从 JSON Android Studio 设置文本单选按钮 - AlertDialog

Question

我有来自 JSON（数据库）的用户名。如果我从 json 设置文本单选按钮该怎么办？

这是我的自定义布局 form_tugas.xml

 <RadioGroup
        android:id="@+id/radiogrup"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:layout_marginLeft="20dp"
        android:layout_marginTop="40dp"
        app:layout_constraintStart_toStartOf="parent"
        app:layout_constraintTop_toBottomOf="@+id/textView2">

        <androidx.appcompat.widget.AppCompatRadioButton
            android:id="@+id/radio1"
            android:layout_width="match_parent"
            android:layout_height="match_parent"
            android:text="Nama"
            android:checked="true">
        </androidx.appcompat.widget.AppCompatRadioButton>
        <androidx.appcompat.widget.AppCompatRadioButton
            android:id="@+id/radio2"
            android:layout_width="match_parent"
            android:layout_height="match_parent"
            android:text="Nama">

        </androidx.appcompat.widget.AppCompatRadioButton>

    </RadioGroup>

这是我的代码

private void formTugas()
{

    getJson("http://192.168.43.144/TA/getUserTeknisi.php");
    dialog = new AlertDialog.Builder(this);
    inflater = getLayoutInflater();
    dialogView = inflater.inflate(R.layout.form_tugas,null);
    dialog.setView(dialogView);
    dialog.setCancelable(true);
    dialog.setPositiveButton("Ya", new DialogInterface.OnClickListener() {
        @Override
        public void onClick(DialogInterface dialog, int which) {

        }
    });
    dialog.setNegativeButton("Tidak", new DialogInterface.OnClickListener() {
        @Override
        public void onClick(DialogInterface dialog, int which) {

        }
    });
    dialog.show();
}

这是我从 JSON 获取用户名时的代码

private void ambilUserTeknisi(String json) throws JSONException
{
    JSONObject jsonObject = new JSONObject(json);
    JSONArray jsonArray  = jsonObject.getJSONArray("data");

    int x = 1;

    for (int u = 0; u < jsonArray.length(); u++)
    {
        JSONObject ob = jsonArray.getJSONObject(u);
        String user  = ob.getString("username");

        users.add(new User(user));

        Log.d("username",""+users.get(u).getUsername());

    }
}

这是我的日志

2020-01-12 16:25:15.510 15549-15549/com.example.adum D/username: Ronaldo
2020-01-12 16:25:15.510 15549-15549/com.example.adum D/username: Velguri

我想在 form_tugas.xml 中为我的单选按钮设置用户名“Ronaldo”和“Velguri”。我应该怎么办 ？

感谢您的帮助:)

Answer 1

将您的 formTugas 方法更改为：

 private void formTugas(){

        getJson("http://192.168.43.144/TA/getUserTeknisi.php");
        //add your json parse method here
        ambilUserTeknisi("your json")

        dialog = new AlertDialog.Builder(this);
        inflater = getLayoutInflater();
        dialogView = inflater.inflate(R.layout.form_tugas,null);
        dialog.setView(dialogView);
        //find your view
        AppCompatRadioButton radio1= (AppCompatRadioButton) dialog.findViewById(R.id.radio1);
        AppCompatRadioButton radio2= (AppCompatRadioButton) dialog.findViewById(R.id.radio2);
        //set text to your view
        radio1.settext(users.get(0).name); //set name from your user object 
        radio2.settext(users.get(1).name);

        dialog.setCancelable(true);
        dialog.setPositiveButton("Ya", new DialogInterface.OnClickListener() {
            @Override
            public void onClick(DialogInterface dialog, int which) {

            }
        });
        dialog.setNegativeButton("Tidak", new DialogInterface.OnClickListener() {
            @Override
            public void onClick(DialogInterface dialog, int which) {

            }
        });
        dialog.show();
    }