我试图将数据从表单发送到数据库。这是我使用的形式:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
典型的方法是提交表单，但这会导致浏览器重定向。使用jQuery和Ajax，是否有可能捕获所有表单的数据并将其提交给PHP脚本(例如，form. PHP)?
当前回答
请核对一下。它是完整的Ajax请求代码。
$('#foo').submit(function(event) {
// Get the form data
// There are many ways to get this data using jQuery (you
// can use the class or id also)
var formData = $('#foo').serialize();
var url = 'URL of the request';
// Process the form.
$.ajax({
type : 'POST', // Define the type of HTTP verb we want to use
url : 'url/', // The URL where we want to POST
data : formData, // Our data object
dataType : 'json', // What type of data do we expect back.
beforeSend : function() {
// This will run before sending an Ajax request.
// Do whatever activity you want, like show loaded.
},
success:function(response){
var obj = eval(response);
if(obj)
{
if(obj.error==0){
alert('success');
}
else{
alert('error');
}
}
},
complete : function() {
// This will run after sending an Ajax complete
},
error:function (xhr, ajaxOptions, thrownError){
alert('error occured');
// If any error occurs in request
}
});
// Stop the form from submitting the normal way
// and refreshing the page
event.preventDefault();
});
其他回答
我使用如下所示的方法。它像文件一样提交所有东西。
$(document).on("submit", "form", function(event)
{
event.preventDefault();
var url = $(this).attr("action");
$.ajax({
url: url,
type: 'POST',
dataType: "JSON",
data: new FormData(this),
processData: false,
contentType: false,
success: function (data, status)
{
},
error: function (xhr, desc, err)
{
console.log("error");
}
});
});
HTML:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" class="inputs" name="bar" type="text" value="" />
<input type="submit" value="Send" onclick="submitform(); return false;" />
</form>
JavaScript:
function submitform()
{
var inputs = document.getElementsByClassName("inputs");
var formdata = new FormData();
for(var i=0; i<inputs.length; i++)
{
formdata.append(inputs[i].name, inputs[i].value);
}
var xmlhttp;
if(window.XMLHttpRequest)
{
xmlhttp = new XMLHttpRequest;
}
else
{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
}
}
xmlhttp.open("POST", "insert.php");
xmlhttp.send(formdata);
}
请核对一下。它是完整的Ajax请求代码。
$('#foo').submit(function(event) {
// Get the form data
// There are many ways to get this data using jQuery (you
// can use the class or id also)
var formData = $('#foo').serialize();
var url = 'URL of the request';
// Process the form.
$.ajax({
type : 'POST', // Define the type of HTTP verb we want to use
url : 'url/', // The URL where we want to POST
data : formData, // Our data object
dataType : 'json', // What type of data do we expect back.
beforeSend : function() {
// This will run before sending an Ajax request.
// Do whatever activity you want, like show loaded.
},
success:function(response){
var obj = eval(response);
if(obj)
{
if(obj.error==0){
alert('success');
}
else{
alert('error');
}
}
},
complete : function() {
// This will run after sending an Ajax complete
},
error:function (xhr, ajaxOptions, thrownError){
alert('error occured');
// If any error occurs in request
}
});
// Stop the form from submitting the normal way
// and refreshing the page
event.preventDefault();
});
你可以使用serialize。下面是一个例子。
$("#submit_btn").click(function(){
$('.error_status').html();
if($("form#frm_message_board").valid())
{
$.ajax({
type: "POST",
url: "<?php echo site_url('message_board/add');?>",
data: $('#frm_message_board').serialize(),
success: function(msg) {
var msg = $.parseJSON(msg);
if(msg.success=='yes')
{
return true;
}
else
{
alert('Server error');
return false;
}
}
});
}
return false;
});
我使用这个简单的一行代码多年没有问题(它需要jQuery):
<script src="http://malsup.github.com/jquery.form.js"></script>
<script type="text/javascript">
function ap(x,y) {$("#" + y).load(x);};
function af(x,y) {$("#" + x ).ajaxSubmit({target: '#' + y});return false;};
</script>
这里ap()表示Ajax页面，af()表示Ajax表单。在表单中，只需调用af()函数即可将表单发送到URL，并在所需的HTML元素上加载响应。
<form id="form_id">
...
<input type="button" onclick="af('form_id','load_response_id')"/>
</form>
<div id="load_response_id">this is where response will be loaded</div>
