如何在JSON中找到特定值并在JavaScript中显示这些值的最小值？

Question

我是JS的新手，并尝试从事任务，但找不到同一任务的精确解决方案。 这是我第一次发表问题，所以请纠正我走在哪里。

这是我的问题，

这是我的JSON文件的一小部分，它超过3000行，我给出了同一JSON文件的确切格式，

{
  "status": "success",
  "data": {
    "resultType": "vector",
    "result": [
      {
        "metric": {
          "ABC": "yes",
          "KPI": "M1",
          "__name__": "m1_status",
          "cloud": "m1",
          "cloud_kpi": "m1 status",
          "instance": "m1.e1.s1",
          "job": "m1",
          "src": "m1.m1.m1",
          "src_vm": "m1",
          "timespan": "2h"
        },
        "value": [
          1653633595.484,
          "100"
        ]
      },
      {
        "metric": {
          "ABC": "yes",
          "KPI": "M2",
          "__name__": "m1_status",
          "cloud": "m2",
          "cloud_kpi": "m1.e1",
          "instance": "m2",
          "job": "m2 services",
          "src": "m2.m3",
          "src_vm": "m2",
          "timespan": "2h"
        },
        "value": [
          1653633595.484,
          "99.919094"
        ]
      },
      {
        "metric": {
          "ABC": "yes",
          "KPI": "M2",
          "__name__": "m1_status",
          "cloud": "m2",
          "cloud_kpi": "m1.e1",
          "instance": "m2",
          "job": "m2 services",
          "src": "m2.m3",
          "src_vm": "m2",
          "timespan": "2h"
        },
        "value": [
          1653633595.484,
          "98.406375"
        ]
      },
      {
        "metric": {
          "ABC": "yes",
          "KPI": "M1",
          "__name__": "m1_status",
          "cloud": "m1",
          "cloud_kpi": "m1 status",
          "instance": "m1.e1.s1",
          "job": "m1",
          "src": "m1.m1.m1",
          "src_vm": "m1",
          "timespan": "2h"
        },
        "value": [
          1653633595.484,
          "100"
        ]
      }
}

我的问题是如何检查完整的JSON文件并搜索云键和其值并显示值的最小值。

例如 我必须在整个JSON中以M1的名称获取云，这可能发生在12次。

“云”：“ M2”， “ cloud_kpi”：“ m1.e1”，

在上行中，我所需的键将是云，值将为m2。 之后，我必须获取“值”的云的值

：[ 1653633595.484， “ 100” ]

我需要的值是100。

以同样的方式，我必须在完整的JSON文件中检查这些值并获得12个值

m1-100 M1-100 M2-33 M1-44 .......因此，

在获得这些值后，我需要找到最小的这些值并显示它们。

以同样的方式，我必须获取所有云的及其最小值 M1-min（100,98,94 ...） M2-min（100,38,99，...） M3-min（100,39,20 ..） 等等。...

我尝试了地图方法，这使我感到困惑，也尝试了每种方法，我只能获取所有云名称，但不要进一步移动，

response.json().then(function (data) {
                        var jsonarray = data["data"]["result"]                    
                        for (var k in jsonarray) {
                            if (jsonarray[k] instanceof Object) {
                                console.log(jsonarray[k].metric.cloud)                                                
                            } 
                        }
                    });

请让我知道我必须尝试的下一步是什么。

Answer 1

1. > 将云ID匹配查询的对象。

2. 在该数组上仅返回您需要的值。

3. 来自 阵列使用 math.min 。

const data={status:"success",data:{resultType:"vector",result:[{metric:{ABC:"yes",KPI:"M1",__name__:"m1_status",cloud:"m1",cloud_kpi:"m1 status",instance:"m1.e1.s1",job:"m1",src:"m1.m1.m1",src_vm:"m1",timespan:"2h"},value:[1653633595.484,"100"]},{metric:{ABC:"yes",KPI:"M2",__name__:"m1_status",cloud:"m2",cloud_kpi:"m1.e1",instance:"m2",job:"m2 services",src:"m2.m3",src_vm:"m2",timespan:"2h"},value:[1653633595.484,"99.919094"]},{metric:{ABC:"yes",KPI:"M2",__name__:"m1_status",cloud:"m2",cloud_kpi:"m1.e1",instance:"m2",job:"m2 services",src:"m2.m3",src_vm:"m2",timespan:"2h"},value:[1653633595.484,"98.406375"]},{metric:{ABC:"yes",KPI:"M1",__name__:"m1_status",cloud:"m1",cloud_kpi:"m1 status",instance:"m1.e1.s1",job:"m1",src:"m1.m1.m1",src_vm:"m1",timespan:"2h"},value:[1653633595.484,"999"]}]}};

function findMinimum(data, key) {

  // `filter` out the objects where the value
  // of `cloud` matches your query
  const result = data.data.result.filter(obj => {
    return obj.metric.cloud?.toLowerCase() === key.toLowerCase();

  // For each object return the second element
  // of the value array
  }).map(obj => {
    return obj.value[1];
  });

  // `spread` out the array and use `Math.min` to
  // find the minimum value
  return Math.min(...result);

}

console.log(`Minimum value: ${findMinimum(data, 'M1')}`);

附加文档

• 
  

• “ nofollow noreferrer”> =“ https://developer.mozilla.org/en-us/docs/web/javascript/reference/referent/operators/spread_syntax” rel =“ nofollow noreferrer”> spread Syntax

Answer 2

您可以使用array.filter（）查找数组的值和math.min（）来查找其参数的最小值

response.json().then(function(data){
    var jsonarray = data.data.result.filter(item => item.metric.cloud == 'm2') // here you will have all of the objects with the cloud m2
    const min = Math.min(...jsonarray.map(item => item.value[1]))
    console.log(min) // the minimum value
})

Answer 3

请检查以下代码：

const data = response.json();
const result = data.data.result;

// mapping
const mappedArray = result.map(x => ({
    cloud: x.metric.cloud,
    value: Number(x.value[1] ?? 0)
}))
const maxSorted = mappedArray.sort((x, y) => y.value - x.value) // sorting
const minSorted = mappedArray.sort((x, y) => x.value - y.value) 
const maxValue = maxSorted[0] ?? null;
const minValue = minSorted[0] ?? null;
console.log(maxValue); // max value
console.log(minValue); // min value

简短样式：

response.json().then((data) => {
    const minValue = data.data.result.map(x => ({
        cloud: x.metric.cloud,
        value: Number(x.value[1] ?? 0)
    })).sort((x, y) => x.value - y.value)
    console.log(minValue)
})

Answer 4

因为您想在数组中所有对象中打印最小值。您只需在单独的数组中获取所有值，然后使用Math.min.min（... array）您可以获取Miminum值。

演示：

const obj = {
  "status": "success",
  "data": {
    "resultType": "vector",
    "result": [
      {
        "metric": {
          "ABC": "yes",
          "KPI": "M1",
          "__name__": "m1_status",
          "cloud": "m1",
          "cloud_kpi": "m1 status",
          "instance": "m1.e1.s1",
          "job": "m1",
          "src": "m1.m1.m1",
          "src_vm": "m1",
          "timespan": "2h"
        },
        "value": [
          1653633595.484,
          "100"
        ]
      },
      {
        "metric": {
          "ABC": "yes",
          "KPI": "M2",
          "__name__": "m1_status",
          "cloud": "m2",
          "cloud_kpi": "m1.e1",
          "instance": "m2",
          "job": "m2 services",
          "src": "m2.m3",
          "src_vm": "m2",
          "timespan": "2h"
        },
        "value": [
          1653633595.484,
          "99.919094"
        ]
      },
      {
        "metric": {
          "ABC": "yes",
          "KPI": "M2",
          "__name__": "m1_status",
          "cloud": "m2",
          "cloud_kpi": "m1.e1",
          "instance": "m2",
          "job": "m2 services",
          "src": "m2.m3",
          "src_vm": "m2",
          "timespan": "2h"
        },
        "value": [
          1653633595.484,
          "98.406375"
        ]
      },
      {
        "metric": {
          "ABC": "yes",
          "KPI": "M1",
          "__name__": "m1_status",
          "cloud": "m1",
          "cloud_kpi": "m1 status",
          "instance": "m1.e1.s1",
          "job": "m1",
          "src": "m1.m1.m1",
          "src_vm": "m1",
          "timespan": "2h"
        },
        "value": [
          1653633595.484,
          "100"
        ]
      }]
  }
};

const values = [];

obj.data.result.forEach(obj => {
    obj.value.forEach(value => {
    values.push(Math.round(value))
  });
});

console.log(Math.min(...values));