如何通过 javascript 中的对象和数组获取对象数组

Question

对于对象数组，使用javascript

与对象和列表数组进行比较

应根据以下条件获取对象数组

itemvalue and idvalue same，从那个检查 arrobj cid has same codevalue return 两个对象数组
itemvalue and idvalue are same，从中检查是否 cid not matching the arraylist 和 codevalue has value that does not exist in arraylist，return 对象数组
itemvalue and idvalue same，从中检查是否 codevalue matching the arraylist 和 cid has value that does not exist in arraylist，return 对象数组

return 空数组[] 如果上述失败

//data1
var arraylist = ["IN","FP", "FN"];
var arrobj1 = [
  {id:1, name: "sun", cid: "IN", itemvalue: "3456"},
  {id:2, name: "mon", cid: "FI", itemvalue: "4567"},
  {id:3, name: "tues", cid: "SP", itemvalue: "4567"},
  {id:4, name: "thurs", cid: "FI", itemvalue: "2345"},
]
var obj1 = { id:5, name: "ben", codevalue: "SG", idvalue:"4567"}

Expected Output
[
{id:2, name: "mon", cid: "FI", itemvalue: "4567"},  
{id:3, name: "tues", cid: "SP", itemvalue: "4567"}
]

***
//data2
var larrylist= ["IN","FI","FR"];
var arrobj2 = [
  {id:1, name: "sun", cid: "IN", itemvalue: "1234"},
  {id:2, name: "mon", cid: "FI", itemvalue: "2468"},
  {id:3, name: "tues", cid: "IN", itemvalue: "2468"},
  {id:4, name: "thur", cid: "FI", itemvalue: "2345"},
]
var obj2 = { id:5, name: "ben", codevalue: "SP", idvalue:"2468"}

Expected Output

[]

***

//data2
var arraylist= ["IN","FI","FR"];
var arrobj3 = [
  {id:1, name: "sun", cid: "IN", itemvalue: "1234"},
  {id:2, name: "mon", cid: "FI", itemvalue: "2468"},
  {id:3, name: "tues", cid: "SG", itemvalue: "2468"},
  {id:4, name: "thur", cid: "FI", itemvalue: "2345"},
]
var obj3 = { id:5, name: "ben", codevalue: "FI", idvalue:"2468"}

Expected Output
[
 {id:2, name: "mon", cid: "FI", itemvalue: "2468"}
]

尝试过

const result = arrobj1.filter((item) => {
  return item.itemvalue === obj.idvalue &&
    (
      !arraylist.includes(item.cid)
      || item.cid === obj.codevalue
    )
})

Answer 1

我想这就是你想要做的:

const result = arrobj.filter((item) =>
    item.itemvalue === obj.idvalue && (arraylist.includes(item.cid) || item.cid === obj.codevalue)
);

Answer 2

清楚地陈述问题通常会让您找到解决方案的大部分方法。尝试准确说出你的意思，然后尝试准确你所说的编码。

在这里使用 OP 的问题陈述和示例进行操作，我们得到：

目标是过滤对象数组 (arrobj1) 以生成满足与对象 (obj1) 和另一个数组 (arraylist) 相关标准的另一个数组.

如果 obj1 的属性 idvalue 与数组元素 的属性 itemvalue 匹配，则过滤谓词为真AND...
谓词还必须 (2a) 在 obj1 的属性 codevalue 和数组元素的属性 cid 之间找到匹配项，BUT (2b) 这种比较取决于 arraylist 中对象的属性（codevalue 或 cid）的存在。具体来说，如果元素的 cid 或对象的 codevalue 在 arraylist 中，则元素的 cid 值必须与 [=13= 的 codevalue 匹配].

使用这种（费力的）精确度进行重述，对谓词进行编码就容易多了。让我们把它编码成 bite-sized 块...

