I/颤振 (8545)：类型"int"不是可迭代类型的子类型

Question

类别详细信息

class AccountInfo
{
final int total;
final int loan;
AccountInfo( this.total,this.loan);
}

我的功能是

Future<List<AccountInfo>> _getUsers() async {
var url = "http://abcwebtest.com/book/accounts.php";
var para = {'mid': "1"};
var response = await http.post(url, body: json.encode(para));
if(response.statusCode ==200) {
var jsonData = json.decode(response.body);
List<AccountInfo> accinfo_list = [];
for (var u in jsonData) {
AccountInfo usr_acc = AccountInfo( u["total"], u["loan"]);
accinfo_list.add(usr_acc);
}

return  accinfo_list;
}else{
throw Exception('Failed to load ..........');
}
}

它将从表余额中获取贷款和总额。这两个字段都是int。但是我得到一个错误作为

I/flutter(8545(：类型"int"不是类型"Iterable"的子类型请让我帮助查找解决方案。

Answer 1

您正在尝试迭代从服务器返回的内容。它不是一个数组：

for (var u in jsonData) {
AccountInfo usr_acc = AccountInfo( u["total"], u["loan"]);
accinfo_list.add(usr_acc);
}

相反，这样做：

Map<string, dynamic> res = json.decode(response.body);
AccountInfo usr_acc = AccountInfo( res["total"], u["loan"]);
accinfo_list.add(usr_acc);