Build a strongly-typed update form where only some values are changed. RSS

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  thorma04

  Member

  26 Points

  51 Posts

  Build a strongly-typed update form where only some values are changed.

  Jan 03, 2013 06:50 PM|LINK

  Hello, I am building for that will let users update a table row. The table has 20+ coloumns (legacy deal) and I only want to allow the user to edit 4 colomns. 

  //Form
          @Html.HiddenFor(model => model.EntityLevel) // x20
  
          <div class="editor-label">
              @Html.LabelFor(model => model.EntityName)
              @Html.EditorFor(model => model.EntityName)
              @Html.ValidationMessageFor(model => model.EntityName)
          </div>
  
          <div class="editor-label">
              @Html.LabelFor(model => model.EntityEmail)
              @Html.EditorFor(model => model.EntityEmail)
              @Html.ValidationMessageFor(model => model.EntityEmail)
          </div>
  
          //Controller
          public ActionResult UpdateEntity(Entity entity)
          {
              TryUpdateModel(entity);
              if (ModelState.IsValid)
              {
                  db.Entry(entity).State = EntityState.Modified;
                  db.SaveChanges();
              }
              return View();
          }

  
  
  

  The standard MVC templates normally work great for this kind of thing, but I don't want to have 20 hidden form inputs (eg @Html.HiddenFor(model => model.EntityLevel) ). If no hidden inputs are there, they are treated blank when TryUpdateModel is called, overwriting the existing values with nulls.

  Is there a way to 'merge' the changes in a few rows with existing row in the db?

  Thanks! 

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  bruce (sqlwo...

  All-Star

  36644 Points

  5432 Posts

  Re: Build a strongly-typed update form where only some values are changed.

  Jan 03, 2013 06:59 PM|LINK

  your code makes little sense. TryUpdateModel, binds the from post values to passed model, but you pass a model that has already been bound via it being a parameter.

  //Controller
          public ActionResult UpdateEntity(int id)
          {
              // passing id in hidden field - should really encrypt
              var entity = db.Entity.Find(e => e.ID == id);
              TryUpdateModel(entity);
              if (ModelState.IsValid)
              {
                ....

  
  

  bruce (sqlwork.com)
  • Edited by bruce (sqlwork.com) on Jan 03, 2013 07:00 PM
  • Marked as answer by Young Yang - MSFT on Jan 15, 2013 05:26 AM