// condition 1
// el is an element from arrobj1, and obj is an instance like obj1
// check if their itemvalue and idvalue (respectively) are equal
const idsMatch = (el, obj) => obj.idvalue === el.itemvalue;

// condition 2b
// el is an element from arrobj1, obj is an instance like obj1 and cidList 
// is the given list of cids (like arraylist)
// check if the element's cid and the obj's codevalue is in the arraylist
const cidFound = (el, obj, cidList) => {
  return cidList.includes(el.cid) || cidList.includes(obj.codevalue);
}

// condition 2a
// el is an element from arrobj1, and obj is an instance like obj1
// check if their cid and codevalue (respectively) are equal
const cidPropMatch = (el, obj) => obj.codevalue === el.cid;

// from 2a and 2b, make condition 2
// cidsMatch if the element isn't in the cidList, or if it is and the objects' props match
const cidsMatch = (el, obj, cidList) => {
  return !cidFound(el, obj, cidList) || cidPropMatch(el, obj);
}

// combining into a predicate
const filterPredicate = (el, obj, cidList) => {
  return idsMatch(el, obj) && cidsMatch(el, obj, cidList);
};

Ta-dah！演示...

// condition 1
const idsMatch = (el, obj) => obj.idvalue === el.itemvalue;

// condition 2b
const cidFound = (el, obj, cidList) => {
  return cidList.includes(el.cid) || cidList.includes(obj.codevalue);
}
// condition 2a
const cidPropMatch = (el, obj) => obj.codevalue === el.cid;

// condition 2
const cidsMatch = (el, obj, cidList) => {
  return !cidFound(el, obj, cidList) || cidPropMatch(el, obj);
}

// predicate
const filterPredicate = (el, obj, cidList) => {
  return idsMatch(el, obj) && cidsMatch(el, obj, cidList);
};

var arraylist = ["IN","FP", "FN"];
var arrobj1 = [
  {id:1, name: "sun", cid: "IN", itemvalue: "3456"},
  {id:2, name: "mon", cid: "FI", itemvalue: "4567"},
  {id:3, name: "tues", cid: "SP", itemvalue: "4567"},
  {id:4, name: "thurs", cid: "FI", itemvalue: "2345"},
]
var obj1 = { id:5, name: "ben", codevalue: "SG", idvalue:"4567"}

const result = arrobj1.filter(el => filterPredicate(el, obj1, arraylist));
console.log(result);

// Expected Output
// [
// {id:2, name: "mon", cid: "FI", itemvalue: "4567"},  
// {id:3, name: "tues", cid: "SP", itemvalue: "4567"}
// ]


arraylist = ["IN","FI","FR"];
var arrobj2 = [
  {id:1, name: "sun", cid: "IN", itemvalue: "1234"},
  {id:2, name: "mon", cid: "FI", itemvalue: "2468"},
  {id:3, name: "tues", cid: "IN", itemvalue: "2468"},
  {id:4, name: "thur", cid: "FI", itemvalue: "2345"},
]
var obj2 = { id:5, name: "ben", codevalue: "SP", idvalue:"2468"}

const result2 = arrobj2.filter(el => filterPredicate(el, obj2, arraylist));
console.log(result2);

// Expected Output
// []


arraylist = ["IN","FI","FR"];
var arrobj3 = [
  {id:1, name: "sun", cid: "IN", itemvalue: "1234"},
  {id:2, name: "mon", cid: "FI", itemvalue: "2468"},
  {id:3, name: "tues", cid: "SG", itemvalue: "2468"},
  {id:4, name: "thur", cid: "FI", itemvalue: "2345"},
]
var obj3 = { id:5, name: "ben", codevalue: "FI", idvalue:"2468"}

const result3 = arrobj3.filter(el => filterPredicate(el, obj3, arraylist));
console.log(result3);

// Expected Output
// [
//  {id:2, name: "mon", cid: "FI", itemvalue: "2468"}
// ]

这里重点是方法。尝试准确说出你的意思，尝试准确编码你所说的。我在这里这样做了，不管你信不信，该片段与第一个运行.

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