Question

尝试从亚马逊复制并粘贴图像时报告此错误，虽然我们检查可转移对象是否支持特定的数据风格，但在尝试使用该数据风格获取网址时失败。我以前没见过这个错误，最近转移到了Java 1.8.0_20所以我假设问题出在Java 8上？

 java.net.MalformedURLException: no protocol: <?xml version="1.0" encoding="UTF-8"?>
    <!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd">
    <plist version="1.0">
    <array>
        <string>http://ecx.images-amazon.com/images/I/41NTPXC8EHL.jpg</string>
        <string></string>
    </array>
    </plist>
    at java.net.URL.<init>(URL.java:586)
    at java.net.URL.<init>(URL.java:483)
    at java.net.URL.<init>(URL.java:432)
    at sun.lwawt.macosx.CDataTransferer.translateBytes(CDataTransferer.java:135)
    at sun.awt.datatransfer.ClipboardTransferable$DataFactory.getTransferData(ClipboardTransferable.java:71)
    at sun.awt.datatransfer.ClipboardTransferable.getTransferData(ClipboardTransferable.java:168)

代码提取

public static DataFlavor imageUrlFlavor =  new DataFlavor("application/x-java-url;class=java.net.URL");
public void getImage(Transferable trans)

            if(trans.isDataFlavorSupported(FileDropTarget.imageUrlFlavor))
            {
                imageUrl  = (URL) trans.getTransferData(FileDropTarget.imageUrlFlavor);
            }

Answer 1

不是一个令人满意的答案 您可以列出放置在剪贴板上的所有DataFlavor trans.getDataFlavors();通常，副本会放置几种风格，如纯文本和样式文本。

替代

public static DataFlavor imageUrlFlavor = 
    new DataFlavor(URL.class, "application/x-java-url");

可能也失败了。

Answer 2

似乎OS X将复制的数据包装到plist / xml中，因此它会破坏您的解析。您是直接从浏览器还是从其他应用程序复制图像？

我试图通过我的Mac（OS X 10.9.5，Java 8u20）上的简单摇摆应用程序从Chrome和Safari复制图像和网址，我总是得到text / plain或image / x-java- image mimetypes，never application / x-java-url或xml-wrapped content。

 Transferable clipData = clipboard.getContents(clipboard);
 if (clipData != null) {
     try {
           for (DataFlavor f : clipData.getTransferDataFlavors()) {
              System.out.println("f = " + f);
           }

           if (clipData.isDataFlavorSupported(imageUrlFlavor)) {
               URL url = (URL) clipData.getTransferData(imageUrlFlavor);

               System.out.println("Got url:");
               System.out.println(url);
           }

      } catch (UnsupportedFlavorException ufe) {
          System.err.println("Flavor unsupported: " + ufe);
      } catch (IOException ioe) {
          System.err.println("Data not available: " + ioe);
      }
 }

我也试图通过拖放来做到这一点并得到相同的结果：

public void drop(DropTargetDropEvent dtde) {
    System.out.println("drop");
    dtde.acceptDrop(DnDConstants.ACTION_COPY_OR_MOVE);
    Transferable trans = dtde.getTransferable();
    boolean gotData = false;
    // try for application/x-java-url flavor
    try {
        if (trans.isDataFlavorSupported(urlFlavor)) {
            URL url = (URL) trans.getTransferData(urlFlavor);
            System.out.println ("got an url: " + url);
            statusField.setText(url.toString());
            statusField.setCaretPosition(0);
            gotData = true;
        } else if (trans.isDataFlavorSupported(DataFlavor.stringFlavor)) {
            String s =
                    (String) trans.getTransferData(DataFlavor.stringFlavor);
            System.out.println("Got text: " + s);
            statusField.setText(s);
            gotData = true;
        } else {
            dumpDataFlavors(trans);

        }
    } catch (Exception e) {
        e.printStackTrace();
    } finally {
        System.out.println("gotData is " + gotData);
        dtde.dropComplete(gotData);
    }
}

为什么OSX上没有协议错误试图使用这个可转移的数据风格？

2 个答案: