想要在代码中打开eclipse xml文件，并参考其IFile导航到特定的行号 [英] Want to open a eclipse xml file in code and navigate to a specific line number with reference to its IFile

问题描述

我通过其IFile实例在eclipse IDE中引用了一个xml文件。我知道想在我的视图上添加一个操作，在xml编辑器中打开文件并导航到特定的行号。任何人对如何解决这个问题都有任何想法？

I have a reference to an xml file in eclipse IDE through its IFile instance. I know want to add an action on my view that opens the file in the xml editor and navigate to a specific line number. Anyone have any ideas on how to go about this?

推荐答案

假设你知道文件的网址：

Assuming you know the file's URL:

IWorkbenchPage page = activeWorkbenchPage();
if (page == null) {
    throw new RuntimeException();
}

IFile file;
IFile[] files = ResourcesPlugin.getWorkspace().getRoot()
            .findFilesForLocationURI(url.toURI());
file = files[0];

IMarker marker;
marker = file.createMarker(IMarker.TEXT);
HashMap<String, Object> map = new HashMap<String, Object>();
map.put(IMarker.LINE_NUMBER, lineNumber);
marker.setAttributes(map);
IDE.openEditor(page, marker);
marker.delete();

当然你也需要捕捉/抛出几个例外，但我在这里省略了为简单起见。

Of course you will need to catch/throw a couple of Exceptions as well, but I omitted this here for simplicity.

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