错误：无法获取某些“类” {«类», «类», ...}

发布于 2024-09-28 22:27:48 字数 833 浏览 1 评论 0原文

我正在编写一个在 iTunes 中使用的 Applescript，其中有时我想从曲目列表中选择任何曲目，但我期望它的工作方式出现错误。代码如下：

tell application "iTunes"
    set thePlaylist to the first playlist whose name is "Missing track count"
    -- ...
    -- populate a list of strings: albumList
    -- ...
    repeat with albumName in the albumList
        set theAlbum to (the tracks of thePlaylist whose album is albumName)
        display dialog "Found " & (count theAlbum) & " tracks in the album"
        set aTrack to some track of theAlbum -- ERROR OCCURS HERE
    end repeat
end tell

当我从 iTunes 中执行脚本时出现的错误是：

无法获取应用程序“iTunes”的“class cSrc” id 65 的“class cUsP” id 15982 的“class cFlT” id 16112 的“class cTrk”，...等}

现在，我真的不明白为什么它不起作用，尽管我猜想这一定与专辑中的项目是来自 iTunes 应用程序源的用户播放列表中的文件曲目而不是“只是”曲目有关。有人可以帮我吗？

原文

I'm writing an Applescript for use in iTunes in which at some point I want to select any track from a list of tracks, but the way I expected it to work gives an error. Here's the code:

tell application "iTunes"
    set thePlaylist to the first playlist whose name is "Missing track count"
    -- ...
    -- populate a list of strings: albumList
    -- ...
    repeat with albumName in the albumList
        set theAlbum to (the tracks of thePlaylist whose album is albumName)
        display dialog "Found " & (count theAlbum) & " tracks in the album"
        set aTrack to some track of theAlbum -- ERROR OCCURS HERE
    end repeat
end tell

The error I get when I execute the script from within iTunes is:

Can't get some «class cTrk» of {«class cFlT» id 16112 of «class cUsP» id 15982 of «class cSrc» id 65 of application "iTunes", ... etc}

Now, I don't really see why it doesn't work, although I guess it must have something to do with the fact that the items in theAlbum are file tracks from a user playlist from the source from the iTunes application instead of 'just' tracks. Can anyone help me out here?

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打小就很酷 2024-10-05 22:27:48

在此示例中，我使用 some item 而不是 some track，效果很好。

tell application "iTunes"
    set thePlaylist to the first playlist
    set x to (the tracks of thePlaylist)
    set aTrack to some item in x
end tell

结果

URL track id 87 of library playlist id 82 of source id 64 of application "iTunes"

由于您示例中的所有项目都继承自 track，所以我不知道为什么它不起作用，但它不起作用。

In this example I use some item instead of some track, which works OK.

tell application "iTunes"
    set thePlaylist to the first playlist
    set x to (the tracks of thePlaylist)
    set aTrack to some item in x
end tell

results in

URL track id 87 of library playlist id 82 of source id 64 of application "iTunes"

Since all the items in your example inherit from track, I don't know why it doesn't work, but it doesn't.

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旧情勿念 2024-10-05 22:27:48

theAlbum 是一个列表，而不是播放列表，因此它没有轨道元素；它只有项目。

关于列表的文档，其中指出“您还可以按类别引用索引列表项”。不完整，因此具有误导性。看来您只能对内置类执行此操作。据我所知，原因如下：

对象说明符 (2) 是基于键值编码。说明符可以标识属性（对象属性或一对一关系）或元素（一对多关系）。在示例中，我们正在处理元素。要处理元素，底层 Objective-C 类必须实现集合访问器模式。也就是说，它必须至少实现 - 或 -countOf 和 -objectInAtIndex：（当然，它可以实现所有这些）。列表类对一定数量的 Applescript 类执行此操作（如果您查看列表类的 ObjC 源代码，您会发现类似 countOfApplication 和 -objectInNumberAtIndex: 的方法）。可以想象，它可以使用适当的 -doesNotRecognizeSelector: 处理程序支持任意元素对象说明符，但列表似乎并未以这种方式实现。由于列表没有 -track、-countOfTrack 或 -objectInTrackAtIndex: 处理程序，因此它们无法处理诸如“< code>trackList 的第一首曲目”。