springmvc防止重复提交_java后端解决重复提交问题

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发布于 2022-11-11 13:55:30

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发布于 2022-11-11 13:55:30

文章被收录于专栏：全栈程序员必看

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表单提交时需要校验数据是否已存在，如果已存在需要防止重复提交，做法比较简单，不再赘述。

1.Employee类

public class Employee
{
        public int Id { set; get; }

        [StringLength(10,MinimumLength=10)]
        public string Name { set; get; }

        [RegularExpression("^[1-9][0-9]{0,1}$")]
        public Nullable<int> Age { set; get; }
}

2.View

@model AspNetMvcModelValidate.Models.Employee
<script src="~/Scripts/jquery-1.10.2.min.js"></script>
<script type="text/javascript">
    $(document).ready(function () {
        $("#btn").click(function () {
            var name = $('#name').val();
            $.post('@Url.Action("CheckEmployeeName")', {
                name: name
            },
            function (data) {
                console.log(data);
                if (data) {
                    console.log("yes");
                    $("#myform").submit();
                } else {
                    alert("Employee with name：" + name + "already exist.");//已有相同
                }
            })
        }); 
    });
</script>

<h2>Index2</h2>
@using (@Html.BeginForm("AddEmployee", "Home", FormMethod.Post, new { Id = "myform" }))
{ 
    <table>
        <tr>
            <td>Name</td>
            <td>@Html.TextBoxFor(s => s.Name, new { ID = "name" })</td>
        </tr>
        <tr>
            <td>Age</td>
            <td>@Html.TextBoxFor(s => s.Age)</td>
        </tr>
        <tr>
            <td colspan="2"><input type="button" id="btn" value="submit" style="width:215px" /></td>
        </tr>
    </table>
}

3.控制器

public JsonResult CheckEmployeeName(string name)
{
            var query = ModelsContainer.Employees.Where(x => x.Name == name).Count<Employee>();
            return Json(query < 1, JsonRequestBehavior.AllowGet);
}

public ActionResult AddEmployee(Employee employee)
{
            ModelsContainer.Employees.Add(employee);
            return View(employee);
}

4.数据源

public class ModelsContainer
{
        public static List<Employee> Employees = new List<Employee>
        {
            new Employee{ Id = 1, Name = "Employee01", Age = 25 },
            new Employee{ Id = 2, Name = "Employee02", Age = 25 }
        };
}

5.Result