早上好,我想通过帖子将我的两个变量发送到外部网站。这两个变量是用户名和密码,以及当我转到另一个url时在浏览器中创建缓存的页面,如果他们是正确的,我想做的是强迫表单发送提交url,问题是一切都很好,提交表单不再做任何事情,你能帮我吗?
谢谢
我的代码:
<script language="javascript" type="text/javascript">
$(document).ready(function () {
$("#submit").click(function () {
var user = $("#user").val();
var pass = $("#pass").val();
// Returns successful data submission message when the entered information is stored in database.
var dataString = 'user=' + user + '&pass=' + pass;
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "index.php",
data: dataString,
cache: false,
success: function (result) {
if (result == 'si') {
$("form").attr('action', 'http://xzc.tv/login');
$("form").submit();
//window.open('http://xzc.tv');
alert('entro if');
return true;
} else {
alert('entro else');
}
}
});
return false;
});
});
</script>
</head>
<body>
<form action="" method="post" id="formoid">
<div><label>Usuario:</label><input name="user" placeholder="Usuario" type="text" id="user"></div>
<div><label>Contraseña:</label><input name="pass" placeholder="Contraseña" type="text" id="pass"></div>
<div><input name="enviar" type="submit" value="Enviar" id="submit"></div>
<div><input name="btnPass" type="submit" value="¿Olvidaste tu contraseña?"></div>
</form>
</body>
只需在点击处理程序中声明事件参数,然后执行event.prventDefault()。这样做,默认的提交操作将被忽略,您的代码将被执行:
$(document).ready(function () {
$("#submit").click(function (e) {
e.preventDefault();
var user = $("#user").val();
var pass = $("#pass").val();
// Returns successful data submission message when the entered information is stored in database.
var dataString = 'user=' + user + '&pass=' + pass;
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "index.php",
data: dataString,
cache: false,
success: function (result) {
if (result == 'si') {
$("form").attr('action', 'http://xzc.tv/login');
$("form").submit();
//window.open('http://xzc.tv');
alert('entro if');
return true;
} else {
alert('entro else');
}
}
});
return false;
});
});
检查结果是否返回"si"如果正确尝试
更改代码从…起$("form").attr("action",'http://xzc.tv/login');$("form").submit();到$("#formoid").attr('操作','http://xzc.tv/login');$("#form").submit();
尝试使用表单的id
要提交,请尝试以下
$("form#formoid").submit();
$("form#formoid")[0].submit();
$("form#formoid").submit(function(e) {});