Making multiple vaults (by category) from one massive Aperture 9 library?

I am having trouble backing up a massive Aperture 9 library to a vault. So I want to separate my libraries into several vaults - one for family, one for projects, one for professional work, etc. So how do I do that now that I have this one huge library?
I've gotten to the point that I can't back up successfull to a vault. (Yes I am using Time Machine, too). But how can I separate what I've go in 9 to individual vaults?
Thanks!
Frank

Frank,
Aperture only has versions up to 3.  What version do you have?  It's not version 9.
There is, in theory, no limit to the size of a library which you can back up via vault.  What kinds of errors are you getting when you try to back it up?  How large is your Aperture library (in terms of images)?
As far as I know, a vault is all-or-none.  That is, you cannot choose what pieces of your libray to back up to a particular vault.  However, if worse comes to worse, you can split your library into multiple libraries and then have a vault for each one.
nathan

Similar Messages

  • How to generate multiple output pdf's from one oracle reports

    how to generate multiple output pdf's from one oracle reports.
    I have a report where I have to generate more than one output files from the same report based on a parameter.
    Each output file is for each parameter.
    Is this possible in oracle reports, is so how ?

    You can better post your question in the reports forum instead of this pl/sql forum.

  • Change Photos from one to another Photo library

    Hello....I Want to change Some Photos from one to another Photo Library, How do I do this, and the quality is the same. Thanks.

    Use iPhoto Library Manager
    Regards
    TD

  • Adding multiple same-name nodes from one xml into another

    Hi,
    Following on from my question the other day (Adding multiple different nodes from one xmltype into another), I now have a slightly more complex requirement that I cannot work out where to start, assuming that it's something that can reuse some/all of yesterday's work (thanks again, odie_63!). ETA: I'm on 11.2.0.3
    So, here's the (slightly amended) xml along with yesterday's solution:
    with sample_data as (select xmltype('<root>
                                           <xmlnode>
                                             <subnode1>val1</subnode1>
                                             <subnode2>val2</subnode2>
                                           </xmlnode>
                                           <xmlnode>
                                             <subnode1>val3</subnode1>
                                             <subnode2>val4</subnode2>
                                           </xmlnode>
                                         </root>') xml_to_update,
                                xmltype('<a>
                                           <b>valb</b>
                                           <c>valc</c>
                                           <d>
                                             <d1>vald1</d1>
                                             <d2>vald2</d2>
                                           </d>
                                           <e>vale</e>
                                           <f>valf</f>
                                           <g>
                                             <g1>valg1</g1>
                                             <g2>valg2</g2>
                                           </g>
                                           <h>
                                             <h1>valh1</h1>
                                             <h2>valh2</h2>
                                           </h>
                                           <multinode>
                                             <name>fred</name>
                                             <type>book</type>
                                             <head>1</head>
                                           </multinode>
                                           <multinode>
                                             <name>bob</name>
                                             <type>car</type>
                                             <head>0</head>
                                           </multinode>                                        
                                         </a>') xml_to_extract_from
                          from   dual)
    select xmlserialize(document
               xmlquery(
                 'copy $d := $old
                  modify (
                    insert node element extrainfo {
                      $new/a/b
                    , $new/a/d
                    , $new/a/f
                    , $new/a/h
                    } as first into $d/root
                  return $d'
                 passing sd.xml_to_update as "old"
                       , sd.xml_to_extract_from as "new"
                 returning content
             indent
    from sample_data sd;
    That gives me:
    <root>
      <extrainfo>
        <b>valb</b>
        <d>
          <d1>vald1</d1>
          <d2>vald2</d2>
        </d>
        <f>valf</f>
        <h>
          <h1>valh1</h1>
          <h2>valh2</h2>
        </h>
      </extrainfo>
      <xmlnode>
        <subnode1>val1</subnode1>
        <subnode2>val2</subnode2>
      </xmlnode>
      <xmlnode>
        <subnode1>val3</subnode1>
        <subnode2>val4</subnode2>
      </xmlnode>
    </root>
    However, I now need to add in a set of new nodes based on information from the <multinode> nodes, something like:
    <root>
      <extrainfo>
        <b>valb</b>
        <d>
          <d1>vald1</d1>
          <d2>vald2</d2>
        </d>
        <f>valf</f>
        <h>
          <h1>valh1</h1>
          <h2>valh2</h2>
        </h>
        <newnode>
          <name>fred</name>
          <type>book</type>
        </newnode>
        <newnode>
          <name>bob</name>
          <type>car</type>
        </newnode>
      </extrainfo>
      <xmlnode>
        <subnode1>val1</subnode1>
        <subnode2>val2</subnode2>
        <type>book</type>
      </xmlnode>
      <xmlnode>
        <subnode1>val3</subnode1>
        <subnode2>val4</subnode2>
        <type>car</type>
      </xmlnode>
    </root>
    If it's easier, I *think* we would be ok with something like:
    <newnode>
      <type name="fred">book</type>
      <type name="bob">car</type>
    </newnode>
    The closest I've come is:
    with sample_data as (select xmltype('<root>
                                           <xmlnode>
                                             <subnode1>val1</subnode1>
                                             <subnode2>val2</subnode2>
                                           </xmlnode>
                                           <xmlnode>
                                             <subnode1>val3</subnode1>
                                             <subnode2>val4</subnode2>
                                           </xmlnode>
                                         </root>') xml_to_update,
                                xmltype('<a>
                                           <b>valb</b>
                                           <c>valc</c>
                                           <d>
                                             <d1>vald1</d1>
                                             <d2>vald2</d2>
                                           </d>
                                           <e>vale</e>
                                           <f>valf</f>
                                           <g>
                                             <g1>valg1</g1>
                                             <g2>valg2</g2>
                                           </g>
                                           <h>
                                             <h1>valh1</h1>
                                             <h2>valh2</h2>
                                           </h>
                                           <multinode>
                                             <name>fred</name>
                                             <type>book</type>
                                             <head>1</head>
                                           </multinode>
                                           <multinode>
                                             <name>bob</name>
                                             <type>car</type>
                                             <head>0</head>
                                           </multinode>                                        
                                         </a>') xml_to_extract_from
                          from   dual)
    select xmlserialize(document
               xmlquery(
                 'copy $d := $old
                  modify (
                    insert node element extrainfo {
                      $new/a/b
                    , $new/a/d
                    , $new/a/f
                    , $new/a/h
                    , element newnode {
                                       $new/a/multinode/name
                                       ,$new/a/multinode/type
                    } as first into $d/root
                  return $d'
                 passing sd.xml_to_update as "old"
                       , sd.xml_to_extract_from as "new"
                 returning content
             indent
             ) fred
    from sample_data sd;
    Which produces:
    <newnode>
      <name>fred</name>
      <name>bob</name>
      <type>book</type>
      <type>car</type>
    </newnode>
    - obviously not right!
    Can anyone provide any hints? I've tried searching for similar examples, but I mustn't be putting in the right search terms or something!

