首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >根据javascript中的条件合并两个嵌套的JSON对象

根据javascript中的条件合并两个嵌套的JSON对象
EN

Stack Overflow用户
提问于 2014-05-31 10:15:30
回答 1查看 970关注 0票数 0

我正在研究如何合并两个JSON对象,并根据条件创建一个合并的对象。

我想循环遍历leadlistArray,获取每个条目的rules.id,并将其与rulesArray匹配。最终的结果应该是finalArray。

代码语言:javascript
复制
leadlistArray = {
 "status": "success",
 "data": {
  "leadlists": [
   {
    "rules": [
     {
      "$oid": "53866d4d1fd21c3cc41f52da"
     }
    ],
    "name": "List 1: only general rule"
   },
   {
    "rules": [
     {
      "$oid": "53866d4d1fd21c3cc41f52da"
     },
     {
      "$oid": "53866d9c1fd21c3cc79bf7ce"
     },
     {
      "$oid": "53866d9c1fd21c3cc79bf2cd"
     }
    ],
    "name": "List 2: general and web-based rule"
   },
   {
    "rules": [
     {
      "$oid": "53866d9c1fd21c3cc79bf7ce"
     }
    ],
    "name": "List 3: only web-based rule"
   }
  ]
 }
}

规则数组:

代码语言:javascript
复制
rulesArray = {
 "status": "success",
 "data": {
  "rules": [
   {
    "description": "optimizely no!",
    "start_time": "",
    "values": [
     "optimizely"
    ],
    "end_time": "",
    "operator": "",
    "_id": {
     "$oid": "53866d4d1fd21c3cc41f52da"
    },
    "type": 0,
    "condition": 1
   },
   {
    "description": "google_plus no!",
    "start_time": "",
    "values": [
     "google_plus"
    ],
    "end_time": "",
    "operator": "",
    "_id": {
     "$oid": "53866d9c1fd21c3cc79bf2cd"
    },
    "type": 1,
    "condition": 1
   },
   {
    "description": "google_plus yes!",
    "start_time": "",
    "values": [
     "google_plus"
    ],
    "end_time": "",
    "operator": "",
    "_id": {
     "$oid": "53866d9c1fd21c3cc79bf7ce"
    },
    "type": 1,
    "condition": 0
   }
  ]
 }
}

对于每个列表,finalArray应该按规则类型对规则描述进行分组:

代码语言:javascript
复制
finalArray = [
    {'name': 'List 1: only general rule'
    ,'rules':
        {0: [
            {'description': 'optimizely no!'}
        ]}
    },
    {'name': 'List 2: general and web-based rule'
    ,'rules': 
        {0: [
            {'description': 'optimizely no!'}
        ], 1: [
            {'description': 'google_plus yes!'},
            {'description': 'google_plus no!'}
        ]}
    },
    {'name': 'List 3: only web-based rule'
    ,'rules': {
        1: [
            {'description': 'google_plus yes!'}
        ]}
    }
]
EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2014-05-31 10:49:57

试试这个http://jsfiddle.net/eN52v/

代码语言:javascript
复制
var rules = rulesArray.data.rules
    .reduce(function(acc, item){
        return acc[item._id.$oid] = {
            description: item.description, 
            type: item.type}, acc; 
    }, {}), //cache rules by id to avoid array lookups.

    finalArray = leadlistArray.data.leadlists.map(function(item) {
        return {
            name: item.name, 
            rules: item.rules.reduce(function(acc, rule){ //Why not an array?
                var ruleDesc = rules[rule.$oid],
                    type = ruleDesc.type;

                (acc[type] || (acc[type] = [])).push({description: ruleDesc.description});
                return acc;}, {})    
        }});
票数 6
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/23968796

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档