Javascript JSON按名称分组，然后显示最新记录

仅使用JavaScript，我需要

按代码分组

获取最新的修改日期

将总分组代码显示为计数

开始JSON结果

[
{"ID":1,"code":"AAA","modifieddate":"2019-06-01","user":"John"},
{"ID":2,"code":"AAA","modifieddate":"2019-06-02","user":"Jane"},
{"ID":3,"code":"AAA","modifieddate":"2019-06-03","user":"Sue"},
{"ID":4,"code":"BBB","modifieddate":"2019-06-10","user":"Rick"},
{"ID":5,"code":"CCC","modifieddate":"2019-06-11","user":"Joe"}
]

[
{"ID":3,"code":"AAA","modifieddate":"2019-06-03","user":"Sue","Count":"3"},
{"ID":4,"code":"BBB","modifieddate":"2019-06-10","user":"Rick","Count":"1"},
{"ID":5,"code":"CCC","modifieddate":"2019-06-11","user":"Joe","Count":"1"}
]

所需的JSON结果集

[
{"ID":1,"code":"AAA","modifieddate":"2019-06-01","user":"John"},
{"ID":2,"code":"AAA","modifieddate":"2019-06-02","user":"Jane"},
{"ID":3,"code":"AAA","modifieddate":"2019-06-03","user":"Sue"},
{"ID":4,"code":"BBB","modifieddate":"2019-06-10","user":"Rick"},
{"ID":5,"code":"CCC","modifieddate":"2019-06-11","user":"Joe"}
]

[
{"ID":3,"code":"AAA","modifieddate":"2019-06-03","user":"Sue","Count":"3"},
{"ID":4,"code":"BBB","modifieddate":"2019-06-10","user":"Rick","Count":"1"},
{"ID":5,"code":"CCC","modifieddate":"2019-06-11","user":"Joe","Count":"1"}
]

尝试使用reduce方法

我没有修改服务器端API代码的权限

我使用的是Aurelia JS

---

您可以使用

Array.reduce

按每个项目的

code

属性对结果集进行分组，根据需要递增

Count

，然后从累积对象中获取值。在此过程中，我们将执行日期比较，以确定结果中要包含的最新条目

const data=[{“ID”：1，“code”：“AAA”，“modifieddate”：“2019-06-01”，“user”：“John”}，{“ID”：2，“code”：“AAA”，“modifieddate”：“2019-06-02”，“user”：“Jane”}，{“ID”：3，“code”：“AAA”，“modifieddate”：“2019-06-03”，“user”：“Sue”}，{“ID”：4，“code”：“BBB”，“modifieddateddate”：“2019-06-10”，“user”：“Rick”}，{“ID”：5，“code”：“CCC”，“modifieddated”：“2019-06-06-11”，“user”：；
const result=Object.values（data.reduce）（a，e）=>{
如果（！a[e.code]）{
a[e.code]={…e，计数：0}；
}
if（Date.parse（e.modifieddate）>Date.parse（a[e.code].modifieddate））{
a[e.code]={…e，Count:a[e.code].Count}；
}
a[e.code].Count++；
返回a；
}, {}));
控制台日志（结果）这应该让您：

let array = [
  {"ID":1,"code":"AAA","modifieddate":"2019-06-01","user":"John"},
  {"ID":2,"code":"AAA","modifieddate":"2019-06-02","user":"Jane"},
  {"ID":3,"code":"AAA","modifieddate":"2019-06-03","user":"Sue"},
  {"ID":4,"code":"BBB","modifieddate":"2019-06-10","user":"Rick"},
  {"ID":5,"code":"CCC","modifieddate":"2019-06-11","user":"Joe"}
]
let result = array.reduce(function(total, currentValue, currentIndex, arr) {
  let index = total.findIndex(function(entry) { return entry.code == currentValue.code; })
  if (index >= 0) { // entry already exists
    // check modified
    if (total[index].modifieddate > currentValue.modifieddate) { // already have most recent of the two
      total[index].Count += 1;
    } else { // need to replace with more recent
      currentValue.Count = total[index].Count + 1;
      total[index] = currentValue;
    }
  } else { // first record for this code
    currentValue.Count = 1;
    total.push(currentValue);
  }
  return total;
}, []);
console.log(result);

这是一张工作票

注意：注释是在代码块中进行的
我在本文中没有看到任何关于
modifieddate
的内容。OP也想要最新的唱片。对不起，我没听进去。我的输出与OP完全匹配，所以您能澄清差异在哪里吗？尝试更改前三条记录的顺序。你可以先点，然后总是得到正确的结果。好吧，我不太清楚OP的意图是什么，因为这个例子很简单。更新以符合该要求。做得很好。谢谢你，戈伦先生。非常感谢，先生！工作完美。谢谢你，沃森先生。非常感谢，先生！没问题@Brian