Question

我尝试使用json登录。我的php返回这个结果：

[{"name":"cynthia","password":"123456"},{"name":"John","password":"123456"}]

这些是mySQL中的数据。

我想用这个结果在android studio中进行比较，但我发现它没有用。它总是转到else语句。

public class LoginActivity extends AppCompatActivity {

    public EditText username;
    public EditText password;
    private ArrayList<String> usernameList = new ArrayList<String>();
    private ArrayList<String> passwordList = new ArrayList<String>();

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);
        username = (EditText) findViewById(R.id.username);
        password = (EditText) findViewById(R.id.password);
    }

    public void Login2(View view) {
        if (username.getText().length() == 0 || password.getText().length() == 0) {
            openDialog(view);
        } else {
            new AsyncRetrieve().execute();
            if(usernameList.contains(username)&&passwordList.contains(password)){
                successLogin();
            }else{
                openDialog(view);
            }

        }
    }

    public void openDialog(View view) {
        new AlertDialog.Builder(this)
                .setTitle(R.string.login_dialog_title)
                .setMessage(R.string.login_dialog_msg)
                .show();
    }

    public void successLogin() {
        Intent intent = new Intent();
        intent.setClass(this, HomePageActivity.class);
        startActivity(intent);
    }

    private class AsyncRetrieve extends AsyncTask<Void, Void, String> {
        @Override
        protected String doInBackground(Void... params) {
            return getVideoList();
        }

        private String getVideoList() {
            try {
                URL url = new URL("http://localhost/soften/login_check1.php");
                HttpURLConnection conn = (HttpURLConnection) url.openConnection();
                InputStream in = new BufferedInputStream(conn.getInputStream());
                BufferedReader reader = new BufferedReader(new InputStreamReader(in));
                String response = reader.readLine();

                //set the response from server as json array
                JSONArray userArray = new JSONArray(response);

                usernameList.clear();
                passwordList.clear();
                Log.d("username", "value = " + userArray.length());
                for (int i = 0; i < userArray.length(); i++) {
                    Log.d("i", "i = " + i);
                    //get json object from the json array
                    JSONObject json = userArray.getJSONObject(i);
                    Log.d("json", "" + json.length());
                    //add object to the list for grid view
                    usernameList.add(json.getString("name"));
                    passwordList.add(json.getString("password"));
                }



                return null;
            } catch (Exception ex) {
                return ex.toString();
            }
        }
    }
}

Answer 1

你long2（）是错误的，是吗

if(usernameList.contains(username)&&passwordList.contains(password))

你应该等于username.getText（）和password.getText（）。我的英语不是很好，但你应该能够理解

Answer 2

问题出在这一行

 new AsyncRetrieve().execute();

当您执行AsyncTask时，它会启动后台线程，但是您可以立即继续操作而无需等待来自服务器的结果，当您比较用户名和密码时，这些列表都是空的。

 private ArrayList<String> usernameList = new ArrayList<String>();
 private ArrayList<String> passwordList = new ArrayList<String>();

<强>解决方案：

第一个解决方案是向AsyncTask Call添加get（）方法，如

new AsyncRetrieve().execute().get();

它将暂停您的AsyncTask，直到从服务器收到结果，然后进行比较。

另一个解决方案是覆盖AsyncTask中的onPostExecute（）并在那里进行比较。

 private class AsyncRetrieve extends AsyncTask<Void, Void, String> {
        @Override
        protected String doInBackground(Void... params) {
            return getVideoList();
        }

        private String getVideoList() {
            try {
                URL url = new URL("http://localhost/soften/login_check1.php");
                HttpURLConnection conn = (HttpURLConnection) url.openConnection();
                InputStream in = new BufferedInputStream(conn.getInputStream());
                BufferedReader reader = new BufferedReader(new InputStreamReader(in));
                String response = reader.readLine();

                //set the response from server as json array
                JSONArray userArray = new JSONArray(response);

                usernameList.clear();
                passwordList.clear();
                Log.d("username", "value = " + userArray.length());
                for (int i = 0; i < userArray.length(); i++) {
                    Log.d("i", "i = " + i);
                    //get json object from the json array
                    JSONObject json = userArray.getJSONObject(i);
                    Log.d("json", "" + json.length());
                    //add object to the list for grid view
                    usernameList.add(json.getString("name"));
                    passwordList.add(json.getString("password"));
                }



                return null;
            } catch (Exception ex) {
                return ex.toString();
            }
        }


        @Override
        protected String onPostExecute(String result) {
            if(usernameList.contains(username.getText())&&passwordList.contains(password.getText())){
                successLogin();
            }else{
                openDialog(view);
            }
        }
    }

从Login2方法中删除这些行:)不要忘记添加getText（）方法。

希望得到这个帮助。

如何修复android studio

2 个答案: