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javascript - 对象在javascript中没有返回值

问题描述

如何根据javascript中的组合获取对象值？

我不知道如何根据提供给函数调用的输入迭代对象并获取对象值。需要一些帮助。

预期输出应如下所示。

getValueCreditToBank(obj, "credit", "bank", "SGD");
getValueDebitToBank(obj, "debit", "bank", "THB");

下面的代码获取值但必须执行两个函数，是否有任何方法可以在单个函数调用中执行，

// getValueCreditToBank(obj, "credit", "bank", "SGD");
function getValueCreditToBank(provider, typein, typeout, source){
  return provider.map(item => {
    if (item.country_from[0].paymentIn[0].type == typein 
        && item.country_from[0].currency.includes(source) 
        && item.country_from[0].paymentOut[0].type == typeout) {
      return {
        paymentIn: typein,
        paymentOut: typeout,
        paymentInFee: item.country_from[0].paymentIn[0].fee.number + "%",
        payInSpeed: item.country_from[0].paymentIn[0].speed.number + "days",
        ...item
      }      
    }
   })
  .map(y=>({  
    ...y,
    targetAmountwithPay: y.targetAmount + y.country_from[0].paymentIn[0].fee.number*y.targetAmount/100
  }))
}

// getValueDebitToBank(obj, "debit", "bank", "SGD");
function getValueDebitToBank(provider, typein, typeout, source){
  return provider.map(item => {
    if (item.country_from[0].paymentIn[1].type == typein 
        && item.country_from[0].currency.includes(source) 
        && item.country_from[0].paymentOut[0].type == typeout) {
      return {
        paymentIn: typein,
        paymentOut: typeout,
        paymentInFee: item.country_from[0].paymentIn[1].fee.number + "%",
        payInSpeed: item.country_from[0].paymentIn[1].speed.number + "days",
        ...item
      }      
    }
   })
  .map(y=>({  
    ...y,
    targetAmountwithPay: y.targetAmount + y.country_from[0].paymentIn[0].fee.number*y.targetAmount/100
  }))
}

对象示例：

var obj = [{
    "id": "identity1",
    "fee": '2',
    "rate": '0.5',
    "targetAmount": '1000',
     "country_from": [{
        "currency": [
            "SGD",
            "USD"
        ],
        "paymentIn": [{
            "type": "credit",
            "speed": {
                "unit": "days",
                "number": "1"
            },
            "fee": {
                "type": "%",
                "number": "1.5"
            }
        },{
            "type": "debit",
            "speed": {
                "unit": "days",
                "number": "1"
            },
            "fee": {
                "type": "%",
                "number": "1"
            }
        }],
       "paymentout":[{
          "type":"bank"
       }]
    }]
},
{
    "id": "identity2",
    "fee": '1',
    "rate": '0.5',
    "targetAmount": '1000',
     "country_from": [{
        "currency": [
            "THB",
            "USD"
        ],
        "paymentIn": [{
            "type": "debit",
            "speed": {
                "unit": "days",
                "number": "1"
            },
            "fee": {
                "type": "%",
                "number": "1"
            }
        }
        ],
       "paymentout":[{
          "type":"bank"
       }]
    }]
}]

预期输出：

//getValue(obj, "credit", "bank", "SGD"); should return object as
 {id: "identity1",
 fee: '2',
 rate: '0.5',
 currency: SGD,
 paymentIn: "credit",
 paymentOut: "bank",
 paymentIn Fee: 1.5%,
 targetAmount: 1000,
 targetAmountwithPay: 506.485 //(((targetamount-fee)*rate)+credit fee))}
//getValue(obj, "debit", "bank", "THB"); should return object as
 {id: "identity2",
 fee: '1',
 rate: '0.5',
 currency: THB,
 paymentIn: "debit",
 paymentOut: "bank",
 paymentIn Fee: 1%,
 targetAmount: 1000,
 targetAmountwithPay: 504.49 //(((targetamount-fee)*rate)+credit fee))}

标签： javascriptjqueryarraysobject

解决方案

我在您的代码中看到的第一个问题是您在所有数组上进行了映射，但您只更改了一些元素。

也许您没有注意到这一点，因为您设置的过滤器正在传递所有数组元素，但如果没有，您将在最终数组中看到许多未定义的元素。

因此，我建议您更新引入过滤器的代码：

function getValueCreditToBank(provider, typein, typeout, source){
  return provider
    .filter(item => (item.country_from[0].paymentIn[0].type == typein 
        && item.country_from[0].currency.includes(source) 
        && item.country_from[0].paymentOut[0].type == typeout))
    .map(item => ({
        paymentIn: typein,
        paymentOut: typeout,
        paymentInFee: item.country_from[0].paymentIn[0].fee.number + "%",
        payInSpeed: item.country_from[0].paymentIn[0].speed.number + "days",
        ...item
      }))
  .map(y=>({  
    ...y,
    targetAmountwithPay: y.targetAmount + y.country_from[0].paymentIn[0].fee.number*y.targetAmount/100
  }))
}

您在两个函数中执行几乎相同的任务，并且您没有正确使用参数，因此例如您可以paymentIn通过typein参数确定索引：

function getValueToBank(provider, typein, typeout, source){
  const paymentInIndex = typein === 'credit' ? 0 : 1;
  return provider
    .filter(item => (item.country_from[0].paymentIn[paymentInIndex].type == typein 
        && item.country_from[0].currency.includes(source) 
        && item.country_from[0].paymentOut[0].type == typeout))
    .map(item => ({
        paymentIn: typein,
        paymentOut: typeout,
        paymentInFee: item.country_from[0].paymentIn[paymentInIndex].fee.number + "%",
        payInSpeed: item.country_from[0].paymentIn[paymentInIndex].speed.number + "days",
        ...item
      }))
}

然后你可以实现你的功能：

function getValueCreditToBank(provider, typeout, source){
    return getValueToBank(provider, 'credit', typeout, source)
    .map(y=>({  
        ...y,
        targetAmountwithPay: y.targetAmount + y.country_from[0].paymentIn[0].fee.number*y.targetAmount/100
  }))
}

function getValueDebitToBank(provider, typeout, source){
    return getValueToBank(provider, 'debit', typeout, source)
}

因此，您从原始函数中删除typein参数，因为它由函数名称确定。

这只是一个例子，你可以用不同的方式重写，一个更易读的方法是给你传递给过滤和映射数组的箭头函数命名。

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