当属性匹配时，如何比较两个对象和更新值？

Question

我有两个阵列。一种是模式，其中包含12个月且第二的模式是从API获取的。 模式：

[
    {
      total: 0,
      month_name: "Jan",
    },
    {
      total: 0,
      month_name: "Feb",
    },
    {
      total: 0,
      month_name: "Mar",
    },
    {
      total: 0,
      month_name: "Apr",
    },
...
  ]

提取：

[
    {
        "total": 4,
        "month_name": "Mar"
    },
    {
        "total": 1,
        "month_name": "Apr"
    }
]

我想将获取的数组与模式进行比较，查找匹配的“ month_name”并更新“总计”。 获取的阵列仅在超过0时才包含几个月的对象。

Answer 1

我建议制作一个查找表（totalbymonth），然后您可以通过查找state来循环，然后通过查找inter in totalbymonth。

const state = [
  {total: 0, month_name: "Jan"},
  {total: 0, month_name: "Feb"},
  {total: 0, month_name: "Mar"},
  {total: 0, month_name: "Apr"}
];

const fetched = [
  {total: 4, month_name: "Mar"},
  {total: 1, month_name: "Apr"}
];

//build totalByMonth object
const totalByMonth = {};
for (let f of fetched) {
  totalByMonth[f.month_name] = f.total;
}

//update state
for (let s of state) {
  const total = totalByMonth[s.month_name];
  if (total) s.total = total;
}

console.log(state);

Answer 2

您可以尝试一下：

let result = months.map(month => {
    let matching_result = fetched.filter(f => f.month_name == month.month_name);
    return matching_result[0] ? {...month, total: matching_result[0].total}: month;
});

console.log(result);

//Output
// [
// {total: 0, month_name: 'Jan'},
// {total: 0, month_name: 'Feb'},
// {total: 4, month_name: 'Mar'},
// {total: 1, month_name: 'Apr'},
// ]

Answer 3

let pattern=[{total:0,month_name:"Jan"},{total:0,month_name:"Feb"},{total:0,month_name:"Mar"},{total:0,month_name:"Apr"}]

let fetched=[{total:4,month_name:"Mar"},{total:1,month_name:"Apr"}];

function updateTotal(pattern,fetched){
    fetched.forEach((e) => {
      let index = pattern.findIndex(p => p.month_name === e.month_name)
      if(index > -1){
        pattern[index].total = e.total
      }
    } )
 }

updateTotal(pattern,fetched)
console.log(pattern)