Question

我正在处理一些角度代码。目标是使用尽可能小的角度1.x功能，因为这将很快得到重构。

我正在比较一个数组

  let subscription = [
  "Cinemax Subscription", 
  "Disney Subscription", 
  "Encore Subscription", 
  "Epix Subscription", 
  "HBO Subscription", 
  "MLB Subscription", 
  "NBA Subscription", 
  "NHL Subscription", 
  "Division"
]

到具有键值关系的对象数组，该关系包含另一个数组。

let profiles = [
    {
        name:"1+ Adults in Household",
        qualifiers: [
            {
                name: 'Number of adults in the household'
            }
        ]
    },
    {
        name: '75k',
        qualifers: [
            {
                name: 'Division'
            },
            {
                name: 'Income'
            }
        ]
    },
    {
        name: 'Time Warner',
        qualifers: [
            {
                name: 'Division'
            }
        ]
    }
]

我在索引和循环中遇到了困难。

let = profilesFiltered = [];

最初我尝试使用过滤器和angular.for 每个来比较数组。

let filteredProfiles = subscription.filter( function (src) {
    angular.forEach(profiles, function (key, index) {
        if(src === key.qualifiers[index].name) {
            profilesFiltered.push(key);
        }
    })
});

我在if语句中看到了

key.qualifiers[index].name // 'Number of adults in the household'

与

相比，启动正确

subscription[0]   // however, it will continue to 1 on the first object which isn't there.

我看到它是如何开始失败的。但我不确定如何正确循环** profiles **数组中的限定符。

所需的结果是那些 Division ，它是索引中的最后一个数组项。

profilesFiltered = [
    {
    name: '75k',
    qualifers: [
      {
        name: 'Division'
      },
      {
        name: 'Income'
      }
    ]
  },
  {
      name: 'Time Warner',
      qualifers: [
          {
              name: 'Division'
          }
      ]
  }
]

此时非常感谢任何反馈。

由于

Answer 1

只需使用filter()和some()：

的组合过滤个人资料

profiles.filter(v => v.qualifiers.some(q => subscription.includes(q.name)));

let subscription = [
  "Cinemax Subscription",
  "Disney Subscription",
  "Encore Subscription",
  "Epix Subscription",
  "HBO Subscription",
  "MLB Subscription",
  "NBA Subscription",
  "NHL Subscription",
  "Division"
]

let profiles = [{
    name: "1+ Adults in Household",
    qualifiers: [{
      name: 'Number of adults in the household'
    }]
  },
  {
    name: '75k',
    qualifiers: [{
        name: 'Division'
      },
      {
        name: 'Income'
      }
    ]
  },
  {
    name: 'Time Warner',
    qualifiers: [{
      name: 'Division'
    }]
  }
]

let res = profiles.filter(v => v.qualifiers.some(q => subscription.includes(q.name)));


console.log(res)

Answer 2

您是否在寻找类似的内容，请查看我的代码

cleaned = arr.reject{ |e| e == :c }
cleaned.each { |e| handle(e) }

Answer 3

这是您的数据结构的工作代码。它基于使用reduce函数。如果您不需要重复 - answer here。

https://jsbin.com/dexagiresa/2/edit?js,output

let profiles = [
{
    name:"1+ Adults in Household",
    qualifiers: [
        {
            name: 'Number of adults in the household'
        }
    ]
},
{
    name: '75k',
    qualifiers: [
        {
            name: 'Division'
        },
        {
            name: 'Income'
        }
    ]
},
{
    name: 'Time Warner',
    qualifiers: [
        {
            name: 'Division'
        }
    ]
}
];



let subscription = [
  "Cinemax Subscription", 
  "Disney Subscription", 
  "Encore Subscription", 
  "Epix Subscription", 
  "HBO Subscription", 
  "MLB Subscription", 
  "NBA Subscription", 
  "NHL Subscription", 
  "Division"
];


var filteredProfiles = profiles.reduce(function (collectedResults, profile) {

    var newResults = profile.qualifiers.reduce(function(foundQualifiers, qualifier) {

        if(subscription.indexOf(qualifier.name) > -1) {
            return foundQualifiers.concat(qualifier.name);
        } else {
            return foundQualifiers;
        }

    }, []);

    return collectedResults.concat(newResults);

}, []);

document.write(filteredProfiles);

javascript将数组与对象数组进行比较

3 个答案: