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Get property from latest child as calculated property in Entity Framework (EF)


Get property from latest child as calculated property in Entity Framework (EF)

By : Suresh Kumar
Date : November 21 2020, 04:03 PM
it helps some times Database background: , If I understood it right, this is how your code should look like:
code :
var latestChild = Parent.ChildCollection.OrderByDescending(c => c.Date).FirstOrDefault();

if(latestChild != null) 
{
    Parent.LastDate = latestChild.Date;
    Parent.LastStatus = latestChild.ChildStatus;
}


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Entity Framework Calculated Property problem

Entity Framework Calculated Property problem


By : CPP
Date : March 29 2020, 07:55 AM
seems to work fine Yes, you can do this. Use the LINQ translations library.
Entity Framework with Calculated Property

Entity Framework with Calculated Property


By : Raif André
Date : March 29 2020, 07:55 AM
wish helps you I would go with pulling the count from the tag repository. Since you are likely to be pulling counts for multiple tags at a time I would change your repository method to pull all at once.
I would create specific viewmodels for your view and then map your Project and Tag to them with Automapper.
code :
public class ProjectView {
    public int ProjectID { get; set; }
    public string Title { get; set; }
    public string Description { get; set; }
    public virtual ICollection<TagView> Tags { get; set; }
}

public class ProjectTagView {
   public int TagID { get; set; }
   public string TagValue { get; set; }
   public int TagCount { get; set; }
}
var model = AutoMapper.Map<Project,ProjectView>(project);
Dictionary<string,int> tagCounts = tagRepository.getCounts(model.Tags.Select(t=>t.TagValue));
foreach(var t in model.Tags){
    t.TagCount = tagCounts[t.TagValue];
}
How to specify a property as calculated in Entity Framework?

How to specify a property as calculated in Entity Framework?


By : Marianne
Date : March 29 2020, 07:55 AM
Any of those help In examples from Microsoft, they implicitly defined a calculated property by providing only accessor without mutator: , It can be done in your DataContext in code first
code :
protected override void OnModelCreating(DbModelBuilder modelBuilder) 
{
   modelBuilder.Entity<Your_Class>().Ignore(x => x.CalculatedProperty); 
}
Calculated Property in Entity Framework

Calculated Property in Entity Framework


By : user3630308
Date : March 29 2020, 07:55 AM
it helps some times You can use computed columns in the database, and decorate your entity with DatabaseGenerated attribute to prevent EF to try to write back its value to the table. After load, EF will read its value when you insert or update:
code :
[DatabaseGenerated(DatabaseGeneratedOption.Computed)]    
public string YourComputedProperty { get; set; }
Entity Framework Where Clause for Calculated property

Entity Framework Where Clause for Calculated property


By : JohnnyBloggs
Date : March 29 2020, 07:55 AM
around this issue Your business logic to calculate that property is currently simple enough, so it can be inlined:
code :
public IEnumerable<Account> Get(QueryClause<Account> queryClause, out ForGetCollectionOptions forGetCollectionOptions)
{
        using (_dbContext)
        {
            var result = (
              from account in _dbContext.Accounts
              select new Account
              {
               Id = account.ID,
               Code = account.Code.Trim(),
               AccountStatus = (account.StartDate != null && account.EndDate == null)
                                  ? "Active"
                                  : (account.StartDate != null && account.StartDate > DateTime.Now) 
                                     ? "Future"
                                     : "Inactive"
               }
           ApplyFilterClause(ref result, queryClause.FilterClause);
        }
}
SELECT  foo
       ,(CASE 
          WHEN t.StartDate IS NOT NULL...
          WHEN t.StartDate IS NOT NULL AND t.StartDate > GetDate()
          ELSE ...
        )
FROM bar
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