Question

我目前正在处理来自我的数据库的客户退货清单。

客户被保存为从两个连接表构建的对象。连接表返回重复项，因为在我的First Normal表单数据库中没有类似于数组的数组。

所以考虑像这样的数组

var customers = [
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Cars'}, 
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Computers'},
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Babes'},
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Food'}, 
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Toys'},
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Cake'}];

我一直在疯狂地尝试，但我只能找到在简单数组中删除重复项的示例，没有对象[1,2,3]等。

最终我想要做的是删除重复但是为了节省兴趣。给定上面的示例数组，这样的对象列表。

var reduced =[
{id : 1, firstname : 'John', lastname : 'Doe' interests : ['Cars', 'Computers', 'Babes']},
{id : 2, firstname : 'Frank', lastname : 'Smith' interests : ['Food', 'Toys', 'Cake']}];

到目前为止，我已尝试删除此类重复项

for(var i = 0; i < newCustomersArray.length; i++){
        console.log(i)
        if(i == 0){
            console.log('i = 0, customerId ' + newCustomersArray[i].firstname)
            data.customers.push(newCustomersArray[i]);
        }
        //else if(newCustomersArray[i-1].customerId != newCustomersArray[i].customerId){
        //    
        //}
        else if( newCustomersArray[i-1].customerId != newCustomersArray[i].customerId){
            //console.log(newCustomersArray[i-1].customerId + " == " + newCustomersArray[i].customerId);
            console.log('i = '+i+', customerId ' + newCustomersArray[i].firstname)
            data.customers.push(newCustomersArray[i]);
        }
    }

Answer 1

您可以使用一些Array方法来构建新数组，例如Array.prototype.forEach()

forEach()方法每个数组元素执行一次提供的函数。

和Array.prototype.some()

some()方法测试数组中的某个元素是否通过了由提供的函数实现的测试。

var customers = [
        { id: 1, firstname: 'John', lastname: 'Doe', interest: 'Cars' },
        { id: 1, firstname: 'John', lastname: 'Doe', interest: 'Computers' },
        { id: 1, firstname: 'John', lastname: 'Doe', interest: 'Babes' },
        { id: 2, firstname: 'Frank', lastname: 'Smith', interest: 'Food' },
        { id: 2, firstname: 'Frank', lastname: 'Smith', interest: 'Toys' },
        { id: 2, firstname: 'Frank', lastname: 'Smith', interest: 'Cake' }
    ], reduced = [];

customers.forEach(function (a) {
    !reduced.some(function (b) {
        if (a.id === b.id) {
            b.interests.push(a.interest);
            return true;
        }
    }) && reduced.push({ id: a.id, firstname: a.firstname, lastname: a.lastname, interests: [a.interest] });
});
document.write('<pre>' + JSON.stringify(reduced, 0, 4) + '</pre>');

Answer 2

对customers对象使用array.reduce方法来消除重复。 https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce

请参考以下代码

＆＃13;

var customers = [
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Cars'}, 
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Computers'},
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Babes'},
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Food'}, 
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Toys'},
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Cake'}];

var temp = customers.reduce(function(o,n){
o = o || {};
if(!o[n.firstname+n.lastname+n.id]){
o[n.firstname+n.lastname+n.id] = n;
o[n.firstname+n.lastname+n.id].interest = []
}
o[n.firstname+n.lastname+n.id].interest.push(n.interest);
return o;
},{});

customers = Object.keys(temp).map(function(k) { return temp[k] })

console.log(customers);

＆＃13;

Answer 3

使用对象（具有唯一键）来消除重复：

var tempObj = {};
for(i=0; i<customers.length;i++){
    if(tempObj[customers[i].id] == undefined){
        tempObj[customers[i].id] = customers[i];
    }
    else{
        tempObj[customers[i].id].interest = tempObj[customers[i].id].interest + ","+customers[i].interest;
    }
}

这里有一个演示https://jsfiddle.net/h2aeeth6/

Answer 4

你可以写简单的逻辑，

var customers = [
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Cars'}, 
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Computers'},
{id : 1, firstname : 'John', lastname : 'Doe', interest : 'Babes'},
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Food'}, 
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Toys'},
{id : 2, firstname : 'Frank', lastname : 'Smith', interest : 'Cake'}];

        for (var i=0;i<customers.length;i++){
            var intrest=[customers[i].interest];
            for (var j=i+1; j<customers.length; j++)
            {
                if (customers[i].id == customers[j].id){
                    intrest.push(customers[j].interest);
                    customers.splice(j,1);
                    j--;
                 }
              }
                customers[i].interest=intrest;
       }

    console.log(JSON.stringify(customers));

对象数组，删除重复但存储值（即分组）

4 个答案: