Dart/Flutter双API请求SocketException无法全部捕获求助

同时调用两个返回Future的API请求（getRandomImage()和getRandomFact()）以并行获取数据，无网络时两个请求都会抛出SocketException，但无论使用顺序await还是Future.wait()，都只能捕获其中一个异常，程序会因未处理的异常崩溃，错误提示：_ClientSocketException (Failed host lookup: 'APIs BASE_URL')。

相关代码 #

home_page.dart #

Future<void> setRandomImageAndFact() async {
  setState(() {
    json = imageBytes = null;
  });

  Future<Uint8List> bytesImageFuture = getRandomImage();
  Future<Map<String, dynamic>> decodedJsonFuture = getRandomFact();

  try{
    imageBytes = await bytesImageFuture;
    json = await decodedJsonFuture;
  } on SocketException {
    socketException = true;
  }

  setState(() {});
}

images.dart #

Future<Uint8List> getRandomImage() async {
  Uri url = Uri.https(BASE_URL, "cat");

  http.Response response;

  try {
    response = await http.get(url);
  } on SocketException {
    rethrow;
  }

  if (response.statusCode == 200) {
    return response.bodyBytes;
  }

  throw Exception("Could not get random image ${response.statusCode}");
}

facts.dart #

Future<Map<String, dynamic>> getRandomFact() async {
  Uri url = Uri.https(BASE_URL, "facts/random");
  Future<http.Response> responseFuture = http.get(url);
  Future<List<String>> rejectedListFuture = loadPrefs(Mode.rejected.value);

  List<String> rejectedList = await rejectedListFuture;
  http.Response response;

  try {
    response = await responseFuture;
  } on SocketException {
    rethrow;
  }

  if (response.statusCode == 200) {
    Map<String, dynamic> decodedJson = jsonDecode(response.body);
    if (rejectedList.every((element) => element != decodedJson['_id'])) {
      return decodedJson;
    } else {
      return getRandomFact();
    }
  }

  throw Exception("Could not get random fact ${response.statusCode}");
}

尝试过的无效方法 #

1. 将两个await分别放在独立的try-catch块中，仍出现崩溃。
2. 使用Future.wait()结合catchError，依旧会因未处理异常崩溃：

Future.wait([
  getRandomImage(),
  getRandomFact(),
]).then((value) {
  imageBytes = value[0] as Uint8List;
  json = value[1] as Map<String, dynamic>;
}).catchError((e) {
  socketException = true;
});

问题原因 #

1. 未处理的异步异常：当第一个Future抛出异常被捕获后，第二个Future已在后台执行，其抛出的异常会成为未处理的异步异常，导致程序崩溃。
2. 递归调用的异常泄漏：getRandomFact()中存在递归逻辑，无网络时每次递归都会抛出SocketException，这些递归产生的Future异常无法被上层的单次catch捕获，最终导致崩溃。

解决方案 #

方案1：为每个Future单独处理异常 #

给每个API请求的Future添加独立的catchError，确保所有异常都能被捕获：

Future<void> setRandomImageAndFact() async {
  setState(() {
    json = imageBytes = null;
    socketException = false;
  });

  // 单独处理图片请求的异常
  final imageFuture = getRandomImage().catchError((e) {
    if (e is SocketException || e.toString().contains('Failed host lookup')) {
      socketException = true;
    }
    return null; // 返回默认值，让Future正常完成
  });

  // 单独处理事实请求的异常
  final factFuture = getRandomFact().catchError((e) {
    if (e is SocketException || e.toString().contains('Failed host lookup')) {
      socketException = true;
    }
    return null;
  });

  imageBytes = await imageFuture;
  json = await factFuture;

  setState(() {});
}

方案2：修复递归逻辑，避免异常泄漏 #

将getRandomFact()中的递归改为循环，避免多次递归抛出未被捕获的异常：

Future<Map<String, dynamic>> getRandomFact() async {
  Uri url = Uri.https(BASE_URL, "facts/random");
  List<String> rejectedList = await loadPrefs(Mode.rejected.value);

  while (true) {
    http.Response response;
    try {
      response = await http.get(url);
    } on SocketException {
      rethrow;
    }

    if (response.statusCode == 200) {
      Map<String, dynamic> decodedJson = jsonDecode(response.body);
      // 不在拒绝列表则返回结果
      if (!rejectedList.contains(decodedJson['_id'])) {
        return decodedJson;
      }
      // 在列表中则继续循环请求
    } else {
      throw Exception("Could not get random fact ${response.statusCode}");
    }
  }
}

方案3：统一捕获所有网络相关异常 #

由于_ClientSocketException是SocketException的子类，直接捕获SocketException可能无法覆盖所有情况，可改为捕获通用Exception并判断错误信息：

Future<void> setRandomImageAndFact() async {
  setState(() {
    json = imageBytes = null;
    socketException = false;
  });

  try {
    final results = await Future.wait([
      getRandomImage(),
      getRandomFact(),
    ]);
    imageBytes = results[0];
    json = results[1];
  } catch (e) {
    // 判断是否为网络相关异常
    if (e is SocketException || e.toString().contains('Failed host lookup')) {
      socketException = true;
    } else {
      // 非网络异常继续抛出
      rethrow;
    }
  }

  setState(() {});
}

内容的提问来源于stack exchange，提问作者Ahmed Gado