2 回答

TA貢獻(xiàn)1883條經(jīng)驗(yàn) 獲得超3個(gè)贊
querySelectorAll如果您使用父 div 包裝每個(gè) orderId 和 name 元素,并且您是父 div,則看起來會更清晰。我認(rèn)為不可能將鍵和值同時(shí)作為 JSON 格式的變量。這是我要做的(假設(shè)您的 orderId 和名稱的長度相同):
const orderIds = document.querySelectorAll(".orderid")
const names = document.querySelectorAll(".name")
let records = []
for(let i=0; i < orderIds.length; i++){
const record = {
orderId: orderIds[i].textContent,
name: names[i].textContent
}
records.push(record)
}
然后您發(fā)送records到后端并從中提取數(shù)據(jù)。

TA貢獻(xiàn)1844條經(jīng)驗(yàn) 獲得超8個(gè)贊
我希望這能幫到您:
let results = {orders:{},names:{}}
let ordersElements = document.querySelectorAll(".orderId");
let namesElements = document.querySelectorAll(".name")
ordersElements.forEach((element,idx) => results["orders"][idx] = element.innerHTML)
namesElements.forEach((element,idx) => results["names"][idx] = element.innerHTML)
console.log(results)
<div class="orderId">1</div>
<div class="orderId">2</div>
<div class="orderId">3</div>
<div class="orderId">4</div>
<div class="orderId">5</div>
<div class="name">first</div>
<div class="name">second</div>
<div class="name">third</div>
<div class="name">fourth</div>
<div class="name">last</div>
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