JavaScript:如何映射两个对象以获取将第一个对象的 Id 映射到另一个对象的名称的输出? [英] JavaScript: How to map two objects to get the output which map the Id of first object to the name of other object?

问题描述

我正在尝试创建一个将 data2 对象值映射到 data1 名称的函数.

I am trying to make a function that maps the data2 object value to the name of the data1.

我尝试用 data1 迭代 data2 object2，但它不能正常工作.我能够映射它们，但当 data2 中没有 data1 时，无法获取它的值.

I tried to iterate data2 object2 with data1 but it is not working properly. I am able to map them but not getting the value of data1 when it is not there in data2.

有什么方法可以正确映射以获得代码中提到的所需输出吗?

Is there any way I can map properly to get the desired output mentioned in the code?

let data1 = {
    attributes: [{
            Id: 'test1',
            Name: 'Test1',
            Type: 'date'
        },
        {
            Id: 'test2',
            Name: 'Test2',
            Type: 'string'
        },
        {
            Id: 'test3',
            Name: 'Test3',
            Type: 'string'
        },
        {
            Id: 'test4',
            Name: 'Test4',
            Type: 'boolean'
        }
    ]
};

let data2 = {
    value: [{
            test1: '10-12-2021',
            test2: '4',
            dummy: 'ignore me'
        },
        {
            test3: '3',
            test4: true,
            abc: 'ignore me'
        },
        {
            test1: '12-12-2023',
            test3: '42',
            dummy1: 'ignore me'
        }
    ]
};

//output

let ouput = {
    rows: [{
            Test1: '10/12/2021',
            Test2: '4',
            Test3: '',
            Test4: ''
        },
        {
            Test1: '',
            Test2: '',
            Test3: '3',
            Test4: 'Y'
        },
        {
            Test1: '12/12/2023',
            Test2: '',
            Test3: '42',
            Test4: ''
        }
    ]
};

//function

function mapper(data1, data2) {
    let formattedValue = [];
    data2.value.forEach(val => {
        let data = {};
        for (let prop in val) {
            let name;
            const filter = data1.attributes.filter(el => el.Id === prop)[0];
            if (filter) {
                name = filter.Name;
                switch (filter.Type) {
                    case 'boolean':
                        data[name] =
                            val[prop] === true ? 'Y' : val[prop] === false ? 'N' : '';
                        break;
                    case 'date':
                        data[name] = new Date(val[prop]).toLocaleDateString();
                        break;
                    default:
                        data[name] = val[prop];
                        break;
                }
            }
        }
        formattedValue.push(data);
    });

    return formattedValue;
}

console.log(mapper(data1, data2));

同样，如果我将 data2 作为空值传递，我希望得到低于输出

same if I pass data2 as empty value I am looking to get below output

let data1 = {
  attributes: [
    {
      Id: 'test1',
      Name: 'Test1',
      Type: 'string'
    },
    {
      Id: 'test2',
      Name: 'Test2',
      Type: 'string'
    },
    {
      Id: 'test3',
      Name: 'Test3',
      Type: 'string'
    },
    {
      Id: 'test4',
      Name: 'Test4',
      Type: 'boolean'
    }
  ]
};

let data2 = {
  value: [
  ]
};

//output

let ouput = {
  rows: [
    {
      Test1: '',
      Test2: '',
      Test3: '',
      Test4: ''
    },
  ]
};

推荐答案

const data1 = {attributes:[{Id:'test1',Name:'Test1',Type:'date'},{Id:'test2',Name:'Test2',Type:'string'},{Id:'test3',Name:'Test3',Type:'string'},{Id:'test4',Name:'Test4',Type:'boolean'}]}
      data2 = {value:[{test1:'10-12-2021',test2:'4'},{test3:'3',test4:true},{test1:'12-12-2023',test3:'42'}]};
      data3 = {value: []}

const mapper = (x, y) => {
  const row = x.attributes.map(e => [e.Id, e.Name])
        format = e => { switch (true) {
            case typeof e === 'boolean':
              return e ? 'Y' : 'N'
            case isNaN(e) && !isNaN(Date.parse(e)):
              return new Date(e).toLocaleDateString()
            default: return e
          }}
  const rows = (y.value.length ? y.value : [1]).map(e => 
     Object.fromEntries(row.map(([k, v]) => [v, format(e[k]) ?? '']))
  )
  return { rows }
} 

console.log(mapper(data1, data2))
console.log(mapper(data1, data3))

.as-console-wrapper { max-height: 100% !important; top: 0; }

说明:

y.value.length ?y.value : [1]

检查 data.value 是否为空，如果不是则返回 data.value.如果 yes 返回带有 1 个元素的 array(可以是任何值，只需要执行 1 次迭代即可满足您对空数据的要求，因此值在这里并不重要.

Checks if data.value is empty, if not returns data.value. If yes returns array with 1 element (can be any value, it is just needed to do 1 iteration to meet your requirement about empty data, so value doesn't really matter here.

其余代码应该很清楚，如果没有，请告诉我.

Rest of code should be pretty clear, if not, let me know.

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