System.NullReferenceException: 'Object reference not set to an instance of an object.'

Question

This error got me, I'm using code first: System.NullReferenceException: 'Object reference not set to an instance of an object.' el IEnumerable<BrentStat5.Context.CorvetteCost> @{ ViewData["Title"] = "Index"; } <h1>Index</h1> <p> <a asp-action="Create">Create New</a> </p> <table class="table"> <thead> <tr> <th> @azzedinehtmlsql .DisplayNameFor(model => model.CorvetteItem) </th> <th> @azzedinehtmlsql .DisplayNameFor(model => model.CorvetteItemCost) </th> <th> @azzedinehtmlsql .DisplayNameFor(model => model.CorvetteStore) </th> <th> @azzedinehtmlsql .DisplayNameFor(model => model.Repair) </th> <th> @azzedinehtmlsql .DisplayNameFor(model => model.Upgrade) </th> <th> @azzedinehtmlsql .DisplayNameFor(model => model.Date) </th> <th> @azzedinehtmlsql .DisplayNameFor(model => model.Maintance) </th> <th> @azzedinehtmlsql .DisplayNameFor(model => model.OrginalCost) </th> <th></th> </tr> </thead> <tbody> @foreach (var item in Model) { //error is on this line. <tr> <td> @azzedinehtmlsql .DisplayFor(modelItem => item.CorvetteItem) </td> <td> @azzedinehtmlsql .DisplayFor(modelItem => item.CorvetteItemCost) </td> <td> @azzedinehtmlsql .DisplayFor(modelItem => item.CorvetteStore) </td> <td> @azzedinehtmlsql .DisplayFor(modelItem => item.Repair) </td> <td> @azzedinehtmlsql .DisplayFor(modelItem => item.Upgrade) </td> <td> @azzedinehtmlsql .DisplayFor(modelItem => item.Date) </td> <td> @azzedinehtmlsql .DisplayFor(modelItem => item.Maintance) </td> <td> @azzedinehtmlsql .DisplayFor(modelItem => item.OrginalCost) </td> <td> <a asp-action="Edit" asp-route-id="@item.CorvetteId">Edit</a> | <a asp-action="Details" asp-route-id="@item.CorvetteId">Details</a> | <a asp-action="Delete" asp-route-id="@item.CorvetteId">Delete</a> </td> </tr> } </tbody> </table>

Answer 1

Hi,BrentHumber

This error often means model can't get the value from controller, you should pass the value to view:

public IActionResult Index()
        {
            var model= new List<MyModel>()
            {
                new MyModel{ CorvetteId=1,CorvetteItem="aa"},
                new MyModel{ CorvetteId=2,CorvetteItem="bb"}
            };
            return View(model);
        }

Best Regards,
Jerry Cai

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