我正在创建一个添加到购物车程序,需要将产品及其属性(颜色、尺寸)添加到购物车,为此我需要同时提交两个表单。我不确定我哪里出错了我已经创建了脚本,但它只提交了使用 jquery 为 submit()
选择的第一个表单,而不是其他表单。
下面是我的代码片段,这是 JSFIDDLE
$(document).ready(function (){
$('#cart').click(function (e1) {
var $form = $('#masterform');
var $formcolor = $('#colorform');
var $checkbox = $('.roomselect');
var $checkboxcolor = $('.colorselect');
if (!$checkbox.is(':checked'))
{
$('#tipdivcontent').css("display", "block");
$("#tipdivcontent").delay(4000).hide(200);
e.preventDefault();
}
else
{
if (!$checkboxcolor.is(':checked')) {
$('#tipdivcontentcolor').css("display", "block");
$("#tipdivcontentcolor").delay(4000).hide(200);
e.preventDefault();
} else {
$form.submit();
$formcolor.submit();
}
}
});
});
#tipdivcontent
{
border:1px solid black;
margin-top:0px;
background-color:white;
height:50px;
width:102px;
display:none;
position:relative;
background-color:red;
color:yellow;
font-weight:bold;
}
#tipdivcontentcolor
{
border:1px solid black;
margin-top:0px;
background-color:white;
height:18px;
width:292px;
display:none;
position:absolute;
background-color:red;
color:yellow;
font-weight:bold;
}
<form action="" method="POST" id="masterform">
<table border="1" cellspacing="0">
<tr>
<th colspan="4">Sizes</th>
</tr>
<tr>
<td>
<label for="2.2">2.2</label>
</td>
<td>
<input class="roomselect" type="radio" id="2.2" name="size" value="twopointtwo">
</td>
<td>
<label for="2.4">2.4</label>
</td>
<td>
<input class="roomselect" type="radio" id="2.4" name="size" value="twopointfour">
</td>
</tr>
<tr>
<td>
<label for="2.6">2.6</label>
</td>
<td>
<input class="roomselect" type="radio" id="2.6" name="size" value="twopointsix">
</td>
<td>
<label for="2.8">2.8</label>
</td>
<td>
<input class="roomselect" type="radio" id="2.8" name="size" value="twopointeight">
</td>
</tr>
<tr>
<td colspan="3" align="center">
<label for="2.10">2.10</label>
</td>
<td>
<input class="roomselect" type="radio" id="2.10" name="size" value="twopointten">
</td>
</tr>
</table>
</form>
<div id="tipdivcontent">Please Select Size.</div>
<input type="submit" value="To Cart" class="cartorcheckoutbutton" id="cart">
<form action="" method="POST" id="masterform">
<table border="1" cellpadding="2">
<tr>
<th colspan="8">COLORS</th>
</tr>
<tr>
<th title='White' style='background-color:white;' height='15' width='20'>
<input type='radio' name='color' class="colorselect" value='white'>
</th>
<th title='Red' style='background-color:red;' height='15' width='20'>
<input type='radio' name='color' class="colorselect" value='red'>
</th>
<th title='Green' style='background-color:green;' height='15' width='20'>
<input type='radio' name='color' class="colorselect" value='green'>
</th>
<th title='Blue' style='background-color:blue;' height='15' width='20'>
<input type='radio' name='color' class="colorselect" value='blue'>
</th>
</tr>
</table>
</form>
<div id="tipdivcontentcolor">Please Select Color.</div>
最佳答案
您可以尝试将辅助表单中的颜色输入分配给您的“主”表单。只需使用 <input form='formID' ...>
在任何输入上都会将该输入分配给其他表单,而不管它在页面上的什么位置。
// When the 'master' form is submitted...
$("#masterForm").on("submit", function(e) {
"use strict";
// Stop the default action
e.preventDefault();
if ($("input[type='radio']:checked").length === 0) {
alert("You must check at least one color option.");
return false;
}
// *for logging*
// write the contents of the submitted data
$("p").html("submitted data: " + $(this).serialize());
console.log("submitted data: " + $(this).serialize());
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form id="masterForm">
<input name="someField" type="text" value="test value">
</form>
<form id="anotherForm">
<label>
<input name="color" type="radio" value="blue" form="masterForm">Blue
</label>
<label>
<input name="color" type="radio" value="red" form="masterForm">Red
</label>
</form>
<!-- Submit button outside of the form -->
<button type="submit" form="masterForm">Submit</button>
<p></p>
如果上述选项(和附加的代码段)对您不起作用,请尝试将您的 formData 附加到相关字段。像这样的东西(未经测试):
// When the 'master' form is submitted...
$("#masterForm").on("submit", function(e) {
"use strict";
// Stop the default action
e.preventDefault();
var submissionData = $(this).serialize();
submissionData += "&color=" + $("#slaveForm").find("input[name='color']:checked").val()
// do stuff ...
});
关于javascript - 使用外部提交按钮提交两种不同的表单无法正常工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31513715/