我有这两个表单，我想一键发送它们。第一个表单将图像加载到服务器中。第二个程序将信息加载到数据库中。由于某些原因，总是只发送"myform"。什么好主意吗?
<form action="uploadpic.php" method="POST" enctype="multipart/form- data" />
<input type="file" name="user_image" id="user_image" />
</form>
<form method = "post" id= "myform" name="myform" action = "<?php $_PHP_SELF ?>">
<table width = "400" border = "0" cellspacing = "1" cellpadding = "2">
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function () {
$("#add").click(function () {
var $form1 = $("#myform");
$.post($form1.attr("action"), $form1.serialize(), function () {
alert('Form 1 elküldve');
});
$('form[name="user_image"]').each(function () {
var $form = $(this);
$.post($form.attr("action"), $form.serialize(), function () {
alert('Form 2 elküldve');
});
})
});
});
</script>
你可以同样使用一个表单并发送两个数据到你想要的地方
<form action="uploadpic.php" method="POST" enctype="multipart/form- data" />
<input type="file" name="user_image" id="user_image" />
<table width = "400" border = "0" cellspacing = "1"
cellpadding = "2">
//... other form details
</form>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function () {
$("#add").click(function () {
var $form1 = $("#myform");
$.post($form1.attr("action"), $form1.serialize(), function () {
alert('Form 1 elküldve');
});
$('form[name="user_image"]').each(function () {
var $form = $(this);
$.post($form.attr("action"), $form.serialize(), function () {
alert('Form 2 elküldve');
});
})
});
});
</script>
您需要稍微修改一下javascript。$('input[name="user_image"]').each(function () {将用name="user_image"迭代输入并得到它的形式。(我猜这就是你想要的)
<script type="text/javascript">
$(document).ready(function () {
$("#add").click(function () {
var $form1 = $("#myform");
$.post($form1.attr("action"), $form1.serialize(), function () {
alert('Form 1 elküldve');
});
$('input[name="user_image"]').each(function () {
var $form = $(this).parent('form');
$.post($form.attr("action"), $form.serialize(), function () {
alert('Form 2 elküldve');
});
})
});
});
</script>
BUT如果您只有一个带有name="user_image"的输入元素，那么这是更合适的:
<script type="text/javascript">
$(document).ready(function () {
$("#add").click(function () {
var $form1 = $("#myform");
$.post($form1.attr("action"), $form1.serialize(), function () {
alert('Form 1 elküldve');
});
var $form = $('input[name="user_image"]').parent('form');
$.post($form.attr("action"), $form.serialize(), function () {
alert('Form 2 elküldve');
});
});
});
</script>