parallel.foreach循環問題：具有相同密鑰的項目已被添加

Question

想要了解您是否有其中一位可以闡明這一點。 獲取的項目使用相同的密鑰已經被添加例外在下面的代碼：

foreach (EmailMessage m in findResultsSentItems) 
{ 
    em.Add(m); 
} 
lstMailContactDeatils.Clear(); 
Parallel.ForEach(em, 
     myItem => 
     { 
      myItem.Load(); 
      foreach (EmailAddress e in myItem.ToRecipients) 
      { 
       try 
       { 
        MailContactDeatils _MailContactDeatils1 = new MailContactDeatils(); 
        _MailContactDeatils1.MailID = e.Address; 
        _MailContactDeatils1.DisplayName = e.Name; 
        _MailContactDeatils1.SentTime = myItem.DateTimeSent.ToString(); 
       // if(lstMailContactDeatils.Contains()) 
        lstMailContactDeatils.Add(_MailContactDeatils1); 
       } 
       catch (Exception e1) { exceptions.Enqueue(e1); } 
      } 
      foreach (EmailAddress e in myItem.CcRecipients) 
      { 
       try 
       { 
        MailContactDeatils _MailContactDeatils2 = new MailContactDeatils(); 
        _MailContactDeatils2.MailID = e.Address; 
        _MailContactDeatils2.DisplayName = e.Name; 
        _MailContactDeatils2.SentTime = myItem.DateTimeSent.ToString(); 
        lstMailContactDeatils.Add(_MailContactDeatils2); 
       } 
       catch (Exception e2) { exceptions.Enqueue(e2); } 
      } 
     }); 

的LINQ語句是從哪裏獲得拋出的ArgumentExcpetion說：具有相同鍵的項已被添加

Answer 1

在使用Parallel時，您需要在添加前鎖定。

object padlock = new object(); 

foreach (EmailMessage m in findResultsSentItems) 
{ 
    em.Add(m); 
} 
lstMailContactDeatils.Clear(); 
Parallel.ForEach(em, 
     myItem => 
     { 
      myItem.Load(); 
      foreach (EmailAddress e in myItem.ToRecipients) 
      { 
       try 
       { 
        MailContactDeatils _MailContactDeatils1 = new MailContactDeatils(); 
        _MailContactDeatils1.MailID = e.Address; 
        _MailContactDeatils1.DisplayName = e.Name; 
        _MailContactDeatils1.SentTime = myItem.DateTimeSent.ToString(); 

        lock(padlock) { 
         if(!lstMailContactDeatils.Contains(_MailContactDeatils1)) 
          lstMailContactDeatils.Add(_MailContactDeatils1); 
        } 
       } 
       catch (Exception e1) { exceptions.Enqueue(e1); } 
      } 
      foreach (EmailAddress e in myItem.CcRecipients) 
      { 
       try 
       { 
        MailContactDeatils _MailContactDeatils2 = new MailContactDeatils(); 
        _MailContactDeatils2.MailID = e.Address; 
        _MailContactDeatils2.DisplayName = e.Name; 
        _MailContactDeatils2.SentTime = myItem.DateTimeSent.ToString(); 
        lock(padlock) { 
         if (!lstMailContactDeatils.Contains(_MailContactDeatils2)) 
          lstMailContactDeatils.Add(_MailContactDeatils2); 
        } 
       } 
       catch (Exception e2) { exceptions.Enqueue(e2); } 
      } 
     });