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json从新排列问题请求帮助大神来

发布于 10 年前作者 wuqing5828 5855 次预览最后一次回复是 10 年前来自问答

[ [{“name”:“123”,“url”:“fdsfds”,“type”:“x”},{“name”:“123”,“url”:“fdsfds”,“type”:“y”},{“name”:“123”,“url”:“fdsfds”,“type”:“z”}], [{“name”:“123”,“url”:“fdsfds”,“type”:“x”},{“name”:“123”,“url”:“fdsfds”,“type”:“y”}] ]

将这个从新排按type分类排列

[ [{“name”:“123”,“url”:“fdsfds”,“type”:“x”},{“name”:“123”,“url”:“fdsfds”,“type”:“x”}], [{“name”:“123”,"ur"l:“fdsfds”,“type”:“y”},{“name”:“123”,“url”:“fdsfds”,“type”:“y”}], [{“name”:“123”,“url”:“fdsfds”,“type”:“z”}] ]

格式已经修改就这个意思大神凑合着看

15 回复

先把上面的数据格式写正确吧。

大神还没来吗

三个数组对应type[x,y,z]，遍历上面的，按照type，分别放入不同的数组。代码就不写了。

jysperm

4楼•10 年前

https://lodash.com/docs#sortBy

wuqing5828

5楼•10 年前

coordcn 3楼•40分钟前
三个数组对应type[x,y,z]，遍历上面的，按照type，分别放入不同的数组。代码就不写了。

那xyz 只是我随便写的可能还有abc 只类的

遍历的但是无法分类我试了好多方法了都无法分类

就是按照json里的一个属性归类一起在排列出来

wuqing5828

6楼•10 年前

https://lodash.com/docs#sortBy 这个我看了还是没解决我的问题

var users = [ { ‘user’: ‘barney’, ‘age’: 36 }, { ‘user’: ‘fred’, ‘age’: 40 }, { ‘user’: ‘barney’, ‘age’: 26 }, { ‘user’: ‘fred’, ‘age’: 30 } ];

.map(.sortByAll(users, [‘user’, ‘age’]), _.values); // → [[‘barney’, 26], [‘barney’, 36], [‘fred’, 30], [‘fred’, 40]]

不是我要的结果我要的结果可以是这样的

// → [[‘barney’, 26], [‘barney’, 36]],[ [‘fred’, 30], [‘fred’, 40]]

alsotang

7楼•10 年前

试试 http://github.com/alsotang/tableman 的 group 方法

xavierchow

8楼•10 年前

lodash的flatten, groupBy, map可以搞定，参考： https://gist.github.com/xavierchow/342be76041980e36e663

ruanyl

9楼•10 年前

一个笨办法

var a = arr.reduce(function(p1, p2) { // flatten
  return p1.concat(p2);
}).reduce(function(v1, v2) { // groupBy

  var eleExists = v1.some(function(e) {
    if(e[0]['type'] === v2['type']) {
      e.push(v2);
      return true;
    } else {
      return false;
    }
  });

  if(!eleExists) v1.push([v2]);
  return v1;
}, []).sort(function(s1, s2) { // sort
  return s1[0]['type'] > s2[0]['type'];
});
console.log(a);

@alsotang benchmark不错，嘻嘻

tulayang

12楼•10 年前

function resort(array) {
    var i = 0, 
        length = array.length, 
        result = [], 
        indexes = {},
        j, row, item, type;
    while ((row = array[i++]) instanceof Array) {
        j = 0; 
        while (typeof (item = row[j++]) === 'object' 
                && item !== null 
                && 'type' in item) { 
            type = item.type;
            if (type in indexes) { 
                result[indexes[type]].push(item); 
            } else { 
                indexes[type] = result.length; 
                result.push([item]);
            }
        }
    }
    return result;
}

console.log('\nThe resort 1 is: \n\n', resort([
    [{"name":"123","url":"fdsfds","type":"x"},
     {"name":"123","url":"fdsfds","type":"y"},
     {"name":"123","url":"fdsfds","type":"z"}],
    [{"name":"123","url":"fdsfds","type":"x"},
     {"name":"123","url":"fdsfds","type":"y"}]
]));

console.log('\nThe resort 2 is: \n\n', resort([
    [{"name":"123","url":"fdsfds","type":"x"},
     {"name":"123","url":"fdsfds","type":"y"},
     {"name":"123","url":"fdsfds","type":"z"},
     {"name":"123","url":"fdsfds","type":"z"}],
    [{"name":"123","url":"fdsfds","type":"x"},
     {"name":"123","url":"fdsfds","type":"y"}],
    [{"name":"123","url":"fdsfds","type":"x"}]
]));

~$ node test

The resort 1 is: 

[ [ { name: '123', url: 'fdsfds', type: 'x' },
    { name: '123', url: 'fdsfds', type: 'x' } ],
  [ { name: '123', url: 'fdsfds', type: 'y' },
    { name: '123', url: 'fdsfds', type: 'y' } ],
  [ { name: '123', url: 'fdsfds', type: 'z' } ] ]
  
The resort 2 is: 

[ [ { name: '123', url: 'fdsfds', type: 'x' },
    { name: '123', url: 'fdsfds', type: 'x' },
    { name: '123', url: 'fdsfds', type: 'x' } ],
  [ { name: '123', url: 'fdsfds', type: 'y' },
    { name: '123', url: 'fdsfds', type: 'y' } ],
  [ { name: '123', url: 'fdsfds', type: 'z' },
    { name: '123', url: 'fdsfds', type: 'z' } ] ]
~$

alsotang

13楼•10 年前

用 tableman 的写法：

var tableman = require('tableman');
var _ = require('lodash');

var source = [
  [{"name":"123","url":"fdsfds","type":"x"},{"name":"123","url":"fdsfds","type":"y"},{"name":"123","url":"fdsfds","type":"z"}],
  [{"name":"123","url":"fdsfds","type":"x"},{"name":"123","url":"fdsfds","type":"y"}]
];

source = _.flatten(source);

var result = [];

tableman.group(source, {
  by: 'type',
  action: function (rows) {
    result.push(rows);
  }});


console.log('%j', result);

shekhei

14楼•10 年前

我没看之前的答案，有点长 :P 如果不想用其他的library，可以这样写（用reduce这些方法会更简单）

var arr = [...] // 你的arr
var obj = {}; // 用来做onlyone
var result = [], tobj;
for ( var i = 0; i < arr.length; i++ ) {
   tobj = arr[i];
   obj[tobj.type] = obj[tobj.type] || [];
   obj[tobj.type].push(tobj[i]);
}
for ( var name in obj ) {
  result.push(obj[name]);
}
//没有跑过，可能会有一些问题 :P

luicfer

15楼•10 年前

看看 lodash 一块钱卖不了吃亏买不了上当