javascript 使用ajax将表单数据发送到php
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Using ajax to send form data to php
提问by
I want to send form data from ajax to php. My ajax retrieves the form data but it doesnt send it i dont see anything wrong with the code maybe i need a much more professional help. Thanks in advance
我想将表单数据从 ajax 发送到 php。我的 ajax 检索了表单数据,但它没有发送它我没有看到代码有任何问题,也许我需要更专业的帮助。提前致谢
HTML5 syntax
HTML5 语法
<div align="center"><a id="hi">Header</a></div>
<a id="signup" data-add-back-btn="true" style="float:right;" data-icon="arrow-r">Sign- In</a>
</div>
<form class="check-user" action="php/sup.php" method="POST">
<label>Username</label>
<input id="Susername" name="username" placeholder="username" type="text" >
</div>
<div align="center" data-role="fieldcontain" style="width:100%;overflow:hidden" data-position="static">
<label>Email</label>
<input id="Semail" name="email" placeholder="email" type="email" >
</div>
<div align="center" data-role="fieldcontain" style="width:100%;overflow:hidden" data-position="static">
<label>Password</label>
<input id="Spassword" name="password" placeholder="password" type="password" >
</div>
<!---input type="submit" style="visibility:hidden;" id="send"/-->
</form>
Ajax syntax
Ajax 语法
$('#signup').live('click', function(){
//var name = document.getElementById('Susername').value;
//var email = document.getElementById('Semail').value;
//var pass = document.getElementById('Spassword').value;
var that = $('form.check-user'),
urls = that.attr('action'),
methods = that.attr('method'),
data = {};
that.find('[name]').each(function(index, element) {
var that = $(this),
name = that.attr('name'),
element = that.val();
alert(name+'='+element+' '+methods);
data[name] = element;
});
$.ajax(
{
url: urls,
type: methods,
data : data,
beforeSend: function(response){alert('Sending');},
success: function(response){ alert('success');},
error: function(response){alert('failed');},
complete: function(response){alert('finished');},
}
);
return false;
});
PHP syntax
PHP语法
session_start();
$name = $_POST['username'];
$email = $_POST['email'];
$password = $_POST['password'];
if(isset($name) && isset($email) && isset($password))
{
echo $name;
$_SESSION['username'] = $name;
}
else
{
die('data not set');
}
回答by Gena Moroz
You can use serialize method on form, it will gather everything correct.
您可以在表单上使用序列化方法,它会收集所有正确的内容。
$('#signup').live('click', function(){
var that = $('form.check-user'),
urls = that.attr('action'),
methods = that.attr('method'),
data = that.serialize();
$.ajax(
{
url: urls,
type: methods,
data : data,
beforeSend: function(response){alert('Sending');},
success: function(response){ alert('success');},
error: function(response){alert('failed');},
complete: function(response){alert('finished');},
}
);
return false;
});
回答by Ronjon
Try this,
试试这个,
$('#signup').live('click', function(){
$.ajax({
url:'',//url to submit
type: "post",
dataType : 'json',
data : {
'Susername' : $('#Susername').val(),
'Semail' : $('#Semail').val(),
'Spassword' : $('#Spassword').val()
},
success: function (data)
{
}
});
return false;
});
回答by Ronjon
I solved it works this way on the php side you do this
我解决了它在 php 方面以这种方式工作,你这样做
$name = isset(json_decode($_POST['username']));//htmlentities($values[0]);
$email = isset(json_decode(($_POST['email'])));//htmlentities($values[1]);
$password = isset(json_decode($_POST['password']));//htmlentities($values[2]);
The Ajax side
阿贾克斯方面
$(document).ready(function(e) {
$('#signup').live('click', function(){
//var name = document.getElementById('Susername').value;
//var email = document.getElementById('Semail').value;
//var pass = document.getElementById('Spassword').value;
var that = $('form.check-user'),
urls = that.attr('action'),
methods = that.attr('method'),
data = {};
data = that.serialize();
console.log(data);
$.ajax(
{
url: urls,
type: methods,
dataType:'json',
data : data,
beforeSend: function(response){$.mobile.showPageLoadingMsg(true);},
success: function(response){ $.mobile.showPageLoadingMsg(false);},
error: function(xhr, textStatus, errorThrown){alert(textStatus);},
complete: function(response){},
}
);
return false;
});
});
});