我正在尝试将一个值传递给我的 php web 服务。我已经使用此代码来传递“名称”值:
private class MyAsyncTask extends AsyncTask<String, Void, Void> {
protected Void doInBackground (String... params)
{
Intent intent = getIntent();
String name = intent.getStringExtra("KEY_NAME");
//HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://10.0.2.2/secure_login/get_data_user.php");
List<NameValuePair> nameValuePair = new ArrayList<NameValuePair>(1);
nameValuePair.add(new BasicNameValuePair("KEY_NAME", name));
DefaultHttpClient hc = new DefaultHttpClient();
// HttpResponse response = hc.execute(httppost);
try {
httppost.setEntity(new UrlEncodedFormEntity(nameValuePair));
} catch (UnsupportedEncodingException e){
// writing error to log
e.printStackTrace();
}
try {
HttpResponse response = hc.execute(httppost);
HttpEntity entity = response.getEntity();
InputStream inStream = entity.getContent();
// writing response to log
Log.d("Http Response:", response.toString());
} catch (ClientProtocolException e) {
// writing exception to log
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return null;
}
这个,用于将 Stream 转换为 String。
protected String convertStreamToString(InputStream inStream)
{
BufferedReader reader = new BufferedReader(new InputStreamReader(inStream));
StringBuilder sb = new StringBuilder();
String line = null;
try
{
while ((line = reader.readLine()) != null)
{
sb.append(line + "\n");
}
}
catch (IOException e)
{
e.printStackTrace();
}
finally
{
try
{
inStream.close();
}
catch (IOException e)
{
e.printStackTrace();
}
}
return sb.toString();
}
}
但我只在 log cat 中得到这个响应:
org.apache.http.message.BasicHttpResponse@43e4c068
我需要传递“名称”值,以便我的 php Web 服务可以检索并执行如下查询:
if (isset($_GET['name'])) {
$name = $_GET['name'];
require_once 'DB_Functions.php';
$db = new DB_Functions();
$result = mysql_query("SELECT name, email from users where name = '$name'");
我应该如何修复它?先感谢您。