我正在尝试使用
XMLHttpRequest对象从表单发送
JSON数据.我可以使用以下函数发送数据. FireBug中没有显示错误,请求中的JSON数据显示为FireBug.
<?PHP print_r($_POST); print_r($_GET); foreach (getallheaders() as $name => $value) { echo "$name: $value\n"; } echo file_get_contents('PHP://input'); ?>
POST数组始终为空,但我可以看到file_get_contents返回的JSON字符串.这是怎么发生的？我究竟做错了什么？
Array
(
)
Array
(
)
Host: localhost
User-Agent: Mozilla/5.0 (X11; Ubuntu; Linux i686; rv:10.0.2) Gecko/20100101 Firefox/10.0.2
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: eo,de-de;q=0.8,de;q=0.6,en-us;q=0.4,en;q=0.2
Accept-Encoding: gzip,deflate
Connection: keep-alive
Content-Type: application/json; charset=utf-8
Referer: http://localhost/form.html
Content-Length: 88
Cookie: {{..to much data..}}
Pragma: no-cache
Cache-Control: no-cache
{"type":"my_type","comment":"commented"}
发送功能：
function submit(){
var data={};
data.type=document.form.type.value;
data.comment=document.form.comment.value;
//get right XMLHttpRequest object for current browsrer
var x=ajaxFunction();
var string = JSON.stringify(data);
x.open('POST','echo.PHP',true);
x.setRequestHeader('Content-type','application/json; charset=utf-8');
x.setRequestHeader("Content-length",string.length);
x.setRequestHeader("Connection","close");
x.onreadystatechange = function(){
if (x.readyState != 4) return;
if (x.status != 200 && x.status != 304) {
alert('HTTP error ' + req.status);
return;
}
data.resp = JSON.parse(x.responseText);
if(data.resp.status=='success'){
alert('That worked!');
}else{
alert('That didn\'t work!');
}
}
x.send(string);
return false; //prevent native form submit
}
