我有一個物件陣列，我想根據另一個陣列過濾它。如果filters物件值只有一個，我可以使用 only filter，但是如果物件中還有一個值filters，則什么也不顯示。所以forEach在過濾器內部使用并希望回傳擬合值。你能幫我看看是什么問題嗎？謝謝
預期結果：
[
{
"tags": [
"madewithout__gluten",
"madewithout__nuts",
],
"metafields": "Side"
}
]
const data = [
{
"tags": [
"madewithout__gluten",
"madewithout__nuts",
],
"metafields": "Side"
},
{
"tags": [
"madewithout__gluten",
],
"metafields": "Side"
},
{
"tags": [
"madewithout__nuts"
],
"metafields": "Side"
}
]
const filters = ['gluten', 'nuts']
const result = data.filter((v) => {
filters.forEach((tag) => {
if(!v.tags.includes(`madewithout__${tag}`))
return;
});
});
console.log(result);uj5u.com熱心網友回復：
根據您提供的資料的語意，您要檢查是否所有filters內容都包含在tags：
const result = data.filter((v) =>
filters.every((tag) => v.tags.includes(`madewithout__${tag}`))
)
const data = [
{
"tags": [
"madewithout__gluten",
"madewithout__nuts",
],
"metafields": "Side"
},
{
"tags": [
"madewithout__gluten",
],
"metafields": "Side"
},
{
"tags": [
"madewithout__nuts"
],
"metafields": "Side"
}
]
const filters = ['gluten', 'nuts']
const result = data.filter((v) =>
filters.every((tag) => v.tags.includes(`madewithout__${tag}`))
)
console.log(result);uj5u.com熱心網友回復：
而不是.forEach，使用.some
const result = data.filter((v) => {
return !filters.some((tag) => !v.tags.includes(`madewithout__${tag}`))
});
顯示代碼片段
const data = [{
"tags": [
"madewithout__gluten",
"madewithout__nuts",
],
"metafields": "Side"
},
{
"tags": [
"madewithout__gluten",
],
"metafields": "Side"
},
{
"tags": [
"madewithout__nuts"
],
"metafields": "Side"
}
]
const filters = ['gluten', 'nuts']
const result = data.filter((v) => {
return !filters.some((tag) => !v.tags.includes(`madewithout__${tag}`))
});
console.log(result);您的代碼有問題
Array.forEach不回傳任何東西。所以你的回報不會打破回圈或回傳值。- 你
Array.filter沒有return宣告。因此，默認情況下一切都是假的（回傳未定義），因此是空陣列
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標籤：javascript