使用JS过滤和搜索2个不同的对象数组
使用JS过滤和搜索2个不同的对象数组
提问于 2017-05-10 04:33:46
我有两个对象数组，我需要找出用另一个数组的对象过滤其中一个数组的最佳方法是什么？
第一个数组称为drinksType
[
{
"id": "GROUP1",
"type": "Water"
},
{
"id": "GROUP2",
"type": "Beer"
},
{
"id": "GROUP3",
"type": "Coke"
},
{
"id": "GROUP4",
"type": "Gin"
}
]第二个是根据drinksType过滤的drinksGroup
[
{
"GROUP1": 1,
"GROUP2": 0,
"GROUP3": 0,
"GROUP4": 0
},
{
"GROUP1": 1,
"GROUP2": 1,
"GROUP3": 0,
"GROUP4": 0
},
{
"GROUP1": 0,
"GROUP2": 0,
"GROUP3": 1,
"GROUP4": 0
},
{
"GROUP1": 0,
"GROUP2": 1,
"GROUP3": 1,
"GROUP4": 1
}
]所期望的结果如下所示
[
{
"name": "Water"
},
{
"name": "Water Beer"
},
{
"name": "Coke"
},
{
"name": "Beer Coke Gin"
}
]因此，您可以看到如果drinksGroup中的键的值为1，那么我需要过滤drinksType来获得类型。在某些情况下，可能有几个键的值为1，在这种情况下，我需要使用所有键来过滤drinksType，并连接这些类型。叹息！
到目前为止我的努力是这样的..。
const dataKeys = Object.keys(drinksGroup);
drinks.name = [...dataKeys
.filter((dataKey) => dataKey.match(/GROUP/))
.filter((dataKey) => data[dataKey] === 1)
].join(' ');这给出了结果
[
{
"name": "GROUP1"
},
{
"name": "GROUP1 GROUP2"
},
{
"name": "GROUP3"
},
{
"name": "GROUP2 GROUP3 GROUP3"
}
]但我不知道从那里该做什么。你知道我该怎么做吗？我正在做的项目也可以访问lodash，如果这会更容易的话？非常感谢
回答 5
Stack Overflow用户
回答已采纳
发布于 2017-05-10 04:41:51
var drinksType = [{"id":"GROUP1","type":"Water"},{"id":"GROUP2","type":"Beer"},{"id":"GROUP3","type":"Coke"},{"id":"GROUP4","type":"Gin"}];
var drinksGroup = [{"GROUP1":1,"GROUP2":0,"GROUP3":0,"GROUP4":0},{"GROUP1":1,"GROUP2":1,"GROUP3":0,"GROUP4":0},{"GROUP1":0,"GROUP2":0,"GROUP3":1,"GROUP4":0},{"GROUP1":0,"GROUP2":1,"GROUP3":1,"GROUP4":1}];
var drinksNames = drinksGroup.map((group) => {
var name = '';
for (var groupName in group) {
if (group[groupName]) {
name += drinksType.find((type) => type.id === groupName).type + ' ';
}
}
return {name: name.slice(0, -1)};
});
console.log(drinksNames);
这样行得通吗？
请注意，如果您的目标是IE，则不支持Array.prototype.find。
Stack Overflow用户
发布于 2017-05-10 04:45:01
另一种可能的解决方案。
var arr = [{"id":"GROUP1","type":"Water"},{"id":"GROUP2","type":"Beer"},{"id":"GROUP3","type":"Coke"},{"id":"GROUP4","type":"Gin"}],
arr2 = [{"GROUP1":1,"GROUP2":0,"GROUP3":0,"GROUP4":0},{"GROUP1":1,"GROUP2":1,"GROUP3":0,"GROUP4":0},{"GROUP1":0,"GROUP2":0,"GROUP3":1,"GROUP4":0},{"GROUP1":0,"GROUP2":1,"GROUP3":1,"GROUP4":1}],
hash = arr2.map(v => Object.keys(v).filter(c => !!v[c])),
res = hash.map(v => ({name: v.map(c => arr.find(b => b.id == c).type).join(', ')}));
console.log(res);
Stack Overflow用户
发布于 2017-05-10 04:50:15
映射，过滤器：
const drinksType = [{"id":"GROUP1","type":"Water"},{"id":"GROUP2","type":"Beer"},{"id":"GROUP3","type":"Coke"},{"id":"GROUP4","type":"Gin"}],
drinksGroup = [{"GROUP1":1,"GROUP2":0,"GROUP3":0,"GROUP4":0},{"GROUP1":1,"GROUP2":1,"GROUP3":0,"GROUP4":0},{"GROUP1":0,"GROUP2":0,"GROUP3":1,"GROUP4":0},{"GROUP1":0,"GROUP2":1,"GROUP3":1,"GROUP4":1}];
const r = drinksGroup.map(e => Object.keys(e).filter(k => e[k] === 1))
.map(a => Object.assign({},
{name: a.map(g => drinksType.filter(({id}) => id === g).pop().type).join(' ')}
));
console.log(JSON.stringify(r, null, 2));
首先我们找到值为1的组，然后过滤类型并创建字符串。
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原文链接：
https://stackoverflow.com/questions/43879349
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