我试图使用php webservice将Json数据从mysql发送到Android应用程序 但我没有从服务器得到任何回复,我确信从我的网络服务的路径。 这是我的PHP代码
<?php
header("Content-Type: application/json;charset=utf-8");
$host = 'localhost';
$uname = 'root';
$pwd = '';
$db = 'test';
$flag = array();
$con = mysqli_connect($host, $uname, $pwd,$db) or die('Connection Failed');
mysqli_query($con,"SET NAME 'utf8'");
mysqli_query($con,'SET CHARACTER SET utf8');
$query = mysqli_query($con,"SELECT BRANCH,SUBSCRIBER,REGION,LOCATION_NO,ORG,TYPE,FOR_MONTH,PHASE_TYPE,NAME,KW_COUNTER_NO
,KW_LAST_READ,MULTIPLY_FACTOR,KV_COUNTER_NO,KV_LAST_READ,KV_MULTIPLY_FACTOR,KW_OPSERVER_1_LAST_READ,OP_MULTIPLY_FACTOR,
KW_OPSERVER_2_LAST_READ,KW_OPSERVER_3_LAST_READ,NOTES,Is_Auto_Pay,Counter_Read_Digit,M_B,AMBIR,IS_NEW_COUNTER FROM last_reads");
while($row = mysqli_fetch_array($query)) {
$flag[] = $row;
}
echo json_encode($flag,JSON_UNESCAPED_UNICODE);
mysqli_close($con);
?>
这里的广告是我的Android代码
private void webservice() {
// TODO Auto-generated method stub
new TheTask().executeOnExecutor(AsyncTask.THREAD_POOL_EXECUTOR);
}
class TheTask extends AsyncTask<String,String,String>
{
@Override
protected void onPostExecute(String result) {
pb.setVisibility(View.GONE);
Toast.makeText(getApplicationContext(), "command sent", Toast.LENGTH_LONG).show();
resumeWebService();
}
protected void onProgressUpdate(Integer... progress){
pb.setProgress(progress[0]);
}
@Override
protected void onPreExecute() {
// TODO Auto-generated method stub
}
@Override
protected String doInBackground(String... params) {
try {
Log.d("1","1");
HttpClient httpClient = new DefaultHttpClient();
Log.d("1", "2");
HttpPost httpPost = new HttpPost(ip);
Log.d("1","3");
HttpResponse response = httpClient.execute(httpPost);
Log.d("1","4");
HttpEntity entity = response.getEntity();
Log.d("","length"+response.getEntity().getContentLength()); //return 0 !
is = entity.getContent();
Log.d("1","5");
}
catch (Exception e) {
Log.e("Webservice 1", e.toString());
}
return "";
}
任何人都可以告诉我这是什么问题吗?
答案 0 :(得分:0)
通过调用doInBackground中的resumeWebService()函数来解决问题,resumeWebService()代码是
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(is, "utf-8" ), 8);
StringBuilder sb = new StringBuilder();
int i = 0;
while((line = reader.readLine()) != null) {
Log.d("readline","rr"+i++);
sb.append(line + "\n");
}
is.close();
resultSb = sb.toString();
Log.d("sb",""+resultSb);
}
catch (Exception e) {
Log.e("Webservice 2", e.toString());
}
它应该在doInBackground而不是onPostExecute()方法