javascript - 需要删除对象数组中的重复项并使用 union USING JS 合并数组
问题描述
需要删除对象数组中的重复项并将数组与 union USING JS 合并。试图过滤数组只是想array["INTERFACE"]
在APP_ID
. 并删除重复记录。
未过滤的未合并数组!
var data = [
{
"APP_ID" : "1001",
"INTERFACE" : [
{
"INTERFACE_ID" : "01",
"NAME" : "CIF OPENNING",
"URL" : "/CusIdInfo",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD",
"URL" : "/Dashboard",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "06",
"NAME" : "SUMMARY COPC",
"URL" : "/SummaryCopc",
"STATUS" : "A"
}
]
},
{
"APP_ID" : "1002",
"INTERFACE" : [
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD",
"URL" : "/Dashboard",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "08",
"NAME" : "BIOMETRIC",
"URL" : "/Biometric",
"STATUS" : "A"
}
]
},
{
"APP_ID" : "1001",
"INTERFACE" : [
{
"INTERFACE_ID" : "01",
"NAME" : "CIF OPENNING",
"URL" : "/CusIdInfo",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "05",
"NAME" : "SUMMARY",
"URL" : "/Summary",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "06",
"NAME" : "SUMMARY COPC 2",
"URL" : "/SummaryCopc2",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD 2",
"URL" : "/Dashboard 2",
"STATUS" : "A"
}
]
},
{
"APP_ID" : "1002",
"INTERFACE" : [
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD",
"URL" : "/Dashboard",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "08",
"NAME" : "BIOMETRIC",
"URL" : "/Biometric",
"STATUS" : "A"
}
]
}
];
想要的结果
[
{
"APP_ID" : "1002",
"INTERFACE" : [
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD",
"URL" : "/Dashboard",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "08",
"NAME" : "BIOMETRIC",
"URL" : "/Biometric",
"STATUS" : "A"
}
]
},
{
"APP_ID" : "1001",
"INTERFACE" : [
{
"INTERFACE_ID" : "01",
"NAME" : "CIF OPENNING",
"URL" : "/CusIdInfo",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "05",
"NAME" : "SUMMARY",
"URL" : "/Summary",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "06",
"NAME" : "SUMMARY COPC",
"URL" : "/SummaryCopc",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD",
"URL" : "/Dashboard",
"STATUS" : "A"
}, {
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD 2",
"URL" : "/Dashboard2",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "06",
"NAME" : "SUMMARY COPC 2",
"URL" : "/SummaryCopc2",
"STATUS" : "A"
}
]
}
]
试图过滤数组只是想array["INTERFACE"]
在APP_ID
. 并删除重复记录。
解决方案
这是一个稍微有弹性的解决方案,它依赖于允许动态聚合的函数生成器。
以下示例遵循的逻辑是,在您的数据输入中,主要对象的唯一键是APP_ID
. 接下来,每个 APP_ID 的聚合规则是它应该遵循另一个聚合规则INTERFACE
。事实上,每个界面都有一个独特的NAME
,解释了为什么您的结果样本中有多个“07”和“06”。
代码解释记录在代码本身中。
var data = [
{
"APP_ID" : "1001",
"INTERFACE" : [
{
"INTERFACE_ID" : "01",
"NAME" : "CIF OPENNING",
"URL" : "/CusIdInfo",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD",
"URL" : "/Dashboard",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "06",
"NAME" : "SUMMARY COPC",
"URL" : "/SummaryCopc",
"STATUS" : "A"
}
]
},
{
"APP_ID" : "1002",
"INTERFACE" : [
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD",
"URL" : "/Dashboard",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "08",
"NAME" : "BIOMETRIC",
"URL" : "/Biometric",
"STATUS" : "A"
}
]
},
{
"APP_ID" : "1001",
"INTERFACE" : [
{
"INTERFACE_ID" : "01",
"NAME" : "CIF OPENNING",
"URL" : "/CusIdInfo",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "05",
"NAME" : "SUMMARY",
"URL" : "/Summary",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "06",
"NAME" : "SUMMARY COPC 2",
"URL" : "/SummaryCopc2",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD 2",
"URL" : "/Dashboard 2",
"STATUS" : "A"
}
]
},
{
"APP_ID" : "1002",
"INTERFACE" : [
{
"INTERFACE_ID" : "07",
"NAME" : "DASHBOARD",
"URL" : "/Dashboard",
"STATUS" : "A"
},
{
"INTERFACE_ID" : "08",
"NAME" : "BIOMETRIC",
"URL" : "/Biometric",
"STATUS" : "A"
}
]
}
];
// Aggregate duplicates with a common uniqueKey, invoking the aggregateExpression callback for each pair.
function* aggregateDuplicates(arr, uniqueKey, aggregateExpression) {
const aggregateGroups = arr.reduce((acc,next) => {
acc[next[uniqueKey]] = acc[next[uniqueKey]] || [];
return acc[next[uniqueKey]].push(next), acc;
}, {});
// loop items.
for (var [_, entries] of Object.entries(aggregateGroups)) {
// Aggregate results following the aggregateExpression.
yield Object.assign({}, entries.reduce((acc, next) => aggregateExpression(acc, next)));
}
}
// Aggregate duplicates of data, whose unique key is APP_ID.
const res = [...aggregateDuplicates(data, 'APP_ID', (a,b) => {
// In order to properly aggregate the INTERFACE property, acquire the set of entires interfaces of two items with the same APP_ID.
var interfacesSet = [...a.INTERFACE, ...b.INTERFACE];
// Finally, spread common values between them, then aggregate the INTERFACE property by its unique NAME key.
return Object.assign(a, b, {
INTERFACE: [...aggregateDuplicates(interfacesSet, 'NAME', (c,d) => {
// For that NAME property, just assign the values of both objects, nothing more nothing less.
return Object.assign(c,d)
})]
});
})];
console.log(res);
旁注:INTERFACE
未应用对属性的排序,这是一个优点,但我认为只要输出数据有效正确,它就不是强制性的。
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