如何从点差运算符中删除属性? [英] How to delete property from spread operator?

问题描述

我想从响应中删除drugName，但是它没有发生任何想法如何从传播操作符中删除属性? main.js

I want to delete drugName from the response but it is not happening any idea how to delete property from spread operator ? main.js

  const transformedResponse = transformResponse(response);
  const loggerResponse = {...transformedResponse};
  delete loggerResponse[drugName];
  console.log("LOGGER>>>>", loggerResponse);
  logger().info('Drug Price Response=', { ...loggerResponse, memberId: memberId, pharmacyId: pharmacyId });

\ 数据

LOGGER>>>> {
    '0': {
        isBrand: false,
        drugName: 'test drug',
        drugStrength: '5 mg 1 5 mg',
        drugForm: 'Tablet',
    }
}

transformResponse

transformResponse

[{
    drugName: 'HYDROCODONE-HOMATROPINE MBR',
    drugStrength: '5MG-1.5MG',
    drugForm: 'TABLET',
    brand: false
}]

推荐答案

您可以使用

You could use Rest syntax in Object Destructuring to get all the properties except drugName to a rest variable like this:

const transformedResponse = [{
    drugName: 'HYDROCODONE-HOMATROPINE MBR',
    drugStrength: '5MG-1.5MG',
    drugForm: 'TABLET',
    brand: false
},
{
    drugName: 'HYDROCODONE ABC',
    drugStrength: '10MG',
    drugForm: 'SYRUP',
    brand: true
}]

const output = transformedResponse.map(({ drugName, ...rest }) => rest)

console.log(output)

此外，当您在{}内展开数组时，会得到一个对象，该对象的数组索引为键，而数组的值为value.这就是为什么在loggerResponse中获得以0作为键的对象的原因:

Also, when you spread an array inside {}, you get an object with indices of the array as key and the values of array as value. This is why you get an object with 0 as key in loggerResponse:

const array = [{ id: 1 }, { id: 2 }]
console.log({ ...array })

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