将2个数组转换为键值对象，其中一个数组包含键，第二个数组是匹配索引值的嵌套数组

Question

将valueArr的内部成员的每个刺激组合到keyArr中的匹配键，索引将始终相互匹配。

我正在寻找的结果是：每个对象的键和id等于"；内在的"；但如果不清楚，我会解决的

const valuesArr = 
["10","9","Item","Bank","2","true","true","Result7","5675"],
["9","1","Expenses","OthExpense","4","true","true","9999","9999"],
["8","8","Expenses","OthAsset","6","false","false","6666","77777"],
["7","8","Expenses","AcctPay","4","true","true","666","7777"],
["6","123","123","123","123","123","123","123","123"],
["5","123","123","123","123","123","123","123","123"],
["4","Test1","Item","OthCurrAsset","2","Result5","Result6","ytrytrytryrty","32432"],

["2","Result1","Result2","Result3","Result4","Result5","Result6","Result7","Result8"],
["1","Test1","Test12","Test13","Test14","Test15","Test16","Test17","Test18"]]

const keysArr = 
["internalid",
"custrecord_st_segment_transaction_type",
"custrecord_st_segment_sublist",
"custrecord_st_segment_account_type",
"custrecord_st_segment_subsidiary",
"custrecord_st_segment_department",
"custrecord_st_segment_class",
"custrecord_st_segment_location",
"custrecord_st_segment_custom_segment"]

result:

{"1":{"id":"1","values":{"internalid":"1","custrecord_st_segment_transaction_type":"Test1","custrecord_st_segment_sublist":"Test12","custrecord_st_segment_account_type":"Test13","custrecord_st_segment_subsidiary":"Test14","custrecord_st_segment_department":"Test15","custrecord_st_segment_class":"Test16","custrecord_st_segment_location":"Test17","custrecord_st_segment_custom_segment":"Test18"}},
"2":{"id":"2","values":{"internalid":"2","custrecord_st_segment_transaction_type":"Result1","custrecord_st_segment_sublist":"Result2","custrecord_st_segment_account_type":"Result3","custrecord_st_segment_subsidiary":"Result4","custrecord_st_segment_department":"Result5","custrecord_st_segment_class":"Result6","custrecord_st_segment_location":"Result7","custrecord_st_segment_custom_segment":"Result8"}},

"4":{"id":"4","values":{"internalid":"4","custrecord_st_segment_transaction_type":"Test1","custrecord_st_segment_sublist":"Item","custrecord_st_segment_account_type":"OthCurrAsset","custrecord_st_segment_subsidiary":"2","custrecord_st_segment_department":"Result5","custrecord_st_segment_class":"Result6","custrecord_st_segment_location":"ytrytrytryrty","custrecord_st_segment_custom_segment":"32432"}},
"5":{"id":"5","values":{"internalid":"5","custrecord_st_segment_transaction_type":"123","custrecord_st_segment_sublist":"123","custrecord_st_segment_account_type":"123","custrecord_st_segment_subsidiary":"123","custrecord_st_segment_department":"123","custrecord_st_segment_class":"123","custrecord_st_segment_location":"123","custrecord_st_segment_custom_segment":"123"}},
"6":{"id":"6","values":{"internalid":"6","custrecord_st_segment_transaction_type":"123","custrecord_st_segment_sublist":"123","custrecord_st_segment_account_type":"123","custrecord_st_segment_subsidiary":"123","custrecord_st_segment_department":"123","custrecord_st_segment_class":"123","custrecord_st_segment_location":"123","custrecord_st_segment_custom_segment":"123"}},
"7":{"id":"7","values":{"internalid":"7","custrecord_st_segment_transaction_type":"8","custrecord_st_segment_sublist":"Expenses","custrecord_st_segment_account_type":"AcctPay","custrecord_st_segment_subsidiary":"4","custrecord_st_segment_department":"true","custrecord_st_segment_class":"true","custrecord_st_segment_location":"666","custrecord_st_segment_custom_segment":"7777"}},
"8":{"id":"8","values":{"internalid":"8","custrecord_st_segment_transaction_type":"8","custrecord_st_segment_sublist":"Expenses","custrecord_st_segment_account_type":"OthAsset","custrecord_st_segment_subsidiary":"6","custrecord_st_segment_department":"false","custrecord_st_segment_class":"false","custrecord_st_segment_location":"6666","custrecord_st_segment_custom_segment":"77777"}},
"9":{"id":"9","values":{"internalid":"9","custrecord_st_segment_transaction_type":"1","custrecord_st_segment_sublist":"Expenses","custrecord_st_segment_account_type":"OthExpense","custrecord_st_segment_subsidiary":"4","custrecord_st_segment_department":"true","custrecord_st_segment_class":"true","custrecord_st_segment_location":"9999","custrecord_st_segment_custom_segment":"9999"}},
"10":{"id":"10","values":{"internalid":"10","custrecord_st_segment_transaction_type":"9","custrecord_st_segment_sublist":"Item","custrecord_st_segment_account_type":"Bank","custrecord_st_segment_subsidiary":"2","custrecord_st_segment_department":"true","custrecord_st_segment_class":"true","custrecord_st_segment_location":"Result7","custrecord_st_segment_custom_segment":"5675"}}]

Answer 1

我认为OP正在询问如何"zip"；两个数组，其中一个有键，另一个有值。如果是这样的话，Object.fromEntries()是非常有用的。

一个简单的拉链，如下所示：

// a simple zip
function zip(keys, values) {
return Object.fromEntries(
keys.map((key, index) => [key, values[index]])
);
}

OP似乎希望值数组中的lead元素是新对象中的一个特殊命名字段，以及结果对象的键。在这里，应用简单的zip。。。

function zip(keys, values) {
return Object.fromEntries(
keys.map((key, index) => [key, values[index]])
);
}
// the value array contains an "id" element at the start
// produce an object that looks like { id: number, values: {...} }
function objectFromValues(keys, values) {
return { id: values[0], values: zip(keys, values.slice(1)) }
}
const valuesArr = [
[1, "A", "B", "C"],
[2, "D", "E", "F"]
];

const keys = ["keyA", "keyB", "keyC"];
const result = valuesArr.reduce((acc, values) => {
acc[values[0]] = objectFromValues(keys, values);
return acc;
}, {});

console.log(result)