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FrageHow tdo I get managers full name using BDC XML

  • Donnerstag, 2. Juli 2009 14:09GGratto TeilnehmermedaillenTeilnehmermedaillenTeilnehmermedaillenTeilnehmermedaillenTeilnehmermedaillen
     

    I can get every thing I need except I cannot figure out how to get the managers full name, I can get the managers domain\name, but cannot get the full name to work.

    I am using this as a base line for the SQL query for the XML:   (The data is contained in two tables, the first is the UserProfile_Full table, which contains the Basic Profile Information. The second is the UserProfileValue table. The two are linked by the RecordID Field)

    select a.Employee,  o.[Last Name],n.[First Name],  a.[Office Phone], d.[Cell Phone], c.Email, h.Location, e.Department, b.Title, g.FAX
    from
    (select
    a.RecordID,
    a.PreferredName as Employee,
    b.PropertyVal as [Office Phone]
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=8 and
    a.RecordID=b.RecordID) a

    left outer join
    (select
    a.RecordID,
    b.PropertyVal as Title
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=13 and
    a.RecordID=b.RecordID) b
    on a.RecordID=b.RecordID

    left outer join
    (select
    a.RecordID,
    b.PropertyVal as Email
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=9 and
    a.RecordID=b.RecordID) c
    on a.RecordID=c.RecordID

    left outer join
    (select
    a.RecordID,
    b.PropertyVal as [Cell Phone]
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=19 and
    a.RecordID=b.RecordID) d
    on a.RecordID=d.RecordID

    left outer join
    (select
    a.RecordID,
    b.PropertyVal as Department
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=14 and
    a.RecordID=b.RecordID) e
    on a.RecordID=e.RecordID

    left outer join
    (select
    a.RecordID,
    b.PropertyVal as FAX
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=20 and
    a.RecordID=b.RecordID) g
    on a.RecordID=g.RecordID

    left outer join
    (select
    a.RecordID,
    b.PropertyVal as Location
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=11 and
    a.RecordID=b.RecordID) h
    on a.RecordID=h.RecordID

    left outer join
    (select
    a.RecordID,
    b.PropertyVal as [First Name]
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=4 and
    a.RecordID=b.RecordID) n
    on a.RecordID=n.RecordID

    left outer join
    (select
    a.RecordID,
    b.PropertyVal as [Last Name]
    from
    UserProfile_Full a, UserProfileValue b
    where
    b.PropertyID=5 and
    a.RecordID=b.RecordID) o
    on a.RecordID=o.RecordID


    where
    a.Employee like @employeelast + '%'
    and cast(n.[First Name] as nvarchar(100)) like @employeeFirst + '%'
    and CAST(h.location as nvarchar(100)) like @location + '%'
    and cast(e.Department as nvarchar(100)) like @department + '%'
    and cast(b.Title as nvarchar(100)) like @Title + '%'
    and c.email != ''

    order by
    o.[Last Name], n.[First Name]