javascript - 如何防止将数据传递到 Controller

如果表单未经验证，如何防止将模型发送到Controller？

剃刀

@using (Html.BeginForm("Login", "Login", FormMethod.Post))
{
    <div>
        <div class="form-group input-group">
            <span class="input-group-addon"><i class="glyphicon glyphicon-user"></i></span>
            @Html.TextBoxFor(model => model.AccountName, new { @id = "accountNameText", @class = "form-control", @placeholder = "Account name" })
        </div>
        <div class="form-group input-group">
            <span class="input-group-addon"><i class="glyphicon glyphicon-lock"></i></span>
            @Html.TextBoxFor(model => model.Password, new { @id = "accountPasswordText", @class = "form-control", @placeholder = "Password" })
        </div>
        <div class="form-group">
            <button type="submit" class="btn btn-success btn-block" onclick="valideLogin()">Login</button>
        </div>
        <span id="loginStatusMessege" style="color: red">

        </span>
    </div>
}

jQuery的

function valideLogin() {
    var accName = $('#accountNameText');
    var accPass = $('#accountPasswordText');

    if (accName.val() === '' || accName.val() === '') {
        $('#loginStatusMessege').html('Account name and password cannot be blank.');
    }
}

在我按下登录按钮并且文本框未填充之后，代码成功进入jQuery并就错误进行了响应，但是仍然每次我都进入Controller。如何防止此动作？

最佳答案

尝试添加return false;：

function valideLogin() {
    var accName = $('#accountNameText');
    var accPass = $('#accountPasswordText');

    if (accName.val() === '' || accName.val() === '') {
        $('#loginStatusMessege').html('Account name and password cannot be blank.');
        return false;
    }
}

关于javascript - 如何防止将数据传递到 Controller ，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/34900781/