41","id":10339,"last_heart_beat_time":"2022-02-11 13:43:40","scanner_id":167,"start_date":"2022-02-11","start_seconds":1644551321.0,"start_total_seconds":6899},{"end_date":"2022-02-11","end_seconds":1644558820.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-11 13:53:40","id":10343,"last_heart_beat_time":"2022-02-11 13:53:40","scanner_id":167,"start_date":"2022-02-11","start_seconds":1644558820.0,"start_total_seconds":0},{"end_date":"2022-02-11","end_seconds":1644563620.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-11 14:03:41","id":10344,"last_heart_beat_time":"2022-02-11 15:13:40","scanner_id":167,"start_date":"2022-02-11","start_seconds":1644559421.0,"start_total_seconds":4199},{"end_date":"2022-02-11","end_seconds":1644586551.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-11 15:55:51","id":10350,"last_heart_beat_time":"2022-02-11 21:35:51","scanner_id":167,"start_date":"2022-02-11","start_seconds":1644566151.0,"start_total_seconds":20400},{"end_date":"2022-02-13","end_seconds":1644759468.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-13 14:47:48","id":10410,"last_heart_beat_time":"2022-02-13 21:37:48","scanner_id":167,"start_date":"2022-02-13","start_seconds":1644734868.0,"start_total_seconds":24600},{"end_date":"2022-02-14","end_seconds":1644846304.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-14 10:15:05","id":10430,"last_heart_beat_time":"2022-02-14 21:45:04","scanner_id":167,"start_date":"2022-02-14","start_seconds":1644804905.0,"start_total_seconds":41399},{"end_date":"2022-02-15","end_seconds":1644911999.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-15 15:30:00","id":10468,"last_heart_beat_time":"2022-02-15 15:59:59","scanner_id":167,"start_date":"2022-02-15","start_seconds":1644910200.0,"start_total_seconds":1799},{"end_date":"2022-02-15","end_seconds":1644932084.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-15 19:44:44","id":10472,"last_heart_beat_time":"2022-02-15 21:34:44","scanner_id":167,"start_date":"2022-02-15","start_seconds":1644925484.0,"start_total_seconds":6600},{"end_date":"2022-02-16","end_seconds":1645020217.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-16 15:13:37","id":10491,"last_heart_beat_time":"2022-02-16 22:03:37","scanner_id":167,"start_date":"2022-02-16","start_seconds":1644995617.0,"start_total_seconds":24600},{"end_date":"2022-02-17","end_seconds":1645089245.0,"end_total_seconds":0,"first_heart_beat_time":"2022-02-17 14:59:06","id":10521,"last_heart_beat_time":"2022-02-17 17:14:05","scanner_id":167,"start_date":"2022-02-17","start_seconds":1645081146.0,"start_total_seconds":8099}]

const newArray = []

List.forEach(item = {

// 第一次循环,repeat定义repet,基于find浅拷贝特性,操作repeat会影响到newArray

// 循环开始,find的条件不成立,就会push一个新对象{data: item.start_date,times: [item]}进去

// 检索条件要注意,find的核对条件需要准确

// 第二次循环,find条件成立,则走else

const repeat = newArray.find(res = res.data === item.start_date)

if (!repeat) {

newArray.push({

data: item.start_date,

times: [item]

})

} else {

// 避免循环紊乱,不用item.times.push(item),选用repeat.times.push(item)

repeat.times.push(item)

}

})

打印数据后可以看到，已经将数据成功照相同时间的start_date重新管理

如果前端需要解决时间归类当天的用户多部操作但是后端没处理，就可以用forEach+find的方法处理

JS json数据重组如何把后面json的转换成前面的json啊?

var jsonA=[];

JSONB.forEach(function(o){

if(o.title){

var oo={title:o.title,item:[]};

JSONB.forEach(function(o){

if(o.title==oo.title){

oo.item.push(o.item);

o.title="";

}

})

jsonA.push(oo);

}

})

console.log(jsonA);

用javascript 将对象按属性名称排序重组

//定义一个对象数组

　var data = [{ name: "jiang", age: 22 }, { name: "AAAAAAAAAAAAAA", age: 21 }, { name: "CCCCCCCCc", age: 25}];

　//定义一个比较器

function compare(propertyName) {

return function (object1, object2) {

var value1 = object1[propertyName];

var value2 = object2[propertyName];

if (value2 value1) {

return -1;

}

else if (value2 value1) {

return 1;

}

else {

return 0;

}

//使用方法

data.sort(compare("name"));

alert(data[0].name);//jiang

//使用方法

data.sort(compare("age"));

alert(data[0].age);//25s

文章题目：javascript重组,js重组数组
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