This code has worked for me before but i am not sure anymore what has been causing this error. My only guess is that when i try to create a Player, the Team Data is being sent back to Team table and its trying to duplicate but since TeamId is unique hence this error.
Error
The INSERT statement conflicted with the FOREIGN KEY constraint "FK_dbo.Players_dbo.Teams_TeamId". The conflict occurred in database "Web", table "dbo.Teams", column 'TeamId'. The statement has been terminated.
Player
public class Player
{
...
...
[HiddenInput(DisplayValue = false)]
[ForeignKey("Team")]
public int TeamId { get; set; }
public virtual Team Team { get; set; }
...
}
Team
public class Team
{
[Key]
[HiddenInput(DisplayValue = false)]
public int TeamId { get; set; }
....
public virtual ICollection<Player> Players { get; set; }
}
Controller
//
// GET: /Player/
[HttpGet]
public ActionResult Create()
{
PopulateTeamsDropDownList();
return View();
}
//
// POST: /Player/
[HttpPost]
public ActionResult Create(Player model)
{
if (ModelState.IsValid)
{
using (var db = new EfDb())
{
var userProfile = db.UserProfiles.Single(u => u.UserName == User.Identity.Name);
if (userProfile != null)
{
var player = new Player
{
UserProfile = userProfile,
....
....
};
db.Players.Add(player);
db.SaveChanges();
}
}
}
PopulateTeamsDropDownList(model.TeamId);
return View(model);
}
private void PopulateTeamsDropDownList(object selectedTeams = null)
{
var teamsQuery = from d in _iService.Teams
orderby d.Name
select d;
ViewBag.TeamID = new SelectList(teamsQuery, "TeamId", "Name", selectedTeams);
}
View
<div class="editor-label">
@Html.LabelFor(model => model.TeamId, "Pick Your Team")
</div>
<div class="editor-field">
@Html.DropDownList("TeamId", String.Empty)
@Html.ValidationMessageFor(model => model.TeamId)
</div>
How can resolve this error?
The reason behind this is when you are inserting a record on the child
table which the value of the referencing column doesn't exist yet on the on the parent
table.
Consider the following scenario:
// PARENT TABLE
CREATE TABLE TableA
(
ID INT PRIMARY KEY
);
// CHILD TABLE
CREATE TABLE TableB
(
ID INT PRIMARY KEY,
TableA_ID INT,
CONSTRAINT tb_FK FOREIGN KEY (TableA_ID) REFERENCES TableA(ID)
);
// RECORDS OF PARENT TABLE
INSERT INTO TableA (ID) VALUES (1);
INSERT INTO TableA (ID) VALUES (2);
// RECORDS OF CHILD TABLE
INSERT INTO TableB (ID, TableA_ID) VALUES (1,1);
INSERT INTO TableB (ID, TableA_ID) VALUES (2,1);
INSERT INTO TableB (ID, TableA_ID) VALUES (3,2);
If you execute the statements above, it will not fail because none of them violates the referential integrity rule.
Try executing the following statement:
INSERT INTO TableB (ID, TableA_ID) VALUES (3,4);
It fails because 4
which is the value of TableA_ID
to be inserted doesn't exist on Table1.ID
. Foreign Keys preserved referential integrity between the records.
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