我希望在发布数据后得到响应,但它失败了.我想创建一个登录系统,我已成功将数据提交到PHP文件,一切正常,我想从同一个函数得到响应,但我不知道问题出在哪里.
Here is the Java function:
public class PostDataGetRes extends AsyncTask<String,String,String> {
protected void onPreExecute() {
super.onPreExecute();
}
@Override
protected String doInBackground(String... strings) {
try {
postRData();
} catch (NullPointerException e) {
e.printStackTrace();
} catch (Exception e) {
e.printStackTrace();
}
return null;
}
@Override
protected void onPostExecute(String lenghtOfFile) {
// do stuff after posting data
}
}
public void postRData() {
String result = "";
InputStream isr = null;
final String email = editEmail.getText().toString();
final String pass = editPass.getText().toString();
// Create a new HttpClient and Post Header
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://website.com/appservice.PHP");
try {
// Add your data
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("id",email));
nameValuePairs.add(new BasicNameValuePair("stringdata",pass));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Execute HTTP Post Request
HttpResponse response = httpclient.execute(httppost);
resultView.setText("Inserted");
HttpEntity entity = response.getEntity();
isr = entity.getContent();
//convert response to string
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(isr,"iso-8859-1"),8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
isr.close();
result=sb.toString();
}
catch(Exception e){
Log.e("log_tag","Error converting result "+e.toString());
}
//parse json data
try {
String s = "";
JSONArray jArray = new JSONArray(result);
for(int i=0; i<jArray.length();i++){
JSONObject json = jArray.getJSONObject(i);
s = s +
"Name : "+json.getString("first_name")+"\n\n";
//"User ID : "+json.getInt("user_id")+"\n"+
//"Name : "+json.getString("first_name")+"\n"+
//"Email : "+json.getString("email")+"\n\n";
}
resultView.setText(s);
} catch (Exception e) {
// TODO: handle exception
Log.e("log_tag","Error Parsing Data "+e.toString());
}
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
} catch (IOException e) {
// TODO Auto-generated catch block
}
resultView.setText("Done");
}
And here is PHP code:
if($id){
$query = MysqL_query("SELECT first_name FROM users where email = '$id' ");
while($row=MysqL_fetch_assoc($query)){
$selectedData[]=$row;
}
print(json_encode($selectedData));
}
解决方法
首先要确保从您的网站获得正确的JSON对象 – 尝试将其打印为Toast.makeText().到目前为止,Web浏览器保留了html注释,android将其作为响应.
AsyncTask对象和类的设计不是按照您提供的方式进行的,也不能在doInBackground()中进行任何UI操作. AsyncTask的制作方式是不阻止GUI.
以下是使用AsyncTask类中的方法的一个不同的示例：
class Logging extends AsyncTask<String,Void>{
JSONObject json=null;
String output="";
String log=StringCheck.buildSpaces(login.getText().toString());
String pas=StringCheck.buildSpaces(password.getText().toString());
String url="http://www.mastah.esy.es/webservice/login.PHP?login="+log+"&pass="+pas;
protected void onPreExecute() {
Toast.makeText(getApplicationContext(),"Operation pending,please wait",Toast.LENGTH_SHORT).show();
}
@Override
protected Void doInBackground(String... params) {
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet(url);
request.addHeader("User-Agent","User-Agent");
HttpResponse response;
try {
response = client.execute(request);
BufferedReader br = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
String line="";
StringBuilder result = new StringBuilder();
while ((line = br.readLine()) != null) {
result.append(line);
}
output=result.toString();
} catch (ClientProtocolException e) {
Toast.makeText(getApplicationContext(),"Connection problems",Toast.LENGTH_LONG).show();
} catch (IOException e) {
Toast.makeText(getApplicationContext(),"Conversion problems",Toast.LENGTH_LONG).show();
}
return null;
}
@Override
protected void onPostExecute(Void w) {
try {
json = new JSONObject(output);
if(json.getInt("err")==1){
Toast.makeText(getApplicationContext(),json.getString("msg"),Toast.LENGTH_LONG).show();
}else{
String id_user="-1";
Toast.makeText(getApplicationContext(),Toast.LENGTH_LONG).show();
JSONArray arr = json.getJSONArray("data");
for(int i =0;i<arr.length();i++){
JSONObject o = arr.getJSONObject(i);
id_user = o.getString("id_user");
}
User.getInstance().setName(log);
User.getInstance().setId(Integer.valueOf(id_user));
Intent i = new Intent(getApplicationContext(),Discover.class);
startActivity(i);
}
} catch (JSONException e) {
}
super.onPostExecute(w);
}
}
$data = array(
'err' => 0,'msg' => "",'data' => array(),);
$MysqLi = new MysqLi($dbhost,$dbuser,$dbpass,$dbname);
if($MysqLi->connect_errno){
$data['err'] = 1;
$data['msg'] = "Brak polaczenia z baza";
exit(json_encode($data));
}
if(isset($_GET['login']) && isset($_GET['pass'])){
$MysqLi->query("SET CHARACTER SET 'utf8';");
$query = $MysqLi->query("SELECT banned.id_user FROM banned JOIN user ON user.id_user = banned.id_user WHERE user.login ='{$_GET['login']}' LIMIT 1;");
if($query->num_rows){
$data['err']=1;
$data['msg']="User banned";
exit(json_encode($data));
}else{
$query = $MysqLi->query("SELECT login FROM user WHERE login='{$_GET['login']}' LIMIT 1;");
if($query->num_rows){
$query = $MysqLi->query("SELECT pass FROM user WHERE pass ='{$_GET['pass']}' LIMIT 1;");
if($query->num_rows){
$data['msg']="Logged IN!";
$query = $MysqLi->query("SELECT id_user FROM user WHERE login='{$_GET['login']}' LIMIT 1;");
$data['data'][]=$query->fetch_assoc();
exit(json_encode($data));
}else{
$data['err']=1;
$data['msg']="Wrong login credentials.";
exit(json_encode($data));
}
}else{
$data['err']=1;
$data['msg']="This login doesn't exist.";
exit(json_encode($data));
}
}
}else{
$data['err']=1;
$data['msg']="Wrong login credentials";
exit(json_encode($data));
}
我为我的应用程序创建了小字典$data.我使用它的err键作为标志来知道是否出现任何错误,msg通知用户有关操作结果和发送JSON对象的数据.
你想要if(response == true),如果它存在的东西类似于我在AsyncTask中的onPostExecute(Void w)方法中使用的构造：
if(json.getInt("err")==1){
//something went wrong
}else{
//everything is okay,get JSON,inform user,start new Activity
}
这也是我使用$data [‘data’]获取JSON响应的方式：
if($query->num_rows){
while($res=$query->fetch_assoc()){
$data['data'][]=$res;
}
exit(json_encode($data));
}
