小编P-R*_*RAD的帖子

当滚动列表时,RecyclerView.getChild(index)显示null(索引搞砸了)

我一直在为我的Android应用程序使用SwipeableRecyclerView来为我的recyclerView启用滑动.RecyclerView包含cardViews列表.

我试图为卡片实现撤消功能,当向左滑动时会被删除(第一次滑动显示撤消,下一次滑动触发器删除)

我正在尝试以下代码(部分工作,我猜)

SwipeableRecyclerViewTouchListener srvTouchListner = new SwipeableRecyclerViewTouchListener(rvTimerList,
            new SwipeableRecyclerViewTouchListener.SwipeListener(){

                @Override
                public boolean canSwipe(int i) {
                    return true;
                }

                @Override
                public void onDismissedBySwipeLeft(RecyclerView recyclerView, int[] ints) {
                    for(int position : ints){
                        View view = recyclerView.getChildAt(position);
                            if (view.getTag(R.string.card_undo) == null) {
                                if(viewStack == null) {
                                    saveToViewStack(position, view);
                                    final ViewGroup viewGroup = (ViewGroup) view.findViewById(R.id.time_card2);
                                    view.setTag(R.string.card_undo, "true");
                                    viewGroup.addView(view.inflate(TimerSummary.this, R.layout.timeslot_card_undo, null));
                                }
                            } else {
                                Log.d(TAG, "Removing Item");
                                deleteTimeSlot(timerInstanceList.get(position));
                                Toast.makeText(TimerSummary.this, "Deleted!", Toast.LENGTH_SHORT).show();
                                timerInstanceList.remove(position);
                                finalSummaryAdapter.notifyItemRemoved(position);
                            }

                    }
                    finalSummaryAdapter.notifyDataSetChanged();
                }
                @Override
                public void onDismissedBySwipeRight(RecyclerView recyclerView, int[] …
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android android-viewholder android-recyclerview

14
推荐指数
2
解决办法
2万
查看次数

未解决的参考:junit4:Android ComposeUI 测试

我试图测试我在 android 中使用 jetpack compose 构建的 UI 并添加这些依赖项来设置它

    androidTestImplementation "androidx.compose.ui:ui-test-junit4:1.1.0"
    debugImplementation "androidx.compose.ui:ui-test-manifest:1.0.5"
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为了测试我刚刚开始写以下内容

class LoginActivityComposeTest {
    @get:Rule
    val composeTestRule = createComposeRule()

    @Test
    fun socialPluginsTest() {
        composeTestRule.setContent {
            SocialLogins()
        }
    }
}
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我尝试导入的 createComposeRule() 不可用。我尝试手动导入import androidx.compose.ui.test.junit4.createComposeRule但是当我尝试运行测试后出现错误

Unresolved reference: junit4

这是我正在使用的完整 build.gradle 依赖项

compose_version ='1.0.1'retrofit_version ='2.9.0'

    implementation "androidx.core:core-ktx:1.7.0"
    implementation "androidx.compose.ui:ui:$compose_version"
    implementation "androidx.compose.material:material:$compose_version"
    implementation "androidx.compose.ui:ui-tooling-preview:$compose_version"
    implementation "androidx.lifecycle:lifecycle-runtime-ktx:2.4.0"
    implementation "androidx.activity:activity-compose:1.4.0"
    implementation "androidx.navigation:navigation-compose:2.5.0-alpha01"
    implementation "com.squareup.retrofit2:retrofit:$retrofit_version"
    implementation "com.squareup.retrofit2:adapter-rxjava2:$retrofit_version"
    implementation "com.squareup.retrofit2:converter-gson:$retrofit_version"
    implementation 'io.reactivex.rxjava2:rxandroid:2.0.1'
    implementation "io.reactivex.rxjava2:rxkotlin:2.4.0"
    implementation "org.jetbrains.kotlinx:kotlinx-coroutines-android:1.6.0"
    implementation 'com.google.android.gms:play-services-auth:20.1.0'
    implementation "io.coil-kt:coil-compose:1.4.0"
    testImplementation "junit:junit:4.13.2"
    androidTestImplementation "androidx.test.espresso:espresso-core:3.4.0" …
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android kotlin android-jetpack-compose

9
推荐指数
2
解决办法
3171
查看次数

如何在扑扑中捕捉异常?

这是我的异常类。异常类已经由flutter的抽象异常类实现。我错过了什么吗?

class FetchDataException implements Exception {
 final _message;
 FetchDataException([this._message]);

String toString() {
if (_message == null) return "Exception";
  return "Exception: $_message";
 }
}


void loginUser(String email, String password) {
  _data
    .userLogin(email, password)
    .then((user) => _view.onLoginComplete(user))
    .catchError((onError) => {
       print('error caught');
       _view.onLoginError();
    });
}

Future < User > userLogin(email, password) async {
  Map body = {
    'username': email,
    'password': password
  };
  http.Response response = await http.post(apiUrl, body: body);
  final responseBody = json.decode(response.body);
  final statusCode = response.statusCode;
  if (statusCode != HTTP_200_OK || …
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model-view-controller dart flutter

7
推荐指数
2
解决办法
9892
查看次数

带有默认参数的ES6:是否传入参数

function myfunc( value1, value2='defaultValue' ) {
  //some stuff
}
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如果参数传入或不传入函数,我怎么知道呢?我知道如果你没有传递任何东西作为参数,将设置默认值.我实际上想要检查用户是否作为第二个参数传递任何东西(即使它与默认值相同)并更改我的返回相应的价值.

我正在检查ES6文档这个SO答案,不是我真正想要的.

谢谢,

javascript default-parameters ecmascript-6

4
推荐指数
2
解决办法
485
查看次数