我很难将我在“$ scope.arr”中推送的多行保存到我的SQL Server数据库中。我在下面有我的代码并且它工作正常但是当我通过按“添加人员”按钮添加/推送一些条目后单击“保存到数据库”按钮时,它将具有空值的行传递给SQL Server数据库。请指导我在哪里犯错:
我也听说过使用angular.forEach循环,但我也很困惑。
我的模型类“TestModel.cs”在这里:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
namespace TestProj.Models
{
public class TestModel
{
public int ID { get; set; }
public string FirstName { get; set; }
public string LastName { get; set; }
}
}
我的MVC控制器在这里命名为TestController的Add方法:
[HttpPost]
public string AddPerson(TestModel test)
using (TestContext db = new TestContext())
{
if (test != null)
{
db.TestModels.Add(test);
db.SaveChanges();
return "Successful";
}
else
{
return "Failed";
}
}
}
我的AngularJS脚本:
var app = angular.module("TestApp", []);
app.controller("TestCtrl", function ($scope, TestService) {
$scope.arr = [];
$scope.addPerson = function () {
var myobj = {
FirstName: $scope.firstname,
LastName: $scope.lastname
}
$scope.arr.push(myobj);
};
$scope.savePerson = function () {
var TestData = TestService.AddPer($scope.arr);
TestData.then(function (msg) {
alert(msg.data);
}, function () {
alert('Error In Adding Person');
});
};
});
app.service("TestService", function ($http) {
this.AddPer = function (person) {
var response = $http({
method: "post",
url: "/Test/AddPerson",
data: JSON.stringify(person),
dataType: "json",
});
console.log(response);
return response;
}
});
我的Index.cshtml文件在这里:
<div ng-controller="TestCtrl">
<form>
FirstName: <input class="form-control" ng-model="firstname" /><br />
LastName: <input class="form-control" ng-model="lastname" /><br />
<button class="btn btn-success" ng-click="addPerson()">Add Person</button>
<button class="btn btn-success" ng-click="savePerson()">Save To DB</button>
<table class="table table-bordered">
<tr>
<th>S. No</th>
<th>First Name</th>
<th>Last Name</th>
</tr>
<tr ng-repeat="a in arr">
<td>{{$index+1}}</td>
<td>{{a.FirstName}}</td>
<td>{{a.LastName}}</td>
</tr>
</table>
</form>
</div>
<script src="~/Scripts/MyAngular/Module.js"></script>
我们将不胜感激。谢谢!
答案 0 :(得分:1)
然后服务器端操作应该期望收集TestModel
,您可以在那里List
。如果您在参数之前[FromBody]
,则在将其发布到服务器之前不需要stringify
数据。
<强>代码强>
[HttpPost]
public string AddPerson([FromBody] List<TestModel> test)
using(TestContext db = new TestContext()) {
test.forEach(t=> {
if (t != null) {
db.TestModels.Add(t);
return "Successful";
} else {
return "Failed";
}
});
db.SaveChanges(); //save context at the end, when no error occurs
}
}
答案 1 :(得分:1)
问题在于我的服务器端代码,即我的C#控制器,这种方法有效:
[HttpPost]
public void AddPerson(List<TestModel> test)
{
using (TestContext db = new TestContext())
{
foreach(var t in test)
{
if (t != null)
{
db.TestModels.Add(t);
Console.WriteLine("Success");
}
}
db.SaveChanges(); //save context at the end, when no error occurs
}
}