颤振-功能-返回阵列

Question

我已经建立了一个有44个位置的列表：

List<String> departmentdes = new List(44);

之后我调用了一个函数：

postDepartment();

该函数是一个api调用：

postDepartment() async {
final response = await http.get('url');
final jsonresponse = json.decode(response.body);
List<Department> departments = [];
for(var d in jsonresponse) {
Department department = Department(
fid: d["fid"].toString(),
);
departments.add(department);
}
int index = departments.length -1;
for(int i = 0; i<=index; i++) {
departmentdes[i] = departments[i].fid;
} 
return departmentdes;
}

在postDepartment();之后，我想打印departmentdes，但它总是返回null。为什么？

Answer 1

i<索引

您已经将索引定义为长度-1

只是一个小小的逻辑错误。

Answer 2

将postDepartment更改为此并查看它是否有帮助：

Future<void> postDepartment() async {
final response = await http.get('url');
final jsonresponse = json.decode(response.body);
List<Department> departments = [];
for(var d in jsonresponse) {
Department department = Department(
fid: d["fid"].toString(),
);
departments.add(department);
}
int index = departments.length -1;
for(int i = 0; i<=index; i++) {
departmentdes[i] = departments[i].fid;
} 
return departmentdes;
}

还要检查您的departments是否为空。