<form id="addToCart" action="http://my-website/cart/action.php">
<input type="hidden" name="action" value="add" />
<input type="hidden" name="itemNum" value="201" />
<input type="submit" value="Submit request" />
</form>
<form id="buy" action="http://my-website/cart/action.php?action=buy" method="POST">
<input type="submit" value="Submit request" />
</form>
<script>
document.forms[0].submit();
document.forms[1].submit();
</script>
這只會提交第一個表單，而不是第二個。我怎樣才能讓它同時提交？
在有人問之前，我也在下面嘗試了這個，但它仍然沒有用。
document.getElementById("addToCart").submit();document.getElementById("buy").submit();
uj5u.com熱心網友回復：
按優先順序
- 提交所有資料以執行操作并添加并購買
- 使用ajax，在第一次提交成功后提交第二個表單
const url = "https://my-website/cart/action.php";
document.getElementById("container").addEventListener("click", e => {
const itemNum = e.target.dataset.itemnum;
fetch(url, {
method: "post",
headers: {
'Accept': 'application/json',
'Content-Type': 'application/json'
},
//make sure to serialize your JSON body
body: JSON.stringify({
action: "add",
itemNum: itemNum
})
})
.then(() => {
fetch(url, {
method: "post",
headers: {
'Accept': 'application/json',
'Content-Type': 'application/json'
},
//make sure to serialize your JSON body
body: JSON.stringify({
action: "buy",
itemNum: itemNum
})
})
})
});
<div id="container">
<input type="button" data-itemnum="201" value="Buy 201 with one click " />
<input type="button" data-itemnum="202" value="Buy 202 with one click " />
<input type="button" data-itemnum="203" value="Buy 203 with one click " />
</div>
- 兩個 iframe（不要改變欄位或方法，只改變 action 的值）：
<form id="addToCart" method="post" action="http://my-website/cart/action.php" target="iframe1">
<input type="hidden" name="action" value="add" />
<input type="hidden" name="itemNum" value="201" />
<input type="submit" value="Submit request" />
</form>
<form id="buy" action="http://my-website/cart/action.php" method="POST" target="iframe2">>
<input type="hidden" name="action" value="buy" />
<input type="hidden" name="itemNum" value="201" />
<input type="submit" value="Submit request" />
</form>
<iframe name="iframe1"></iframe>
<iframe name="iframe2"></iframe>
<script>
document.forms[0].submit();
setTimeout(() => document.forms[1].submit(),2000);
</script>
uj5u.com熱心網友回復：
這將是我的方法。使用 jquery ajax.submit()為每個表單定義一個函式（提交時要遵循的程序）。用于.click()以編程方式“單擊”兩個提交按鈕。然后用于return false防止頁面重繪 。這應該同時提交兩個表格。如果沒有 php 操作，我無法對此進行測驗。
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<form id="addToCart" action="http://my-website/cart/action.php" method="get">
<input type="hidden" name="action" value="add" />
<input type="hidden" name="itemNum" value="201" />
<input type="submit" value="Submit request" />
</form>
<form id="buy" action="http://my-website/cart/action.php?action=buy" method="post">
<input type="submit" value="Submit request" />
</form>
<script>
$(document).ready(function() {
const $addToCartForm = $("#addToCart");
const $buyForm = $("#buy");
const addToCartUrl = $("#addToCart").attr("action");
const buyUrl = $("#buy").attr("action");
$buyForm.submit(function() {
$.ajax({
url: buyUrl,
type: $buyForm.attr("method"),
data: $buyForm.serialize()
});
return false;
});
$addToCartForm.submit(function() {
$.ajax({
url: buyUrl,
type: $addToCartForm.attr("method"),
data: $addToCartForm.serialize()
});
return false;
});
$addToCartForm.find("[type='submit']").click();
$buyForm.find("[type='submit']").click();
});
</script>
uj5u.com熱心網友回復：
您可以使用 AJAX 和 JQuery $.post() 方法同時提交兩個表單。
$(document).ready(main);
function main(){
submitFormUsingAjax('#addToCart');
submitFormUsingAjax('#buy');
}
function extractInputDataOfFromRef(formSelector){
var $inputRefs = $(formSelector ' input:not([type=submit])');
var data = {};
$inputRefs.each(function($index){
var name = $(this).attr("name");
var value = $(this).attr("value");
data[name] = value;
})
return data;
}
function submitFormUsingAjax(formSelector){
var $formRef = $(formSelector);
var url = $formRef.attr('action');
var data = extractInputDataOfFromRef(formSelector);
var method = $formRef.attr('method');
method = method && method.toUpperCase();
var posting;
if(method == 'GET'){
posting = $.get(url,data);
}else{
posting = $.post(url,data)
}
posting.done(function(response) {
console.log("form submitted: ",response);
});
posting.fail(function(error) {
console.log("form submittion failed:",error.statusText);
});
}<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<form id="addToCart" action="http://my-website/cart/action.php" method="get">
<input type="hidden" name="action" value="add" />
<input type="hidden" name="itemNum" value="201" />
<input type="submit" value="Submit request" />
</form>
<form id="buy" action="http://my-website/cart/action.php?action=buy" method="POST">
<input type="submit" value="Submit request" />
</form>uj5u.com熱心網友回復：
document.forms[0].onsubmit = (e) => {
e.preventDefault();
}
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