显示带有变量 true 的文档 - Flutter Firebase
Display document with variable true - Flutter Firebase
我必须在屏幕上仅向用户显示 ACTIVE 变量为真。但是我无法实现这个条件,它总是显示在 Firebase 中注册的所有用户。如何针对这段有问题的代码执行此操作?
Expanded(
child: StreamBuilder<QuerySnapshot>(
stream: getListaPacientes(),
builder: (context, snapshot){
switch(snapshot.connectionState){
case ConnectionState.none:
case ConnectionState.waiting:
return Center(
child: Column(
mainAxisAlignment: MainAxisAlignment.center,
children: <Widget>[
Icon(Icons.error_outline),
Text("Usuário não encontrado")
],
),
);
default:
List<DocumentSnapshot> documentos =
snapshot.data.documents;
return ListView.builder(
itemCount: documentos.length,
itemBuilder: (context, index){
return ListTile(
title: Text(items[index].nome,
style: TextStyle(fontSize: 16)),
subtitle: Text('Quarto: ${items[index].quarto}',
style: TextStyle(fontSize: 16)),
leading: CircleAvatar(
backgroundImage: NetworkImage(items[index].foto),
),
onTap: ()=> _navegarParaPerfil(context, items[index]),
);
}
);
}
}
),
)
Stream<QuerySnapshot> getListaPacientes(){
return Firestore.instance.collection('pacientes').snapshots();
}
你有什么建议来修复这个缺陷?非常感谢!
您可以使用 where API 从 firebase 检索活跃用户。
final CollectionReference collection =
FirebaseFirestore.instance.collection('employees');
Stream<QuerySnapshot> getStream() {
return collection.where('ACTIVE', isEqualTo: true).snapshots();
}
我必须在屏幕上仅向用户显示 ACTIVE 变量为真。但是我无法实现这个条件,它总是显示在 Firebase 中注册的所有用户。如何针对这段有问题的代码执行此操作?
Expanded(
child: StreamBuilder<QuerySnapshot>(
stream: getListaPacientes(),
builder: (context, snapshot){
switch(snapshot.connectionState){
case ConnectionState.none:
case ConnectionState.waiting:
return Center(
child: Column(
mainAxisAlignment: MainAxisAlignment.center,
children: <Widget>[
Icon(Icons.error_outline),
Text("Usuário não encontrado")
],
),
);
default:
List<DocumentSnapshot> documentos =
snapshot.data.documents;
return ListView.builder(
itemCount: documentos.length,
itemBuilder: (context, index){
return ListTile(
title: Text(items[index].nome,
style: TextStyle(fontSize: 16)),
subtitle: Text('Quarto: ${items[index].quarto}',
style: TextStyle(fontSize: 16)),
leading: CircleAvatar(
backgroundImage: NetworkImage(items[index].foto),
),
onTap: ()=> _navegarParaPerfil(context, items[index]),
);
}
);
}
}
),
)
Stream<QuerySnapshot> getListaPacientes(){
return Firestore.instance.collection('pacientes').snapshots();
}
你有什么建议来修复这个缺陷?非常感谢!
您可以使用 where API 从 firebase 检索活跃用户。
final CollectionReference collection =
FirebaseFirestore.instance.collection('employees');
Stream<QuerySnapshot> getStream() {
return collection.where('ACTIVE', isEqualTo: true).snapshots();
}