Question

鉴于以下成员数据数组，我要为会员身份不是最新的会员返回formEl : Model -> E.Element MyStyles variation Msg formEl model = E.column None [] <| [ Input.email FormField [ Attr.width (Attr.px 300) ] { onChange = FormChange , value = "" , label = Input.labelAbove <| E.text "Email" , options = [ Input.errorBelow <| E.el Error [] (E.text "Fill in an email address") ] } , Input.text FormField [ Attr.width (Attr.px 300) ] { onChange = FormChange , value = "" , label = Input.labelAbove <| E.text "Distance" , options = [ Input.errorBelow <| E.el Error [] (E.text "Fill in a distance") ] } -- <DATE PICKER HERE> ]数组。

这是会员数据：

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memberID

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以下是我获取该数据的代码：

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var members = [
  {
    firstName: 'Howard',
    lastName: 'Lee',
    membershipIsCurrent: true,
    MemberID: 235
  },
  {
    firstName: 'James',
    lastName: 'Icharis',
    membershipIsCurrent: false,
    MemberID: 236
  },
  {
    firstName: 'Thomas',
    lastName: 'Cronquist',
    membershipIsCurrent: true,
    MemberID: 237
  },
  {
    firstName: 'Philip',
    lastName: 'Grover',
    membershipIsCurrent: false,
    MemberID: 238
  },
  {
    firstName: 'Eric',
    lastName: 'Broadstone',
    membershipIsCurrent: true,
    MemberID: 239
  },
  {
    firstName: 'Hunter',
    lastName: 'Gonzales',
    membershipIsCurrent: true,
    MemberID: 240
  }];

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结果如下。我不明白为什么它会为会员资格尚未失效的每位会员返回 function lapsedIDs (array ) { return array.map( function ( member ) { if ( member.membershipIsCurrent === false ) { return member.MemberID; } }); } lapsedIDs(members); 。任何指针??

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undefined

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Answer 1

map()将为原始数组中的每个现有元素创建一个新元素。

如果你没有return所有条件的东西，函数会自动返回undefined，这就是你所看到的

您可以先filter()删除不需要的内容，然后map()

 return array.map( function ( member ) {
    return member.membershipIsCurrent
  }).map(function(member){
     return member.MemberID;
  });

Answer 2

为什么会出现这个问题： - 因为 .map 每次都应该返回一个值，如果条件为false，则函数返回undefined，因为你没有提供任何返回值。该怎么办？首先使用过滤器，然后使用地图查看下面的解决方案

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var members = [
  {
    firstName: 'Howard',
    lastName: 'Lee',
    membershipIsCurrent: true,
    MemberID: 235
  },
  {
    firstName: 'James',
    lastName: 'Icharis',
    membershipIsCurrent: false,
    MemberID: 236
  },
  {
    firstName: 'Thomas',
    lastName: 'Cronquist',
    membershipIsCurrent: true,
    MemberID: 237
  },
  {
    firstName: 'Philip',
    lastName: 'Grover',
    membershipIsCurrent: false,
    MemberID: 238
  },
  {
    firstName: 'Eric',
    lastName: 'Broadstone',
    membershipIsCurrent: true,
    MemberID: 239
  },
  {
    firstName: 'Hunter',
    lastName: 'Gonzales',
    membershipIsCurrent: true,
    MemberID: 240
  }];
function lapsedIDs (array ) {
  return array.filter( function ( member ) {
    return member.membershipIsCurrent === false 
      
  }).map(function(member){
      return member.MemberID;
});
}


console.log(lapsedIDs(members));

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Answer 3

Array.map()在源数组中的每个元素上运行，并将其映射到其他内容。您没有从当前成员返回成员ID，因此它将它们映射到undefined。您可能只想修改函数以使用let lapsedMembers = [];开始，然后使用forEach而不是map，并在该循环中将ID推送到lapsedMembers数组。 / p>

var members = [
  {
    firstName: 'Howard',
    lastName: 'Lee',
    membershipIsCurrent: true,
    MemberID: 235
  },
  {
    firstName: 'James',
    lastName: 'Icharis',
    membershipIsCurrent: false,
    MemberID: 236
  },
  {
    firstName: 'Thomas',
    lastName: 'Cronquist',
    membershipIsCurrent: true,
    MemberID: 237
  },
  {
    firstName: 'Philip',
    lastName: 'Grover',
    membershipIsCurrent: false,
    MemberID: 238
  },
  {
    firstName: 'Eric',
    lastName: 'Broadstone',
    membershipIsCurrent: true,
    MemberID: 239
  },
  {
    firstName: 'Hunter',
    lastName: 'Gonzales',
    membershipIsCurrent: true,
    MemberID: 240
  }];
  
function lapsedIDs ( array ) {
  let lapsedMembers = [];
  
  array.forEach(function ( member ) {
    if ( member.membershipIsCurrent === false ) {
      lapsedMembers.push(member.MemberID);
    }
  });
  
  return lapsedMembers;
}

console.log(lapsedIDs(members));

Answer 4

即使您的map测试失败，membershipIsCurrent也希望将某些放入数组中。所以它会引发undefined。

您可以在地图之前filter来避免这种情况。

function lapsedIDs(array) {
    return array.filter(member => {
        return !member.membershipIsCurrent;
    }).map(member => member.MemberID);
}

lapsedIDs(members);

Answer 5

我试图解决同样的问题，并在看了这篇文章后意识到最简单的解决方案是

function lapsedIDs (array ) {
    return array.map( function ( member ) {
      if ( member.membershipIsCurrent === false ) {
        return member.MemberID;
      }
    }).filter(y => y !== undefined);
}

就我而言，我不知道输入是什么，我也不在乎，我只希望地图返回所有定义的值。

JavaScript映射为某些元素返回undefined

5 个答案: