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社区首页 >问答首页 >如何将颤振api json数据实现为第三级类别的列表视图?

如何将颤振api json数据实现为第三级类别的列表视图?
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Stack Overflow用户
提问于 2022-05-29 04:08:45
回答 1查看 77关注 0票数 -1

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.

  • Doctors有很多
  1. 列表,按部门分类,例如:心内科,神经科。

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。

EN

回答 1

Stack Overflow用户

发布于 2022-06-17 07:18:13

不知道它能不能帮到你

代码语言:javascript
复制
- 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,
            };
        }
票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/72420846

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