初めてモデルを挿入しようとしたときにデータベースからエラーが返された後、モデルを挿入できないのはなぜですか。
Report report = null;
using (var session = SessionFactory.OpenSession()) {
try {
using (var transaction = session.BeginTransaction()) {
report = new Report();
session.SaveOrUpdate(report);//Exception: Name field required
transaction.Commit();
}
}
catch { }
try {
using (var transaction = session.BeginTransaction()) {
report.Name = "theName";
session.SaveOrUpdate(report);
//Causes Exception:
//Row was updated or deleted by another transaction (or unsaved-value
//mapping was incorrect): [ReportViewer.DataAccess.Models.Report#22]
transaction.Commit();
}
}
catch { }
}
しかし、既存のモデルを更新しているときにエラーが発生した場合は、修正を行い (この場合は名前を設定)、更新を再試行することができます。
Report report = null;
using (var session = SessionFactory.OpenSession()) {
using (var transaction = session.BeginTransaction()) {
report = new Report();
report.Name = "theName";
session.SaveOrUpdate(report);
transaction.Commit();
}
}
using (var session = SessionFactory.OpenSession()) {
//get entity saved from previous session
report = session.Get<Report>(report.Id);
try {
using (var transaction = session.BeginTransaction()) {
report.Name = null;
session.SaveOrUpdate(report);//Exception: Name field required
transaction.Commit();
}
}
catch { }
try {
using (var transaction = session.BeginTransaction()) {
//updates and does not give an error
report.Name = "theName";
session.SaveOrUpdate(report);
transaction.Commit();
}
}
catch { }
}