下面的 dart 代碼允許您解碼來自 resultValue 中的后端的 json，我有下面顯示的 json 的其余呼叫的回應，當我去創建具有以下錯誤的值的串列時，到什么到期？我如何解決它？
JSON 鏈接
錯誤：
//Line: return data.map((jsonObject) => new Manutenzione(
flutter: Errore: type 'List<String>'
is not a subtype of type 'String' in type cast
飛鏢代碼：
var urlmod = await Storage.leggi("URL") "/services/rest/v3/processes/PreventiveMaint/instances";
final resultValue = await apiRequest(Uri.parse(urlmod), {}, UrlRequest.GET, true);
Map<String, dynamic> map = json.decode(resultValue);
List<dynamic> data = map["data"];
try {
return data
.map((jsonObject) => new Manutenzione(
["_id"] as String,
["_user"] as String,
["__user_description"] as String,
["Description"] as String,
["ShortDescr"] as String,
["_Site_description"] as String,
["_Team_description"] as String,
["_status_description"] as String,
["_Company_description"] as String,
))
.toList();
} catch (err) {
print("Errore: " err.toString());
}
return List.empty();
}
JSON：
{
...
data: [
{
_id: value1,
_user: value2,
..
},
{
_id: value1,
_user: value2,
..
},
]
}
uj5u.com熱心網友回復：
您忘記jsonObject在方括號之前指定。
var urlmod = await Storage.leggi("URL") "/services/rest/v3/processes/PreventiveMaint/instances";
final resultValue = await apiRequest(Uri.parse(urlmod), {}, UrlRequest.GET, true);
Map<String, dynamic> map = json.decode(resultValue);
List<dynamic> data = map["data"];
try {
return data
.map((jsonObject) => new Manutenzione(
jsonObject["_id"] as String,
jsonObject["_user"] as String,
jsonObject["__user_description"] as String,
jsonObject["Description"] as String,
jsonObject["ShortDescr"] as String,
jsonObject["_Site_description"] as String,
jsonObject["_Team_description"] as String,
jsonObject["_status_description"] as String,
jsonObject["_Company_description"] as String,
))
.toList();
} catch (err) {
print("Errore: " err.toString());
}
return List.empty();
}
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