    odie_63 wrote:
    or, similarly, to get the alternate output :
    copy $d := $old
    modify (
      insert node element extrainfo {
        $new/a/b
      , $new/a/d
      , $new/a/f
      , $new/a/h
      , element newnode {
          for $i in $new/a/multinode
          return element type {
            attribute name {data($i/name)}
          , data($i/type)
      } as first into $d/root
    return $d
    So we're going with the second method, but I've discovered that the "$i/name" node is not always present. When that happens, I would like to use a default value (for example, "george"). I promise I've searched and searched, but I'm completely failing to turn up anything that sounds remotely like what I'm after (seriously, I can't believe my google-fu sucks this badly!).
    Is there a simple way of doing it? The only thing that I've found that looks vaguely relevant is "declare default namespace...." but I'm not sure that that's the correct thing to use, or if it is, how I'm supposed to reference it when populating the attribute value.

  • Multiple account assignment category for one line item in PO

    Dear Experts,
    We are on SRM 5.0 ECS
    One line item of a PO has multiple account assignment category with the cost distribution for
    1. Cost Center - 50%
    2. Order          - 50%
    Is it possible?
    As per my understanding one line item in a PO can have only one account assignment category
    Regards
    Mick

    Hi,
    Yes it is possible by Po line item.
    Just go to transaction bbp_poc through the web interface.
    Select your PO
    Select your line item PO.
    goto the account assignment tab (item overview)
    you get a cost distribution field in percentage
    then you get a cost distribution button : click it and you get several lines where now you can split your 100% in as many lines as you want.
    BR,
    Disha.
    Pls reward points for useful answers.

  • Streaming to Multiple Apple TV's from One MacBook

    Hey everyone! So I am wondering if I can operate 5 different Apple TV's from one single MacBook Pro at once. We are having a convention and want to have 5 TV's constantly streaming video involving our product. Is it possible to do this? The TV's will be positioned so people can walk up and watch a looping video whenever they want. Please let me know if you need more information!

    Not possible via AirPlay.
    You can stream the same content on the ATV via HomeSharing although it will not be in sync.

  • How do you copy photos from one iMac's iPhoto library to another iMac's iPhoto library??

    I have 20K+ photos on my original iMac and enjoy the desktop shuffle of pictures.  I have a newer iMac that has its own library of pictures but I would really like to take all of the pictures from the original iMac and put them on the newer one.  I have a wireless network with an Airport Extreme that both machines are on.  Both machines are running Leopard.  Same versions of iPhoto 11, 9.2.2.  I have an iPad that has all of the pictures on it and that I sync on the older iMac.  When I tried to connect my iPad on the newer iMac in hopes that it would upload the pictures, iTunes gave me warnings about the device not being synced, erasing the library, etc., so I immediately ejected the iPad.  I have all these options for copying the data from one machine to the other but can't seem to make any of them work.  Any ideas?

    Thanks, Larry, I'll look into the iPhoto Library Manager.  I've downloaded the free trial to get a feel for it.  My large photo library is 85GB and I want to avoid having to buy an external drive just to make this one move.  My iDisk is about 20 GB so I would be doing 5 exports to get it all done.  I guess what I really need to do is understand my network a bit better and see how I can transfer this across my lines.  I appreciate your quick response!

  • Transferring from iPhoto to Aperture, Deleted Library Too Early

    I'm currently moving my entire 20GB photo library from iPhoto to Aperture. I imported all of the photos to Aperture, checked that they all worked right. After playing around with the program some, I decided I really liked it, and (stupidly) deleted my iPhoto library because I figured I was done with it. I later realized that there's one more step I needed to take, that of "Consolidating Masters." Now when I click this, it tells me there's no reference files, of course.
    Here's the thing. All of my pictures are actually in the Aperture library. What does "consolidating masters" do, and how do I get around it and get the pictures to actually appear in my Aperture library, instead of just their current limbo state? Any help is appreciated!

    When you imported from iPhoto, did you choose to store the files in their current location, or did you choose to store them in the Aperture library? If you chose the latter, then yes, they are in Aperture library package so there are no referenced files because they are all in your library already. You would have only had to consolidate if you had chosen to store the files in their current location which would have been in the iPhoto library.
    Check out this part of the Aperture manual.

  • Upgrade from iPhoto or Aperture 2 library?

    I have Aperture 2 installed alongside iPhoto. The libraries are nearly identical with some recent shots since yearend in iPhoto but not yet transferred to Aperture 2, but they easily could be. I have purchased Aperture 3, and I am wondering am I better off upgrading from iPhoto or Aperture 2? I am running the latest software for OS X and for both applications. Any thoughts? Thanks!

    Either would do but Aperture is more capable overall - a pro app vs the consumer iPhoto). Since you bought Aperture it would seem sensible to use that. No advantage I can think of to using both which I think is a recipe for confusion.

  • After "consolidating masters" to transition from iPhoto to Aperture, iPhoto library is still 36 GB

    I have, I believe completed the process of moving everything from iPhoto to Aperture (after having a mix of managed and referenced files) by the "consolidate masters" process.
    I notice now that my iPhoto library is still 36 GB in size.  When I "show package contents", I see there are still a number of folders (named by year) each containing subfolders which contain a number of photographs. (See attachment).
    I've got about 25,000 photos in Aperture now and I believe these seem to never have been imported (either as managed or referenced photos) into Aperture.
    Are they located within the iPhoto Library package in such a way as to have been hidden from Aperture when I originally elected to reference iPhoto files?
    Just trying to figure out what happened...

    When you consolidated, did you elect to move or copy the files to the new location?
    Regards
    TD

  • Move existing Aperture photos (from iPhoto) to Aperture's library's library

    Hi,
    I recently moved from iPhoto to Aperture. I did something wrong tho, since now most of my photos in Aperture are still in iPhoto's library (all of these show the "iPhoto Original" keyword).
    I want to have them all in my Aperture's library, but I put many time on Aperture's organization and don't want to have to import all from the very beginning. Is there any way I can move those files still in iPhoto's library folder to Aperture's library? I plan to uninstall iPhoto after that, that's why I want all photos in Aperture's library.
    Thank you

    If the image is shown in Aperture, the image is in Aperture's Library.
    The question is, where are the Master files on which the images in your Aperture Library are based.  (The keyword "iPhoto Original" tells you nothing about the location of the Master.)
    Right-click an image.  Does "Show in Finder" appear in the pop-up menu?  If so, then your image has a Referenced Master (Referenced Masters are located outside the Aperture Library (the image is in the Library and has a pointer to its Master which is outside the Library)).  If not, your image has a Managed Master (the image, the pointer, and the Master are all inside the Aperture Library.
    You can filter for images with Referenced or Managed Masters.  Also, most of the default Metadata Overlays include a Badge which will show you which images have Referenced Masters (as well as the status of the Master).
    Since you (seem to) want to have all of your images' Masters be Managed (= inside the Aperture Library), simply select all images and execute the command "File→Consolidate Masters".  You can confirm that this has been done by using a Filter set to show all images with Referenced Masters.  It should show no images.
    Once all of your Masters have been consolidated into your Aperture Library, you can safely delete your iPhoto Library.  You can also uninstall iPhoto is you wish.

