Flutter：如何过滤List<Map<String, dynamic>>及SkillsCard列表的completedBy字段

1. 在getData中安全处理类型并完成过滤 #

Firestore返回的e["completedBy"]是dynamic类型，直接调用contains会报错，先做类型校验与转换，确保它是List<String>后再执行过滤逻辑：

Future<List<Map<String, dynamic>>> getData({required bool completed, required String userEmail}) async {
  QuerySnapshot snapshot = await FirebaseFirestore.instance.collection('your_collection').get();
  
  return snapshot.docs.map((doc) => doc.data() as Map<String, dynamic>)
    .where((item) {
      // 安全转换字段类型，避免类型错误
      final completedBy = item["completedBy"];
      if (completedBy is! List<String>) return false;
      
      // 根据参数过滤数据
      return completed ? completedBy.contains(userEmail) : !completedBy.contains(userEmail);
    })
    .toList();
}

2. 修复readData的返回类型不匹配问题 #

readData需要返回Future<List<SkillsCard>>，先获取过滤后的Map列表，再逐个转换为SkillsCard对象：

Future<List<SkillsCard>> readData({required bool completed, required String userEmail}) async {
  // 获取已过滤的原始数据
  final filteredMaps = await getData(completed: completed, userEmail: userEmail);
  
  // 转换为SkillsCard列表，假设SkillsCard有对应构造方法
  return filteredMaps.map((map) => SkillsCard(
    id: map["id"],
    title: map["title"],
    completedBy: map["completedBy"] as List<String>,
    // 补充其他字段
  )).toList();
}

额外优化建议 #

• 若Firestore中completedBy可能存在非字符串元素，可增加一层校验：

  final completedBy = (item["completedBy"] as List).where((element) => element is String).cast<String>().toList();
  

• 若必须在readData内过滤，需确保SkillsCard的completedBy是List<String>类型：

  Future<List<SkillsCard>> readData({required bool completed, required String userEmail}) async {
    final rawMaps = await getData(); // 未过滤的原始数据方法
    final cards = rawMaps.map((map) => SkillsCard.fromMap(map)).toList();
    
    return cards.where((card) => completed 
      ? card.completedBy.contains(userEmail) 
      : !card.completedBy.contains(userEmail)
    ).toList();
  }
  

  这里要求SkillsCard实现fromMap构造方法，将Map转为实体类。

内容的提问来源于stack exchange，提问作者FlutterBuilder