＆QUOT;过滤和QUOT; JSON独特的钥匙，并得到所有相关的值 [英] "Filter" JSON for unique key and get all related values

问题描述

什么是找到所有与一组可能的相关值的最佳方式。

  VAR表= [
    {组：一，东东：新}，
    {组：一，东东：老}，
    {组：B的东西：newOld}，
    {组：B的东西：旧}，
    {组：C的东西：新}，
    {组：C的东西：旧}，
    {组：C的东西：newOld}，
];
 

我要填充包含独特的组值的下拉列表。并选择我想用所有相关的的东西作进一步处理。同时也希望添加它包含了所有的东西 A组。例如：

 所有的选择 - ＆GT;新，老，newOld
                一个 - ＆GT;新旧
                b  - ＆GT; newOld，老
                C  - ＆GT;新，老，newOld
 

解决方案

短期和precise，没有全阵列中的每个时间查找条目的开销：

  VAR组= {所有：{}};
table.forEach（函数（）{
    如果（！组[a.group]）{组[a.group] = {}; }
    组[a.group] [a.stuff] =组[所有] [a.stuff] = 1;
}）;
 

在对象列表中的东西，这样你就没有重复的条目（这就是为什么，而多余的 = 1 ）。但是你可以很容易地扩展它来算重复：

  table.forEach（函数（）{
    如果（！组[a.group]）{组[a.group] = {}; }
    VAR东西=组[所有] [a.stuff];
    组[所有] [a.stuff] =！东西？ 1：++的东西;
    东西=组[a.group] [a.stuff];
    组[a.group] [a.stuff] =！东西？ 1：++的东西;
}）;
 

结果将如下所示：

  //组拥有所有组元素和自己的东西值
组= {全：{新：2，老：3，newOld：2}，
           一：{新：1，旧：1}，
           B：{newOld：1，老：1}，
           C：{新：1，老：1，newOld：1}
         }
 

要检索组的值，简单地说：

  VAR组名=A; //任何你需要的组
Object.keys（集团[群组名称]）;
//会给你：
[新旧]
 

演示

^{得留意支持 Object.keys ，当然还有 Array.prototype.forEach 。}

What would be the best way to find all the possible related values with a group.

var table = [
    {group:"a", stuff:"new"},
    {group:"a", stuff:"old"},
    {group:"b", stuff:"newOld"},
    {group:"b", stuff:"old"},
    {group:"c", stuff:"new"},
    {group:"c", stuff:"old"},
    {group:"c", stuff:"newOld"},
];

I want populate a Dropdown containing Unique group values. and on selection I want to use all related stuff for further processing. and also want to add a group which contains all stuff. for example

on selection of all -> new, old, newOld
                a -> new, old
                b -> newOld, old
                c -> new, old, newOld

解决方案

short and precise, without the overhead of looking up the entry each time in the whole array:

var groups = {all:{}};
table.forEach(function(a){
    if (!groups[a.group]){ groups[a.group] = {}; }
    groups[a.group][a.stuff] = groups["all"][a.stuff] = 1;
});

List the stuff in an object, thus you have no duplicate entries (that's why the rather redundant =1). But you could easily extend it to count the duplicates:

table.forEach(function(a){
    if (!groups[a.group]){ groups[a.group] = {}; }
    var stuff = groups["all"][a.stuff];
    groups["all"][a.stuff] = !stuff ? 1 : ++stuff;
    stuff = groups[a.group][a.stuff];
    groups[a.group][a.stuff] = !stuff ? 1 : ++stuff;
});

The result will look like the following:

// "groups" holds all group elements and their stuff values
groups = { "all": {"new":2,"old":3,"newOld":2},
           "a" : {"new":1,"old":1},
           "b" : {"newOld":1,"old":1},
           "c" : {"new":1,"old":1,"newOld":1}
         }

To retrieve the values of a group, simply say:

var groupname = "a"; // whatever group you need
Object.keys(groups[groupname]);
// will give you:
["new","old"]

Demo

^{Gotta watch out for support of Object.keys and Array.prototype.forEach of course.}

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