Nar*_*n S 5 javascript object ecmascript-6 spread-syntax
我有3个具有相同数据的对象,但内部数组具有单独的服务和提供ID,因此我尝试获得如下所述的预期结果, 请在此处检查我的尝试。提前致谢
对象1:
const obj1 = {
bid : 1,
mobile : 9533703390,
services : [
{
service_id : 5,
offer_id : 10,
count : 1
}
]
}
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对象2:
const obj2 = {
bid : 1,
mobile : 9524703390,
services : [
{
service_id : 8,
offer_id : 12,
count : 1
}
]
}
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对象3:
const obj3 = {
bid : 1,
mobile : 9524703390,
services : [
{
service_id : 5,
offer_id : 10,
count : 1
}
]
}
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最终结果-每个对象具有单独的服务和要约,然后,如果要约的要约和服务要约相同,则需要加计数+ 1,否则返回数据
const result = {
bid : 1,
mobile : 9524703390,
services : [
{
service_id : 5,
offer_id : 10,
count : 2
},
{
service_id : 8,
offer_id : 12,
count : 1
}
]
}
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您可以使用array#reduce将所有对象合并为单个对象和array#concat值services。然后使用array#reduce将所有基于的服务对象合并service_id到一个对象中,并将该对象的值重新分配给services。
const obj1 = { bid : 1, mobile : 9533703390, services : [ { service_id : 5, offer_id : 10, count : 1 } ] },
obj2 = { bid : 1, mobile : 9524703390, services : [ { service_id : 8, offer_id : 12, count : 1 } ] },
obj3 = { bid : 1, mobile : 9524703390, services : [ { service_id : 5, offer_id : 12, count : 1 } ] };
var combined = [obj1, obj2, obj3].reduce((r,o) => Object.assign({}, o, {services : r.services.concat(o.services)}));
combined.services = Object.values(combined.services.reduce((res, services) => {
if(res[services.service_id])
res[services.service_id].count += services.count;
else
res[services.service_id] = Object.assign({}, services);
return res;
},{}));
console.log(combined)Run Code Online (Sandbox Code Playgroud)