PHP返回无效的请求错误

时间:2012-08-23 09:20:57

标签: php android

我正在将数据从Android设备保存到MySQL。我的代码使用Localhost工作正常,但是当我使用托管域尝试它时。它给了我和错误。

错误返回

ERROR
The requested URL could not be retrieved

While trying to process the request:

POST /insert.php HTTP/1.1
Transfer-Encoding: chunked
Content-Type: application/x-www-form-urlencoded
Host: mydomain.com
Connection: Keep-Alive
User-Agent: Apache-HttpClient/UNAVAILABLE (java 1.4)

The following error was encountered:
Invalid Request

AsyncTask中的Java代码

.
.
.
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("user_id", userId));
HttpPost httppost = new HttpPost("http://mydomain.com/insert.php");
HttpClient httpclient = new DefaultHttpClient();

UrlEncodedFormEntity urlEncodeFormEntity = new UrlEncodedFormEntity(nameValuePairs);
urlEncodeFormEntity.setChunked(true);

httppost.setEntity(urlEncodeFormEntity);
HttpResponse response = httpclient.execute(httppost);
.
.
.

请帮我解决这个问题。它再次使用Localhost。

编辑:

PHP代码:这只是为了测试我的设备是否可以访问指定的网址

<?php
    echo "Ive been reached.";
    //Codes to connect DB and To Insert goes here
?>

2 个答案:

答案 0 :(得分:0)

试试这段代码

 HttpClient httpclient = new DefaultHttpClient();
 HttpPost httppost = new HttpPost("http://exampple.com/insert-data.php");
 List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
 nameValuePairs.add(new BasicNameValuePair("id", "1"));
 try{
  httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
  ResponseHandler<String> responseHandler = new BasicResponseHandler();

  String response = httpclient.execute(httppost, responseHandler);

  String reverseString = response;

  Log.d("Mishu Complain",reverseString);
 } catch(Exception e){ // check error here  }

在日志中找到任何错误你可以把这段代码

catch (Exception anyError) {
             Log.e("Mishu-Cn-Error", "exception",anyError);

        }

答案 1 :(得分:0)

我通过删除行

解决了这个问题
urlEncodeFormEntity.setChunked(true);