java - 从 HttpPost Android 中提取 URL

Question

我有一个使用 nameValuePairs 构建的长 URL，我目前正试图弄清楚为什么该帖子在某些设备上导致 500 错误，而在某些设备上却出现 200其他设备。 我需要从 httppost 中提取完整的 URL，尽管据我所知它应该只是以下格式:

http://xx.com/site.asmx?var1=blah?var2=blah

同样的构建可以在手机上运行但不能在平板电脑上运行。我尝试查看我的代码并找出平板电脑可能无法定义此类变量的任何地方，例如这种情况:

final TelephonyManager tm = (TelephonyManager) getBaseContext()
                .getSystemService(Context.TELEPHONY_SERVICE);
        String tmDevice;
        try{
        tmDevice = "" + tm.getDeviceId().toString();
        }
        catch (NullPointerException e)
        {
            tmDevice = "null";
        }

下面是一些 nameValuePairs 的样子以及代码发布部分的摘录:

List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
    nameValuePairs
            .add(new BasicNameValuePair("CurrentOwnerUsername", name));
    nameValuePairs.add(new BasicNameValuePair("OSVersion",
            Build.VERSION.RELEASE));
    nameValuePairs.add(new BasicNameValuePair("hash",
            "xxx"));
    nameValuePairs.add(new BasicNameValuePair("OSName", "Android"));
    nameValuePairs.add(new BasicNameValuePair("Model", Build.MODEL));
    nameValuePairs.add(new BasicNameValuePair("Manufacturer",
            Build.MANUFACTURER));
    nameValuePairs.add(new BasicNameValuePair("IMEI", id));
    try {
        nameValuePairs.add(new BasicNameValuePair("SerialNumber",
                Build.SERIAL));
    } catch (NoSuchFieldError e) {
        nameValuePairs.add(new BasicNameValuePair("SerialNumber",
                "NotAvailable"));
    }
    nameValuePairs.add(new BasicNameValuePair("CarrierName", carrier));
    // .add(new BasicNameValuePair("FriendlyName", Build.DEVICE));
    nameValuePairs.add(new BasicNameValuePair("Type", " "));
    nameValuePairs.add(new BasicNameValuePair("VisitorID", VisitorID));
    nameValuePairs.add(new BasicNameValuePair("PhoneNumber", phoneNumber));
    nameValuePairs.add(new BasicNameValuePair("SSID", ssid));
    nameValuePairs.add(new BasicNameValuePair("MACAddress", macAddr));
    nameValuePairs.add(new BasicNameValuePair("UserDepartment",
            department_val));
    nameValuePairs.add(new BasicNameValuePair("BatteryLevel", batteryText));

    int toastDuration = Toast.LENGTH_SHORT;
    try {
        HttpPost httppost = new HttpPost(URL); // post object
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost); // execution
        int result = response.getStatusLine().getStatusCode();
        if (result == 200) {
            CharSequence toastText = "Success";
            Toast.makeText(context, toastText, toastDuration).show();
        } else {
            CharSequence toastText = "Failure";
            Log.d("checkin", String.valueOf(response.getStatusLine()
                    .getStatusCode()));
            Toast.makeText(context, toastText, toastDuration).show();
        }
    } catch (ClientProtocolException e) {
        CharSequence toastText = "ClientProtocolException";
        Toast.makeText(context, toastText, toastDuration).show();
    } catch (IOException e) {
        CharSequence toastText = "IOException";
        Toast.makeText(context, toastText, toastDuration).show();
    }

所以我需要的是一种从 httppost 中提取完整 URL 的方法。当前传递的字符串 URL 只是一个基本 url，例如:http://xx.com/ nameValuePairs 紧随其后。 想法？

Answer 1

对于 HttpPost，值作为请求的主体传递，因此 URL 将与您传入的相同。但是，您可以使用 EntityUtils.toString(HttpEntity entity)使用您的 UrlEncodedFormEntity 打印出将在 HttpPost 中传递的值。