使用JQuery的AJAX中最简单的脚本不起作用
[英]easiest script in AJAX with JQuery doesn't work
问题描述
im tring to learn using ajax but no one of my script work: 
我想学习使用ajax，但我的脚本没有工作：
<script>
function test()
{
var par = document.getElementById("nome").value;
alert ("i m here!"+par);
$.Ajax({
type: "POST",
url: "ProvaAJAX.php",
data: "par="+par,
success: function(msg){
alert( "Data Saved: " + msg );
}
});
}
</script>
Select Item: 选择物品：
<select id="nazione_pr" class="select-registrazione" onchange="prova()" name="nazione_pr">
Input item: 输入项目：
<input id="nome" class="input-text" type="text"name="nome">
ProvaAJAX.php ProvaAJAX.php
<html>
<?php
echo "Test success". $_POST['par'];
?>
</html>
.... i think this script should be ok but my server doesn't think this... when i change the "select" value the "im here" alert box appear but after nothing happen... ....我认为这个脚本应该没问题，但我的服务器不认为这...当我更改“选择”值时，“我在这里”警告框出现但是什么都没发生......
3 个解决方案
解决方案1
2 已采纳 2013-09-03 11:56:12
Many issues 很多问题
It's lowercase a in $.ajax
它是$ .ajax中的小写字母 data: {"par":par},
数据：{“par”：par}， you use jQuery so use var par = $("#nome").val();
你使用jQuery所以使用var par = $（“＃nome”）。val（）; your function is called something else than what you call it
你的函数被称为别的东西而不是你所说的
Hit F12 in IE or Chrome or install firebug in Firefox and hit F12 there too to see the console 在IE或Chrome中点击F12或在Firefox中安装firebug并在那里点击F12以查看控制台
<script>
$(function() { // when the page has loaded
$("#nazione_pr").on("change",function() {
var nazione = $(this).val(); // the select's value
var par =$("#nome").val(); // the textfield's value
$.ajax({
type: "POST",
url: "ProvaAJAX.php",
data: {"par":par,"nazione":nazione},
success: function(msg){
alert( "Data Saved: " + msg );
}
});
});
});
</script>
解决方案2
0 2013-09-03 11:54:40
please change onchange="prova()" to onchange="test()" 请将onchange =“prova（）”更改为onchange =“test（）”
<script>
function test()
{
var par = document.getElementById("nome").value;
$.ajax({
type: "POST",
url: "ProvaAJAX.php",
data: { par:par,fun:"test" }
}).done(function( data )
{
alert( "Data Saved: " + data );
});
}
</script>
and php file:- 和PHP文件： -
if($_REQUEST['fun'] == "test")
{
echo $_REQUEST['par'];
}
解决方案3
0 2013-09-03 11:59:55
// Launch AJAX request.
$.ajax(
{
// The link we are accessing.
url: jLink.attr( "href" ),
// The type of request.
type: "get",
// The type of data that is getting returned.
dataType: "html",
error: function(){
ShowStatus( "AJAX - error()" );
// Load the content in to the page.
jContent.html( "<p>Page Not Found!!</p>" );
},
beforeSend: function(){
ShowStatus( "AJAX - beforeSend()" );
},
complete: function(){
ShowStatus( "AJAX - complete()" );
},
success: function( strData ){
ShowStatus( "AJAX - success()" );
// Load the content in to the page.
jContent.html( strData );
}
}
);
// Prevent default click.
return( false );
}
);
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