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首次尝试jsonp后脚本无法正常工作

[英]Script not working after first attempt with jsonp

So I'm learning jsonp right now and I can't currently get this test request to work. 因此,我现在正在学习jsonp,并且目前无法使该测试请求正常工作。

I have a file that where the main script occurs looking like this. 我有一个文件,其中主脚本出现如下所示。

(function(){
var jQuery;

if (window.jQuery==undefined || window.jQuery.fn.jquery!=='1.8.1'){
var script_tag=document.createElement('script');
script_tag.setAttribute("type", "text/javascript");
script_tag.setAttribute("src","http://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js");

if(script_tag.readyState){
script_tag.onreadystatechange=function(){
    if(this.readyState=='complete' || this.readyState=='loaded'){
    scriptLoadHandler();
    }
    };
}else{
    script_tag.onload=scriptLoadHandler;
    }
    (document.getElementsByTagName("head")[0] || document.documentElement).appendChild(script_tag);
}else{
jQuery=window.jQuery;
main();
}

function scriptLoadHandler(){
jQuery=window.jQuery.noConflict(true);
main();
}

function main(){
$(document).ready(function($){
     var jsonp_url = "http://www.reflap.com/beta/assets/js/atest.php?callback=theresponse";
        $.getJSON(jsonp_url, 'name=Michael', function(data) {
          alert (data.fullname);
        });
});
}
})();

And on the server at atest.php I have this 在服务器上atest.php我有这个

<?php
function theresponse(){
$fname= $_GET['name'];

if($fname=='Michael'){
echo  $_GET['callback']. '(' . "{'fullname':'Michael Yeah'}" . ')';
}
else
echo "Your not allowed here";
}
?>

However when I go on jsfiddle.net and execute 但是当我进入jsfiddle.net并执行时

<script src="http://www.reflap.com/beta/assets/js/widget2.js"></script>

It doesn't start the alert box. 它不会启动警报框。 What's wrong? 怎么了? I don't really see where I made a mistake. 我真的看不出我在哪里犯了错误。

$.getJSON is for ordinary JSON, not JSONP. $.getJSON用于普通JSON,而不是JSONP。 Try: 尝试:

var jsonp_url = "http://www.reflap.com/beta/assets/js/atest.php';
$.ajax({
    url: jsonp_url,
    dataType: 'jsonp',
    data: { name: 'Michael' }
}).done(function(data) {
    alert(data.fullname);
});

You don't need to put the callback=? 您不需要放置callback=? in the URL, jQuery will do that itself. 在URL中,jQuery会自行执行。

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