如何在颤动飞镖中使用LINQ "where select into“语法

Question

我有一个对象要发布到我的api，这是我的请求类。

public class MyItem
{
    
    public int CustomerId{ get; set; }
    public string Desc{ get; set; }
    public List<ReceiptDetailItem> receiptDetails{ get; set; }
}

public class ReceiptDetailItem
{
    public int UnitId { get; set; }
    public decimal Amount { get; set; }

}

我需要在OrderDetails中使用where子句，因为我只希望发布匹配的详细信息。

  await generalService.POSTAPI(
          url: "POSTMyItem",
          body: {
            "CustomerId": receipt[0]['CUSTOMERID'].toString(),
            "DESC": receipt[0]['DESC'].toString(),
            "OrderDetails": receiptDetailItem
                .where((element) =>
                    element["LOCALRECEIPTID"] ==
                    receipt[0]['LOCALRECEIPTID'])
                .toSet()
                .toList()
                .map((e) => 
                receiptDetails(
                    unitId: e["unitId"],
                    amount: e["amount"]
                   ))
          },
        ).then((value) async {...

基本上我想在flutter中使用where select toList并在list类型中创建receiptDetails，但它不起作用，return< type 'Null‘不是type 'int’的子类型

Answer 1

我这样解决了我的问题：

  await generalService.POSTAPI(
              url: "POSTMyItem",
              body: {
                "CustomerId": receipt[0]['CUSTOMERID'].toString(),
                "DESC": receipt[0]['DESC'].toString(),
                "OrderDetails": receiptDetailItem
                    .where((element) =>
                        element["LOCALRECEIPTID"] ==
                        receipt[0]['LOCALRECEIPTID'])
                    .toList()
                    .map((e) => 
                   return {
                        unitId: e["unitId"].toString(),
                        amount: e["amount"].toString()
                        };
              }).toList()
            }).then((value) async {