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Home/ Questions/Q 7959721
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T04:37:04+00:00 2026-06-04T04:37:04+00:00

This is a code snippet of my my Create.cshtml in my ASP.NET MVC3.. I

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This is a code snippet of my my Create.cshtml in my ASP.NET MVC3.. I don’t know how to save the DateAdded as DateTimeNow. Im using CRUDE to Create a new Record. I already tried to remove the code below and replace it with DateTimeNow (system time) without displaying the textbox and label from the code below:

<div class="editor-label">
    @Html.LabelFor(model => model.DateAdded)
</div>
<div class="editor-field">
    @Html.EditorFor(model => model.DateAdded)
    @Html.ValidationMessageFor(model => model.DateAdded)
</div>

Here’s my snippet of Create.cshtml

@model PhoneBook.Models.Contact

@{
    ViewBag.Title = "Create";
}

<h2>Create</h2>

<script src="@Url.Content("~/Scripts/jquery.validate.min.js")" type="text/javascript"></script>
<script src="@Url.Content("~/Scripts/jquery.validate.unobtrusive.min.js")" type="text/javascript"></script>

@using (Html.BeginForm()) {
    @Html.ValidationSummary(true)
<fieldset>
    <legend>Contact</legend>
    <div class="editor-label">
        @Html.LabelFor(model => model.DateAdded)
    </div>
    <div class="editor-field">
        @Html.EditorFor(model => model.DateAdded)
        @Html.ValidationMessageFor(model => model.DateAdded)
    </div>

  <p>
        <input type="submit" value="Create" />
    </p>
</fieldset>
}

<div>
    @Html.ActionLink("Back to List", "Index")
</div>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T04:37:04+00:00Added an answer on June 4, 2026 at 4:37 am

    You should give the DateTime.Now value in the controller:

    [HttpPost]
    public ActionResult Foo(Contact model)
    {
        model.DateAdded = DateTime.Now;
        ...
        ...
    }
    

    Or if you want the datetime to be in the page when it is rendered:

    [HttpGet]
    public ActionResult Foo()
    {
        Contact model = new Contact{ DateAdded = DateTime.Now};
    
        return View(model);
    }
    
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