如何查看从httppost发送的字符串？

Question

下面是一段试图发布到本地服务器上的PHP脚本的代码片段.我正在将数据传回服务器的MySQL表,因为数据量非常小,我想我会使用HttpClient将数据嵌入到URL参数中.

但是,即使我从服务器获得的响应是​​正常的(HTTP代码= 200),很明显服务器端PHP脚本没有获得正确格式化的数据(对MySQL表没有影响).但是,当我在浏览器中手动输入URL + args时,如下所示:

http://10.0.0.13/jobs_returndata_test.php?jobnum=189193&pnum=3&entime=13:00&extime=14:00 

例如,一切正常(PHP脚本正确写入MySQL表).

我的问题:我有没有办法真正查看HttpClient发送的内容？(即HttpPost对象的整个.toString内容？).这是Android端代码段:

// Create a new HttpClient and Post Header, send data as args to PHP file
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://10.0.0.13/jobs_returndata.php");
try {
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("jobnum",convertSimple(jobNum)));
    nameValuePairs.add(new BasicNameValuePair("pnum",convertSimple(rowNum)));
    nameValuePairs.add(new BasicNameValuePair("entime",enteredInTime));
    nameValuePairs.add(new BasicNameValuePair("extime",enteredOutTime));

    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    Log.i("HTTP Post", "Execute Post sending jobnum="+ convertSimple(jobNum) +      
    "&pnum="+ convertSimple(rowNum) + "&entime=" + enteredInTime
                    + "&extime=" + enteredOutTime);
    // Execute HTTP Post Request
    HttpResponse response = httpclient.execute(httppost);
    Log.i("HTTP Post", "Response from server = " + response.getStatusLine().getReasonPhrase() + "  Code = " + convertSimple(response.getStatusLine().getStatusCode()));

        } catch (ClientProtocolException e) {
            Log.e("HTTP Post", "Protocol error = " + e.toString());
        } catch (IOException e) {
            Log.e("HTTP Post", "IO error = " + e.toString());
        }

我的服务器端PHP文件是:

?php
$jobnum=$_GET['jobnum'];
//echo "jobnum = $jobnum <br />";
$pnum=$_GET['pnum'];
//echo "pnum = $pnum <br />";

$entime=$_GET['entime'];
//echo "entime = $entime <br />";

$extime=$_GET['extime'];
//echo "extime = $extime <br />";


$dbh=mysql_connect ("localhost", "Steve", "pluggin");
mysql_select_db ("DriverJobs");

$result = mysql_query("UPDATE jobs SET EntryTime = '$entime' WHERE JobNumber = '$jobnum' AND PointNum = '$pnum' ");  
$result = mysql_query("UPDATE jobs SET ExitTime = '$extime' WHERE JobNumber = '$jobnum' AND PointNum = '$pnum' ") ;  

//echo "done <br />";
?>

相关部分的logcat:

06-22 13:33:11.341:I/HTTP Post(27788):执行发送jobnum = 189193&pnum = 3&entime = 9:00&extime = 9:05 06-22 13:33:11.581:I/HTTP Post(27788):来自server的响应= OK Code = 200

权限很好(在INTERNET的Manifest文件中,对于PHP文件,Apache2正在运行,API = 8,使用带有Android模拟器的Eclipse).convertSimple()是一个2线程,返回int原语的String等效项.

提前谢谢你看这个!

Answer 1

使用此方法

private String convertStreamToString(InputStream is) throws Exception {
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();
    String line = null;
    while ((line = reader.readLine()) != null) {
        sb.append(line);
    }
    is.close();
    return sb.toString();
}   

并调用这样来查看post方法中包含的实体

try {
    HttpPost postMethod = new HttpPost("your url");
    Log.i(TAG,"HTTP Entiry : " + convertStreamToString(postMethod.getEntity().getContent()));
}catch (Exception e){
    e.printStackTrace();
}