onActivityResult的intent.getPath()没有给我正确的文件名

问题描述

我正在尝试以这种方式获取文件:

I am trying to fetch a file this way:

final Intent chooseFileIntent = new Intent(Intent.ACTION_GET_CONTENT);
    String[] mimetypes = {"application/pdf"};
    chooseFileIntent.setType("*/*");
    chooseFileIntent.addCategory(Intent.CATEGORY_OPENABLE);
    if (chooseFileIntent.resolveActivity(activity
                        .getApplicationContext().getPackageManager()) != null) {
        chooseFileIntent.putExtra(Intent.EXTRA_MIME_TYPES, mimetypes);
        activity.startActivityForResult(chooseFileIntent, Uploader.PDF);
    }

然后在onActivityResult中:

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
}

根据许多线程，我应该使用data.getData().getPath()从意图中获取文件名，我期望的文件名是 my_file.pdf ，但是我得到的却是这个:

According to many threads I'm supposed to fetch the file name from the intent with data.getData().getPath(), the file name I'm expecting is my_file.pdf, but instead I'm getting this :

那该怎么办?感谢您的帮助.

So what to do? Thanks for your help.

推荐答案

没有该代码.该代码要求用户选择内容.这可能是文件，也可能不是.

Not with that code. That code is asking the user to pick a piece of content. This may or may not be a file.

尽管它倾向于在较旧的Android版本上运行，但这永远是不正确的.

That was never correct, though it tended to work on older versions of Android.

嗯，这取决于.

(更新2019-04-06 :由于Android Q禁止大多数文件系统访问，因此该解决方案不再可行)

(UPDATE 2019-04-06: since Android Q is banning most filesystem access, this solution is no longer practical)

如果您愿意允许用户使用ACTION_GET_CONTENT选择一段内容，请理解它不必是文件，也不必具有类似于文件名的内容.您将获得的最接近的数字:

If you are willing to allow the user to pick a piece of content using ACTION_GET_CONTENT, please understand that it does not have to be a file and it does not have to have something that resembles a filename. The closest that you will get:

• 如果Uri中的getScheme()返回file，则您原来的算法将起作用

• If getScheme() of the Uri returns file, your original algorithm will work

如果Uri的getScheme()返回content，请使用DocumentFile.fromSingleUri()创建DocumentFile，然后在该DocumentFile上调用getName()—这应该返回一个显示名称"，用户应该可以识别

If getScheme() of the Uri returns content, use DocumentFile.fromSingleUri() to create a DocumentFile, then call getName() on that DocumentFile — this should return a "display name" which should be recognizable to the user

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