我有以下物件陣列；但是，這可能是任何未知的鍵/值并且可以無限嵌套，現在這是一個測驗示例：
[
{
"reference_id": "R123",
"customer": "Person 1",
"customer_email": "[email protected]",
"location": "UK",
"bookings": [
{
"product": "Product 1",
"provider": "Company 1",
"cancellable": true
},
{
"product": "Product 2",
"provider": "Company 2",
"cancellable": true
},
{
"product": "Product 3",
"provider": "Company 1",
"cancellable": true
}
]
},
{
"reference_id": "R1234",
"customer": "Person 2",
"customer_email": "[email protected]",
"location": "USA",
"bookings": [
{
"product": "Product 1",
"provider": "Company 1",
"cancellable": true
},
{
"product": "Product 3",
"provider": "Company 1",
"cancellable": true
}
]
},
{
"reference_id": "R12345",
"customer": "Person 3",
"customer_email": "[email protected]",
"location": "UK",
"bookings": [
{
"product": "Product 2",
"provider": "Company 2",
"cancellable": true
},
{
"product": "Product 3",
"provider": "Company 1",
"cancellable": true
}
]
}
]
我目前的實作如下：
const selected = [
{
term: 'Company 1',
column: 'provider',
},
{
term: 'Person 1',
column: 'customer',
},
];
const recursivelyFilterByValue = () => (value) => selected.every((item) => {
if (!value) return false;
if (typeof value === 'string') {
// console.log('value', value === item.term);
return value === item.term;
}
if (Array.isArray(value)) {
return value.some(this.recursivelyFilterByValue());
}
if (typeof value === 'object') {
return Object.values(value).some(this.recursivelyFilterByValue());
}
return false;
});
const results = data.filter(recursivelyFilterByValue());
基本上，我將添加到“選定”陣列中，然后使用它來過濾資料陣列。我確實想確保密鑰與“列”匹配，但我還沒有添加。
對于上面的輸入，我希望輸出以下內容：
[
{
"reference_id": "R123",
"customer": "Person 1",
"customer_email": "[email protected]",
"location": "UK",
"bookings": [
{
"product": "Product 1",
"provider": "Company 1",
"cancellable": true
},
{
"product": "Product 2",
"provider": "Company 2",
"cancellable": true
},
{
"product": "Product 3",
"provider": "Company 1",
"cancellable": true
}
]
},
]
但是輸出陣列是空的。如果我只搜索一個術語（從所選陣列中洗掉除一個術語之外的所有術語），則該術語的輸出是正確的，但是任何后續術語都會回傳一個空白陣列。
我想知道我對 .some() 的使用是否是問題，但是改變它會導致太多的遞回錯誤。
Essentially, I want to return the original parent object so long as there is a key:value match for all my conditions in the selected array, at any level of its children.
Any guidance would be much appreciated, thank you.
uj5u.com熱心網友回復：
我不太確定這是否是你要找的。它假設我在評論中的猜測是正確的：
我有這個權利嗎？您有一個（可能是動態的）條件，表明一個物件要么具有
provider具有值的屬性，要么具有具有值"Customer 1"的（遞回）后代物件。并且您有關于customerand的第二個條件"Person 1"，并且您正在尋找同時滿足兩個（或所有）這些條件的物件。這是否描述了您正在嘗試做的事情？
這里我們有兩個相當簡單的輔助函式，testRecursive以及makePredicates主函式recursivelyFilterByValue：
const testRecursive = (pred) => (obj) =>
pred (obj) || Object (obj) === obj && Object .values (obj) .some (testRecursive (pred))
const makePredicates = (criteria) =>
criteria .map (({term, column}) => (v) => v [column] == term)
const recursivelyFilterByValue = (criteria, preds = makePredicates (criteria)) => (xs) =>
xs .filter (obj => preds .every (pred => testRecursive (pred) (obj)))
const selected = [{term: 'Company 1', column: 'provider'}, {term: 'Person 1', column: 'customer'}]
const input = [{reference_id: "R123", customer: "Person 1", customer_email: "[email protected]", location: "UK", bookings: [{product: "Product 1", provider: "Company 1", cancellable: true}, {product: "Product 2", provider: "Company 2", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}, {reference_id: "R1234", customer: "Person 2", customer_email: "[email protected]", location: "USA", bookings: [{product: "Product 1", provider: "Company 1", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}, {reference_id: "R12345", customer: "Person 3", customer_email: "[email protected]", location: "UK", bookings: [{product: "Product 2", provider: "Company 2", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}]
console .log (recursivelyFilterByValue (selected) (input)).as-console-wrapper {max-height: 100% !important; top: 0}testRecursive檢查謂詞對于物件或嵌套在其中的任何物件是否為真。makePredicates將 -objects陣列{term, column}轉換為謂詞函式，用于測驗物件在列命名的屬性中是否具有適當的術語。recursivelyFilterByValue結合這些，呼叫makePredicates將所選專案轉換為謂詞函式，然后通過測驗每個謂詞是否為真來過濾輸入。
這不是可以想象的最有效的代碼。它重新掃描每個謂詞的層次結構。我相信我們可以找出一個版本來只進行一次掃描，但我認為它會產生更復雜的代碼。因此，您可能希望在生產規模的資料中測驗它是否足夠快滿足您的需求。
uj5u.com熱心網友回復：
這個解決方案不像公認的答案那么優雅，但為什么不展示我的努力。也許有人會覺得這種方式更容易理解。
const selected = [{term: 'Company 1', column: 'provider'}, {term: 'Person 1', column: 'customer'}]
const input = [{reference_id: "R123", customer: "Person 1", customer_email: "[email protected]", location: "UK", bookings: [{product: "Product 1", provider: "Company 1", cancellable: true}, {product: "Product 2", provider: "Company 2", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}, {reference_id: "R1234", customer: "Person 2", customer_email: "[email protected]", location: "USA", bookings: [{product: "Product 1", provider: "Company 1", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}, {reference_id: "R12345", customer: "Person 3", customer_email: "[email protected]", location: "UK", bookings: [{product: "Product 2", provider: "Company 2", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}]
const iter = (obj, sel) =>
Object.entries(obj).some(([key, value]) => {
if (Array.isArray(value))
return value.some((obj) => iter(obj, sel));
if (typeof value === 'object' && value !== null)
return iter(value, sel);
return (key === sel.column) && (value === sel.term);
});
const deepFilter = (arr, sels) =>
arr.filter((obj) =>
sels.every((sel) => iter(obj, sel)));
console.dir(deepFilter(input, selected), {depth: null});.as-console-wrapper {max-height: 100% !important; top: 0}轉載請註明出處，本文鏈接：https://www.uj5u.com/shujuku/411301.html
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