Question

您好，我正在尝试从EditText获取数据并将其作为JSON对象发布，但它返回没有数据的空对象返回的对象只包含键，而不是映射到EditText中输入的值，如屏幕所示注意：我只需要在Toast

中检索JSON对象

public class MainActivity extends AppCompatActivity {

    EditText userName;
    EditText password;
    EditText email;
    EditText phone;
    Button submit;

    Person person;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        person = new Person();

        userName = (EditText) findViewById(R.id.user_name);
        password = (EditText) findViewById(R.id.password);
        email = (EditText) findViewById(R.id.email);
        phone = (EditText) findViewById(R.id.phone);
        submit = (Button) findViewById(R.id.submit);

        person.setUserName(userName.getText().toString());

        person.setPassword(password.getText().toString());

        person.setEmail(email.getText().toString());

        person.setPhone(phone.getText().toString());
        submit.setOnClickListener(new View.OnClickListener() {@Override
            public void onClick(View v) {
                new JsonDataConverter().execute(person);
            }
        });
    }
    private class JsonDataConverter extends AsyncTask < Person,
    Void,
    String > {

        @Override
        protected String doInBackground(Person...params) {

            try {
                JSONObject jsonObject = new JSONObject();
                jsonObject.put("userName", person.getUserName());
                jsonObject.put("password", person.getPassword());
                jsonObject.put("email", person.getEmail());
                jsonObject.put("phone", person.getPhone());
                return jsonObject.toString();
            } catch(JSONException ex) {
                ex.printStackTrace();
            }
            return null;
        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            try {
                JSONObject jsonUser = new JSONObject(s);
                jsonUser.get("userName");
                jsonUser.get("password");
                jsonUser.get("email");
                jsonUser.get("phone");

                Toast.makeText(MainActivity.this, jsonUser.toString(), Toast.LENGTH_SHORT).show();

            } catch(JSONException e) {
                e.printStackTrace();
            }

        }
    }
}

Person.java

public class Person implements Serializable {
    String userName;
    String password;
    String email;
    String phone;

    public Person() {}

    public String getUserName() {
        return userName;
    }

    public void setUserName(String userName) {
        this.userName = userName;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public String getPhone() {
        return phone;
    }

    public void setPhone(String phone) {
        this.phone = phone;
    }

}

Screen picture

Answer 1

你刚刚在错误的地方使用了Setters。在ClickListener中使用它，如下所示..

protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);


person = new Person();

userName = (EditText) findViewById(R.id.user_name);
password = (EditText) findViewById(R.id.password);
email = (EditText) findViewById(R.id.email);
phone = (EditText) findViewById(R.id.phone);
submit = (Button) findViewById(R.id.submit);

submit.setOnClickListener(new View.OnClickListener() {
    @Override
    public void onClick(View v) {
       person.setUserName(userName.getText().toString());

       person.setPassword(password.getText().toString());

       person.setEmail(email.getText().toString());

       person.setPhone(phone.getText().toString());

        new JsonDataConverter().execute(person);
    }
 });
}

修改你使用的是更高级别的成员，而不是param [0]

Java Android Post JSON数据

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