为什么函数不能从android中获取数据？
我想在发布数据后得到回复，但失败。 我想创建一个登录系统，我已经成功地将数据提交到php文件，一切工作正常，现在我想从相同的功能得到响应，但我无法知道问题出在哪里。
这是Java函数：
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")+"nn";
//"User ID : "+json.getInt("user_id")+"n"+
//"Name : "+json.getString("first_name")+"n"+
//"Email : "+json.getString("email")+"nn";
}
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");
}
这里是php代码：
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));
}
请帮助我我已经尝试过，但无法取得任何结果。 请帮助我如何在查询执行后从php文件获得响应。
首先请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，网页浏览器保留了html注释，android会对此做出回应。
AsyncTask对象和类不是按照您提供的方式设计的，也不能在doInBackground()进行任何UI操作。 AsyncTask是以不阻止GUI的方式进行的。 这是一个没有太多不同的例子，它使用AsyncTask类中的方法：
class Logging extends AsyncTask<String,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(), json.getString("msg"), 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);
}
}
PHP文件内容：
$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$data 。 我用它的err键作为标志来知道是否有错误出现， msg通知用户有关操作结果和data发送JSON对象。
你想要if(response == true)如果它存在的事情是类似于我在AsyncTask onPostExecute(Void w)方法中使用的构造：
if(json.getInt("err")==1){
//something went wrong
}else{
//everything is okay, get JSON, inform user, start new Activity
}
另外这里是我使用$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));
}