  • Insert multiple rows into dB from one dynamic HTML form problem

    Hi,
    - I have form which has dynamically generated text fields so I don't want to hard code the field names into the servlet
    - A user will fill in one or more of the text boxes with a number and submit to the servlet (the name of the field represents an item ID, the value represents a quantity the user wants of that item).
    - The servlet needs to insert a new record into a table for each field that is not null.
    My question is how!
    If I send as an array or string, how will the servlet know that each submitted field needs to be inserted as a new record as opposed to one long record?
    I know I'll need to use a loop of somekind; how will I know how long to loop for if the number of 'not null' fields is not static? I need to get a value for the number of 'not null' fields from the form before the loop starts I think but don't know how...
    Also, should I call an SQL procedure that has an insert-loop OR should I have the servlet loop a call to a single insert statement? (am I making sense!?)
    Anyway, I've seen many examples where a submitted form updates/inserts one record into a table but never any for multiple records at one time. I'm using a Tomcat/Oracle set up, and I'm "New to Java"...
    Many thanks in advance.

    sorry, but I dont' understand very well what do you want to do!
    In any case if you want to retrieve all parameters (and you don't know what they are), you can use a general cosde like this:
    Enumeration names= request.getParameterNames();
    while(names.hasMoreElements()) {
    // in 'name' you store the name of the parameter
    String name = (String) names.nextElement();
    // in 'value' you store it's value
    String value = request.getParameter(name);
    remember that ALL parameter values can't be null (if they are void their value is "").
    The only case that you can get a null value is when you try to access a non existing parameter (like request.getParameter("pwqjsak"));
    I hope this helps! Else try to give me more info about your problem

  • How to create multiple TYPES of objects from one menu?

    Q: How can I create a single class to create objects of multiple 'object classes' in a way that is not a huge switch statement?
    Explaination:
    Let's say that I have an application that I am building, that manages five hundred object types. A properly-built object subclassing tree is created, and I want to be able to create objects of any 'leaf node' of this subclassing tree using a single 'objectCreate()' method in a 'factory object'. The purpos of this method will be to create an instance of the correct object, pass a handle to a few collections for properly sorting and storing these objects in groups.
    Usually, one could create a switch in this function, testing for the type of object that the user wants to create from the menu. But in the case of having hundreds of possible object choices, this becomes harder and harder code to maintain (let alone performance).
    Any suggestions?

    But if my menu has:
    1. German Shepard
    2. Doberman Pinscher
    3. Malamut
    4. Persian Long-hair
    5. Siamese
    6. Tabby
    And my object class tree goes:
                                  [ Animal ]
                 [ Cat ]                              [ Dog ]
      [ various breeds ]                         [ various breeds ]How do I code the menu class to respond to the input, so that it runs the correct [breed] object's constructor?
    The line:
    Animal choice = new xxxxxxxx();
    I can't use a variable to replace 'xxxxxxxx' in run-time, but having a ton of choices in code sounds/looks unreasonable.
    if (choice == "Doberman Pinscher")
    Animal choice = new doberman();
    else if (choice == "Tabby")
    Animal choice = new tabby();
    Do you see what I am trying to avoid? I am not experienced enough to instantly realize how to avoid the latter, and instead, do a single instantiation command for the correct constructor.

  • How to create multiple reports and dashboards from one dataset?

    Hi there,
    I've centralized the data to an excel workbook from multiple data sources and added as a dataset to my new Power BI site. How can I build multiple reports and dashboards out of one dataset?

    Hi Kuber,
    In the Datasets section on the Navigation pane (left side of your Dashboards page), you can click "..." next to each dataset. One of the options in the contextual menu is Explore. When you explore a dataset, you can create reports and save
    them.
    You can find more about this and other topics in our Getting Started content:
    https://go.microsoft.com/fwlink/?LinkID=512054
    Thanks,
    M.

  • How do you create multiple events (different dates) from one large event?

    I recently downloaded a large number of photos from my iphone to iphoto. All of the photos are now under event 4/9/2011 (the date and times are correct, just all in one event). How do I change/break up photos and add/change event dates? I don't want to scroll up hundreds (half a year) of photos to see photos I took yesterday.

    Select the Event and go Events -> Autosplit
    Regards
    TD

Maybe you are looking for