MembershipUsers IsLockedOut List By UserName?

I cannot find an easy solution to simply fill a listbox with a list of users and each users IsLockedOut status. I was using the very adequate Membership.GetAllUsers() method to list just userNames. Here's my unsuccessful attempt to lists all users AND IsLockedOut status:

 void Page_Init(object sender, EventArgs e)
    {        
        lbxAllUsers.DataSource = GetAllUsersAndStatus(); // Membership.GetAllUsers();
        lbxAllUsers.DataMember = "UserName";
        lbxAllUsers.DataBind();

        lbxLockedOutStatus.DataSource = GetAllUsersAndStatus();
        lbxLockedOutStatus.DataMember = "LockedOut";
        lbxLockedOutStatus.DataBind();
    }
       
    protected DataSet GetAllUsersAndStatus()
    {
        DataSet ds = new DataSet();
        DataTable dt = new DataTable("Members");
        ds.Tables.Add(dt);
        DataColumn dcUserName = new DataColumn("UserName");
        DataColumn dcLockedOut = new DataColumn("LockedOut");
        dt.Columns.Add(dcUserName);
        dt.Columns.Add(dcLockedOut);
        DataRow row;

        MembershipUserCollection allUsers = Membership.GetAllUsers();

        foreach (MembershipUser user in allUsers)
        {
            row = dt.NewRow();
            row[dcUserName] = user.UserName;
            row[dcLockedOut] = user.IsLockedOut.ToString();
            dt.Rows.Add(row);
        }
        return ds;
    }
pointemanAsked:
Who is Participating?
 
David H.H.LeeCommented:
Hi pointeman,
Quick fix, i've noticed the error that raised from the declared column name.
eg:
These lines:
 row[dcUserName] = user.UserName;
 row[dcLockedOut] = user.IsLockedOut.ToString();
           
Should returned as
 row["UserName"] = user.UserName;
 row["LockedOut"] = user.IsLockedOut.ToString();


0
 
pointemanAuthor Commented:
Two heads are better than one, thanks.....
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.