Javascript 从对象数组中过滤唯一值
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Filter unique values from an array of objects
提问by bob.mazzo
How can I use Array.filter() to return unique idwith name?
如何使用Array.filter（）返回唯一id有name？
My scenario is slightly different than the solutions I have researched in that I have an array of objects. Every example I find contains a flat array of single values.
我的场景与我研究的解决方案略有不同，因为我有一组对象。我找到的每个示例都包含一个单一值的平面数组。
data=[
{id: 555, name: "Sales", person: "Jordan" },
{id: 555, name: "Sales", person: "Bob" },
{id: 555, name: "Sales", person: "John" },
{id: 777, name: "Accounts Payable", person: "Rhoda" },
{id: 777, name: "Accounts Payable", person: "Harry" },
{id: 888, name: "IT", person: "Joe" },
{id: 888, name: "IT", person: "Jake" },
];var unique = data.filter(
function (x, i) {
return data[i].id.indexOf(x.id) === i
});Thanks in advance.
提前致谢。
回答by Mamun
I think forEach()is better to achieve what you are looking for:
我认为forEach()更好地实现您的目标：
var data=[
{id: 555, name: "Sales", person: "Jordan" },
{id: 555, name: "Sales", person: "Bob" },
{id: 555, name: "Sales", person: "John" },
{id: 777, name: "Accounts Payable", person: "Rhoda" },
{id: 777, name: "Accounts Payable", person: "Harry" },
{id: 888, name: "IT", person: "Joe" },
{id: 888, name: "IT", person: "Jake" },
];
var resArr = [];
data.forEach(function(item){
var i = resArr.findIndex(x => x.name == item.name);
if(i <= -1){
resArr.push({id: item.id, name: item.name});
}
});
console.log(resArr);If you really want to use filter()try the following way:
如果您真的想使用，请filter()尝试以下方式：
var data=[
{id: 555, name: "Sales", person: "Jordan" },
{id: 555, name: "Sales", person: "Bob" },
{id: 555, name: "Sales", person: "John" },
{id: 777, name: "Accounts Payable", person: "Rhoda" },
{id: 777, name: "Accounts Payable", person: "Harry" },
{id: 888, name: "IT", person: "Joe" },
{id: 888, name: "IT", person: "Jake" },
];
var resArr = [];
data.filter(function(item){
var i = resArr.findIndex(x => x.name == item.name);
if(i <= -1){
resArr.push({id: item.id, name: item.name});
}
return null;
});
console.log(resArr);