javascript - 如何将成功函数名称作为函数参数传递给 AJAX 调用？

Question

在这里，我尝试将 AJAX 调用作为单个函数进行，因为我将成功函数名称作为函数参数传递给 AJAX 调用函数。

我尝试编写以下函数:

function ApiCallFunction(Datatext, ApiName, FunctionName) {
  $.ajax({
    url: Apiurl + ApiName,
    type: "POST",
    data: Datatext,
    contentType: "application/json",
    dataType: "json",
    success: function(data) {
      var funname = FunctionName + '("' + data + '")';
      eval(funname);
    },
    error: function(error) {
      jsonValue = jQuery.parseJSON(error.responseText);
      ErrorWhileSave(jsonValue.Message);
    },
    failure: function(response) {
      ErrorWhileSave("");
    }
  });
}

函数调用:

var datatext = {
  BillChild: {},
  BillDate: "2018-07-23T08:35:32.319Z",
  EntryTime: "2018-07-23T08:35:32.319Z",
  ExitTime: "2018-07-23T08:35:32.319Z",
  TotalTime: "2018-07-23T08:35:32.319Z",
  Total: 0,
  OtherCharges: 0,
  Discount: 0,
  TaxableAmount: 0,
  TotalTax: 0,
  GrandTotal: 0,
  RoundOff: 0,
  NetAmount: 0,
  ByCash: 0,
  ByBank: 0,
  CashReceived: 0,
  BalanceReceivable: 0,
  FirmId: 0,
  UserId: 0,
  BillId: 35,
  CustomerId: 0,
  BranchId: 0,
  BillType: "string",
  BillNo: "string",
  PaymentType: "string",
  Notes: "string",
  TaxType: "string",
  BankId: "string",
  CreatedBy: "string",
  HostIp: "string",
  BranchTransfer: "string",
  ConsultId: 0,
  SearchKey: "string",
  Flag: "SELECTONE"
};

var Datatext = (JSON.stringify(datatext));
ApiCallFunction(Datatext, "Bill_master", "ReturnFunction");

我尝试使用的 Success 函数是:

function ReturnFunction(ReturnValue) {
  alert(data.data.Table1[0].BillId);
}

当我尝试 alert(ReturnValue) 时，它显示为 object object。我还尝试了 ReturnValue.data.data.Table1[0].BillId 仍然无法使用这些值。 AJAX 调用成功，我从中获得了值(value)，但我无法将结果 JSON 对象传递给其他函数。

如何将 JSON 对象传递给其他函数？请帮助我。

Answer 1

您可以通过不同的方式实现您想要做的事情。

Method 1 simply assigning the function object as the parameters

function ApiCallFunction(Datatext, ApiName, onSucess,onError) {
    $.ajax({
        url: Apiurl + ApiName,
        type: "POST",
        data: Datatext,
        contentType: "application/json",
        dataType: "json",
        success: onSucess,
        error: onError,
        failure: function (response) {
            ErrorWhileSave("");
        }
    });
}

并将函数的实现作为:

function ReturnFunction(response){
 //assuming that response is of JSON type
 alert(response.data.Table1[0].BillId);
}

function myError(response){
 console.log(JSON.parse(response.responseText).Message);
}

调用:

ApiCallFunction(DataText,"Bill_master",ReturnFunction,myError);

Method 2 if you happen to have a string instead of the function object

function ApiCallFunction(Datatext, ApiName, FunctionName) {
    $.ajax({
        url: Apiurl + ApiName,
        type: "POST",
        data: Datatext,
        contentType: "application/json",
        dataType: "json",
        success: function (data) {
             window[FunctionName].apply(this,data);
        },
        error: function (error) {
            jsonValue = jQuery.parseJSON(error.responseText);
            ErrorWhileSave(jsonValue.Message);
        },
        failure: function (response) {
            ErrorWhileSave("");
        }
    });
}

调用:

ApiCallFunction(DataText,"Bill_master","ReturnFunction");