我得到了一个表格,其中包含一些需要发送的数据,我想用ajax来完成。我有一个功能,响应按钮的onclick事件。当我点击按钮时,我在firebug中获得了一些帖子数据,但它没有达到我的PHP脚本。有谁知道什么是错的?
JS:
function newItem() {
var dataSet = $("#createItem :input").serialize();
confirm(dataSet); //Below this code box is the output of this variable to check whether it is filled or not
var request = $.ajax({
type: "POST",
url: "/earnings.php",
data: dataSet,
dataType: "json"
});
request.done(function(){
$('.create_item').children(".row").slideUp('100', 'swing');
$('.create_item').children("h2").slideUp('100', 'swing');
confirm("succes");
});
request.fail(function(jqXHR, textStatus) {
confirm( "Request failed:" + textStatus );
});
}
完全填写表单时的 dataSet结果:
id=123&date=13-09-2013&amount=5&total=6%2C05&customer=HP&invoicenumber=0232159&quarter=1&description=Test
PHP:
<?php
include('includes/dbconn.php');
function clean_up($string){
$html = htmlspecialchars($string);
$string = mysql_real_escape_string($html);
return $string;
}
if($_POST){
$date = clean_up($_POST['date']);
$amount = clean_up($_POST['amount']);
$total = clean_up($_POST['total']);
$customer = clean_up($_POST['customer']);
$invoicenumber = clean_up($_POST['invoicenumber']);
$quarter = clean_up($_POST['quarter']);
$description = clean_up($_POST['description']);
$sql = ("INSERT INTO earnings (e_date, e_amount, e_total, e_customer, e_invoicenumber, e_quarter, e_description)
VALUES ($date, '$amount', '$total', '$customer', $invoicenumber, $quarter, '$description')");
echo $sql;
if($mysqli->query($sql) === true){
echo("Successfully added");
}else{
echo "<br /> \n" . $mysqli->error;
}
}
?>
表单在没有ajax的情况下工作正常,但是它只是不起作用。
感谢您的帮助!
答案 0 :(得分:1)
尝试使用此代码段代码...
<form id="F_login">
<input type="text" name="email" placeholder="Email">
<input type="password" name="password" placeholder="Password">
<button id="btn_login" type="submit">Login</button>
</form>
$("#btn_login").click(function(){
var parm = $("#F_login").serializeArray();
$.ajax({
type: 'POST',
url: '/earnings.php',
data: parm,
success: function (data,status,xhr) {
console.info("sukses");
},
error: function (error) {
console.info("Error post : "+error);
}
});
});
如果你试试这个,请回复我......
答案 1 :(得分:0)
阻止您的表单提交并使用这样的ajax:
<form id="createItem">
<input id="foo"/>
<input id="bar"/>
<input type="submit" value="New Item"/>
</form>
$('#createItem).on("submit",function(e){
e.preventDefault;
newItem();
});
答案 2 :(得分:0)
试试这个
<input type="text" id="foo"/>
<input type="text" id="bar"/>
<input type="button" id="btnSubmit" value="New Item"/>
<script type="text/javascript">
$(function() {
$("#btnSubmit").click(function(){
try
{
$.post("my php page address",
{
'foo':$("#foo").val().trim(),
'bar':$("#bar").val().trim()
}, function(data){
data=data.trim();
// alert(data);
// this data is data that the server sends back in case of ajax call you
//can send any type of data whether json or json array or any other type
//of data
});
}
catch(ex)
{
alert(ex);
}
});
});
</script>