我正在尝试开发一个颤振应用程序,这个想法是:
医院和医院的doctors.
。
这样做的目的是从api json数据中获得医院列表,单击特定医院后,应该有部门列表(api json data ),然后从列表中选择一个特定的部门--它将加载来自该部门的列表中的所有医生。
有人能帮忙吗?我用Laravel 9作为后端。
发布于 2022-06-17 07:18:13
不知道它能不能帮到你
- Json code
{
"hospital":[
{
"name": "hospital1",
"department":[
{
"name": "department1",
"doctors":[
{
"name": "doctor1"
},
{
"name": "doctor2"
}
]
},
{
"name": "department2",
"doctors":[
{
"name": "doctor1"
},
{
"name": "doctor2"
}
]
}
]
},
{
"name": "hospital2",
"department":[
{
"name": "department1",
"doctors":[
{
"name": "doctor1"
},
{
"name": "doctor2"
}
]
},
{
"name": "department2",
"doctors":[
{
"name": "doctor1"
},
{
"name": "doctor2"
}
]
}
]
}
]
}
- Dart code
class DataResponse{
DataResponse({
this.hospital,
});
List<Hospital> hospital;
factory DataResponse.fromJson(Map<String, dynamic> json) => Welcome(
hospital: List<Hospital>.from(json["hospital"].map((x) => Hospital.fromJson(x))),
);
Map<String, dynamic> toJson() => {
"hospital": List<dynamic>.from(hospital.map((x) => x.toJson())),
};
}
class Hospital {
Hospital({
this.name,
this.department,
});
String name;
List<Department> department;
factory Hospital.fromJson(Map<String, dynamic> json) => Hospital(
name: json["name"],
department: List<Department>.from(json["department"].map((x) => Department.fromJson(x))),
);
Map<String, dynamic> toJson() => {
"name": name,
"department": List<dynamic>.from(department.map((x) => x.toJson())),
};
}
class Department {
Department({
this.name,
this.doctors,
});
String name;
List<Doctor> doctors;
factory Department.fromJson(Map<String, dynamic> json) => Department(
name: json["name"],
doctors: List<Doctor>.from(json["doctors"].map((x) => Doctor.fromJson(x))),
);
Map<String, dynamic> toJson() => {
"name": name,
"doctors": List<dynamic>.from(doctors.map((x) => x.toJson())),
};
}
class Doctor {
Doctor({
this.name,
});
String name;
factory Doctor.fromJson(Map<String, dynamic> json) => Doctor(
name: json["name"],
);
Map<String, dynamic> toJson() => {
"name": name,
};
}https://stackoverflow.com/questions/72420846
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