我有一些将由用户填写的对象。我将这些对象解析为json并将该json添加到列表中以传递请求正文。但是我做不到。
incrementListPaymentSlipes(PaymentSlipes objPayment) async {
objPayment.name = "Douglas";
objPayment.personalId = "00000000000";
Map<String, dynamic> json = objPayment.toJson();
listPaymentSlipes.add(jsonEncode(json));
}
var response = await http.post(url, body: {
"payment_slips": listPaymentSlipes,
}
正确的 body 示例:
"payment_slips": [
{
"personal_id": "01888728680",
"name": "Fulano da Silva"
}
]
{"error":"'{{personal_id: 00000000000, name: Douglas}}' é invalido como 'payment_slips'","code":"payment_slips_invalid"}```
最佳答案
因此,看来您没有获得所需的JSON。我整理了一些代码,向您展示如何获得所需的 body 。
链接以在DartPad中运行https://dartpad.dartlang.org/3fde03078e56efe13d31482dea8e5eef
class PaymentSlipes {
String name;
String personaId;
ObjPayment({this.name, this.personaId});
//You create this to convert your object to JSON
Map<String, dynamic> toJson() => {'name': name, 'personaId': personaId};
}
// This method is needed to convert the list of ObjPayment into an Json Array
List encondeToJson(List<PaymentSlipes> list) {
List jsonList = List();
list.map((item) => jsonList.add(item.toJson())).toList();
return jsonList;
}
// This is an example and this code will run in DartPad link above
void main() {
PaymentSlipes objPayment = PaymentSlipes(name: "Douglas", personaId: "123425465");
PaymentSlipes objPayment2 = PaymentSlipes(name: "Dave", personaId: "123425465;
PaymentSlipes objPayment3 = PaymentSlipes(name: "Mike", personaId: "123425465");
var list = [objPayment, objPayment2, objPayment3];
// This is the mapping of the list under the key payment_slips as per your example and the body i would pass to the POST
var finalJson = {"payment_slips": encondeToJson(list)};
print(finalJson);
}
关于json - 如何在Flutter中将json列表传递给http请求正文(post)?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/58084921/