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使用js的数组映射-将值与另一个数组进行比较并从第二个数组返回值

dictionary arrays javascript

sogeniusio · 技术社区 · 8 年前

我想把这张桌子的 chapter_id 和 brother_id 与下面的兄弟和章节表一起返回 brothername 和 name 分别是菲尔德。使用js或jquery。我使用Vuejs返回分钟数组作为计算属性。见下文。

在 sql 有点像

select brothername from brothers where minute.brother_id = brothers.id …然后把妓院作为新的价值观 大哥

同样的事情 第四章 以下内容:

从brothers中选择brothername,其中minute.brother_id=brothers.id …然后把妓院作为新的价值观 大哥

结果数组或对象应为:

期望的数组

[
   {
      "location":"UCLA",
      "chapter_id":"Beta", 
      "brother_id":"Golpher", 
      "created_at":"2008-05-15 22:23:00",
      "status":"Approved"
   },
   { ... },
   {
      "location":"John's Deli",
      "chapter_id":"Beta", notice the change in the array based on the ids
      "brother_id":"Sheera", notice the change in the array based on the ids
      "created_at":"2008-05-15 22:23:00",
      "status":"Approved"
   }
]

会议记录表 (原始数组)

[
   {
      "location":"UCLA",
      "chapter_id":2,
      "brother_id":1,
      "created_at":"2008-05-15 22:23:00",
      "status":"Approved"
   },
   { ... },
   {
      "location":"John's Deli",
      "chapter_id":2,
      "brother_id":4,
      "created_at":"2008-05-15 22:23:00",
      "status":"Approved"
   }
]

章表

[
   {
      "id":1,
      "letter_representation":"A",
      "name":"Alpha",
      "founded_at":"UCLA",
      ...
   },
   { ... }
]

兄弟的桌子

[
   {
      "id":1,
      "profile_id":1,
      "chapter_id":1,
      "brothername":"Golpher",
      "firstname":"Jack",
      ...
   },
   { ... },
   {
      "id":4,
      "profile_id":4,
      "chapter_id":1,
      "brothername":"Sheera",
      "firstname":"Jake",
      ...
   }
]

Vue.js公司

computed: {
    brothers () {
        return this.$store.state.brothers
    },
    chapters () {
        return this.$store.state.chapters
    },
    minutes () {
        return this.$store.getters.model
    }
},

3 回复 | 直到 8 年前

Damon 8 年前

我假设您不想用这个操作改变原始数组中的对象。

注意你可能想处理这个案子 brother_id 或 chapter_id 不存在于相应的表中。在下面的示例中,它只是将属性值设置为 undefined

const minutesTable = [{
  "location": "UCLA",
  "chapter_id": 2,
  "brother_id": 1,
  "created_at": "2008-05-15 22:23:00",
  "status": "Approved"
}, {
  "location": "John's Deli",
  "chapter_id": 2,
  "brother_id": 4,
  "created_at": "2008-05-15 22:23:00",
  "status": "Approved"
}]
const chapterTable = [{
  "id": 1,
  "letter_representation": "A",
  "name": "Alpha",
  "founded_at": "UCLA",
}]
const brotherTable = [{
  "id": 1,
  "profile_id": 1,
  "chapter_id": 1,
  "brothername": "Golpher",
  "firstname": "Jack",
}, {
  "id": 4,
  "profile_id": 4,
  "chapter_id": 1,
  "brothername": "Sheera",
  "firstname": "Jake",
}]

// your result
const result = minutesTable.map(m => {
  const brother = brotherTable.find(b => b.id === m.brother_id)
  const chapter = chapterTable.find(c => c.id === m.chapter_id)

  return Object.assign({}, m, {
    brother_id: brother && brother.brothername,
    chapter_id: chapter && chapter.name,
  })
})

console.log(result)

Manually Overridden 8 年前

这应该是你需要的

const minutesTable = [
   {
      "location":"UCLA",
      "chapter_id":2,
      "brother_id":1,
      "created_at":"2008-05-15 22:23:00",
      "status":"Approved"
   },
   {
      "location":"John's Deli",
      "chapter_id":2,
      "brother_id":4,
      "created_at":"2008-05-15 22:23:00",
      "status":"Approved"
   }
]
const chapterTable = 
[
   {
      "id":1,
      "letter_representation":"A",
      "name":"Alpha",
      "founded_at":"UCLA",
   }
]
const brotherTable = [
   {
      "id":1,
      "profile_id":1,
      "chapter_id":1,
      "brothername":"Golpher",
      "firstname":"Jack",
   },
   {
      "id":4,
      "profile_id":4,
      "chapter_id":1,
      "brothername":"Sheera",
      "firstname":"Jake",
   }
]
/* code starts here */
let newMinutesTable = JSON.parse(JSON.stringify(minutesTable)).map(a => {
  let brother = brotherTable.find(id => id.id === a.brother_id);
  let chapter = chapterTable.find(id => id.id === a.chapter_id)
  if (brother) a.brother_id = brother.brothername
  if (chapter) a.chapter_id = chapter.name;
  return a;
})
console.log([minutesTable,newMinutesTable]);

Kaique Garcia 8 年前

我认为你应该先准备好这些价值观,以便更好地理解。所以我做了这个,让我分块解释。

您的输入信息:

var minutesTable = [{
  "location": "UCLA",
  "chapter_id": 2,
  "brother_id": 1,
  "created_at": "2008-05-15 22:23:00",
  "status": "Approved"
}, {
  "location": "John's Deli",
  "chapter_id": 2,
  "brother_id": 4,
  "created_at": "2008-05-15 22:23:00",
  "status": "Approved"
}],
    chapterTable = [{
  "id": 1,
  "letter_representation": "A",
  "name": "Alpha",
  "founded_at": "UCLA",
}],
    brotherTable = [{
  "id": 1,
  "profile_id": 1,
  "chapter_id": 1,
  "brothername": "Golpher",
  "firstname": "Jack",
}, {
  "id": 4,
  "profile_id": 4,
  "chapter_id": 1,
  "brothername": "Sheera",
  "firstname": "Jake",
}];

不知怎么的,你会被迫把这些信息当作变量。我们会处理的。

准备数据

在处理对象数组时,如果需要从不同的数组中查找每个对象的唯一信息,特别是如果希望多次运行此操作,则需要稍微复杂一点。因此,我们不必使用对象数组,而是可以保存我们的生命,将其更改为对象的对象,其中每个项索引必须是唯一的id。看:

var chapters = {},
    brothers = {};
chapterTable.map(function(el, i) {
    chapters[el.id] = el;
});
brotherTable.map(function(el, i) {
    brothers[el.id] = el;
});

现在你可以很容易地找到一章一章或者一个兄弟一章,对吧?然后你可以像这样完成这个问题:

var output = [];
minutesTable.map(function(el, i) {
    var item = {
        "location": el.location, // note that values are just default values, just in case
        "chapter_id":"-", 
        "brother_id":"-", 
        "created_at": el.created_at,
        "status": el.status
    };
    // PS: you need to check if that brother_id really exists!
    if(brothers[el.brother_id] != undefined) {
        item.brother_id = brothers[el.brother_id].brothername;
    }
    // PS: the same with chapters
    if(chapters[el.chapter_id] != undefined) {
        item.chapter_id = chapters[el.chapter_id].name;
    }
    output.push(item);
});

就这样。无论如何,如果您可以更改sql查询,那么最好使用sql连接并在那里准备值。