从 json 字符串中获取匹配范围键的值

Question

我在 db 中存储了 json 格式的数据，问题是它的值或属性没有引号。

"{1000:{AB DATA: NR, FC MATADATA: CR},
4000:{NK HX: I, NK FY: I},
8000:{XL LONG: I, NY CITY: I}}"

object 就在它下面是一个 object 的数组。

[{
        "current_milage": 0,
        "MasterVehicle": {
            "recommended_services": 
             "{1000:{AB DATA: NR, FC MATADATA: CR},
               4000:{NK HX: I, NK FY: I},
               8000:{XL LONG: I, NY CITY: I}}",
            "make": "Bajaj",
            "model": "Discover"
        }
    }
..........
],

我想从recommended_services中获取与current_milage匹配范围的值
我想从 json 中找到键的值，它是current_milage值的上限。
exp： for current_milage=3500, 4000: {...props}, for current_milage=7800, 8000: {..props}
结束我想要一个像下面这样的结构：

data = [{
"Bajaj Discover": { 
        "FUEL LINE4000": "I",
        "FUEL LEVEL": "I",
        "FUEL STRAINER SCREEN": "C",
        "THROTTLE OPERATION": "I"
        } //{AB DATA: NR, FC MATADATA: CR
 },
...
}]

到目前为止，这是我尝试过的。

    const data = {}
    _.forEach(userVehicles, obj => {
       const current_milage = obj.current_milage
       const vehicle = obj.MasterVehicle.make + ' ' + obj.MasterVehicle.model
       const recommended_services = JSON.parse(JSON.stringify(obj.MasterVehicle.recommended_services))
       const kms = _.keys(recommended_services)
       loop_1:
       for(let i = 0 ; i < kms.length ; i++)
       {
           if(current_milage < kms[i])
            {
                data[vehicle] = recommended_services[kms[i]]
                break loop_1
            }
       }
   })

它不起作用。 它给我

{
    "Bajaj Discover": "1",
    "TVS Star Sport": "1"
}

和_.keys(recommended_services)给出从 0 开始的数字，例如0,1,2,3.....

Answer 1

我建议尝试修复无效服务 json，我创建了一个 function fixServiceJson 来执行此操作。

然后，您可以为车辆执行下一个服务逻辑。

 function fixServiceJson(json) { return json.replace(/(['"])?(\w+\s?\w+)(['"])?:([^\/])/g, '"$2": $4').replace(/\:\s{2,}(\w+)/g, ': "$1"'); } let userVehicles = [{ "current_milage": 0, "MasterVehicle": { "recommended_services": `{1000:{AB DATA: NR, FC MATADATA: CR}, 4000:{NK HX: I, NK FY: I}, 8000:{XL LONG: I, NY CITY: I}}`, "make": "Bajaj", "model": "Discover" } }, { "current_milage": 15000, "MasterVehicle": { "recommended_services": `{10000:{AB DATA: NR, FC MATADATA: CR}, 20000:{NK HX: I, NK FY: I}, 30000:{XL LONG: I, NY CITY: I}}`, "make": "Honda", "model": "Insight" } } ]; const data = userVehicles.reduce( (acc, vehicle) => { const vehicleKey = vehicle.MasterVehicle.make + ' ' + vehicle.MasterVehicle.model const recommended_services = JSON.parse(fixServiceJson(vehicle.MasterVehicle.recommended_services)); acc[vehicleKey] = Object.entries(recommended_services).find(([km, service]) => vehicle.current_milage < km)[1]; return acc; }, {}) console.log("Result:",data);

只是为了解释这条线

acc[vehicleKey] = Object.entries(recommended_services).find(([km, service]) => vehicle.current_milage < km)[1];

Object.entries(recommended_services) 调用为我们提供了如下数组：

[
  [
    "1000",
    {
      "AB DATA": "NR",
      "FC MATADATA": "CR"
    }
  ],
  [
    "4000",
    {
      "NK HX": "I",
      "NK FY": "I"
    }
  ],
  [
    "8000",
    {
      "XL LONG": "I",
      "NY CITY": "I"
    }
  ]
]

然后我们在数组上调用 find 并像这样解构：

.find(([km, service]) => vehicle.current_milage < km)

这将返回下一个服务，例如

[
    "1000",
    {
      "AB DATA": "NR",
      "FC MATADATA": "CR"
    }
],

然后我们访问 [1] 中的第二项以返回

{
    "AB DATA": "NR",
    "FC MATADATA": "CR"
}

服务 object。