我想重复发送用户名和密码的值到php
脚本。我该怎么做呢 ?喜欢将值发送到动作脚本,我们使用submit
按钮,但是如何将值自动发送到脚本并且连续发送?
<form method="post" action="processor.php">
<input type="username" value="suhail" />
<input type="password" value="secret_code" />
<input type="submit" />
</form>
答案 0 :(得分:1)
使用jQuery form plugin,您可以执行以下操作:
setInterval(function() {
$('form').ajaxSubmit();
}, 1000);
另一个解决方案是将表单定位到iframe,因此如果您提交表单,则不会重新加载页面:
HTML:
<form id="myform" method="post" action="processor.php" target="frm">
<input type="username" value="suhail" />
<input type="password" value="secret_code" />
<input type="submit" />
</form>
<iframe name="frm" id="frm"></iframe>
JS:
var form = document.getElementById('myform');
setInterval(function() {
form.submit();
}, 1000);
答案 1 :(得分:0)
尝试这样的事情
<强> JAVASCRIPT 强>
<script language=javascript>
var int=self.setInterval(function(){send_data()},1000);
function send_data()
{
document.getElementById('my_form').submit()
}
</script>
<强> HTML 强>
<form method="post" id="my_form" action="processor.php">
<input type="username" value="suhail" />
<input type="password" value="secret_code" />
</form>
答案 2 :(得分:0)
<form id="myform" method="post" action="processor.php">
<input type="username" value="suhail" />
<input type="password" value="secret_code" />
<input type="submit" />
</form>
<script type="text/javascript">
var count=100,i=0;
for(i=0;i<count;i++) {
document.getElementById('myform').submit();
}
</script>
这将提交表格100次
答案 3 :(得分:0)
使用Ajax,使用jQuery非常简单。要将表单数据发送到processor.php
脚本:
var sendForm = function () {
$.ajax({
type: 'post',
url: 'processor.php',
dataType: 'JSON',
data: {
username: $('#username').val(),
password: $('#password').val()
},
success: function (data) {
// do something with the answer from server?
},
error: function (data) {
// handle error
}
});
}
因此,sendForm
是一个将表单数据发送到服务器的函数。现在,我们需要设置一个将重复调用它的计时器:
window.setInterval(sendForm, 1000); // sends form data every 1000 ms
答案 4 :(得分:0)
您可以反复向$ .post或$ .get或$ .ajax请求发送连续请求。
$(document).ready(function(){
setInterval(function() {
var username = $("#username").val();
var password = $("#password").val();
var dataString = 'username='+username+"&password="+password;
$.post('login.php',dataString,function(response){
//your code what you want to do of response
alert(response);
});
}, 1000);
});
和html代码如下所示
<form method="post" action="processor.php">
<input type="username" value="suhail" id="username"/>
<input type="password" value="secret_code" id="password"/>
<input type="submit" />
</form>
答案 5 :(得分:0)
这是一个完整的HTML文件,可以满足您的需求,阅读评论。
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<form method="post" action="processor.php">
<input type="username" id="username" value="suhail" />
<input type="password" id="password" value="secret_code" />
<input type="submit" />
</form>
<script>
function send_request(username, password) {
var dataString = 'username='+username+"&password="+password;
$.post('login.php',dataString,function(response){
// You can check if the login is success/fail here
console.log(response);
// Send the request again, this will create an infinity loop
send_request(username, password);
});
}
// Start sending request
send_request($('#username').val(), $('#password').val());
</script>
答案 6 :(得分:0)
试试这个,
<强> JS:强>
$(document).ready(function(){
var int=self.setInterval(function(){statuscheck()},1000);
function statuscheck()
{
var username = $("#username").val();
var password = $("#password").val();
$.ajax({
type:"post",
url:"processor.php",
dataType: "html",
cache:false,
data:"&username="+username+"&password="+password,
success:function(response){
alert(response);
}
});
}
});
<强> HTML:强>
<form method="post" action="processor.php">
<input type="username" value="suhail" id="username"/>
<input type="password" value="secret_code" id="password"/>
<input type="submit" />
</form>