从Uri读取文件给出java.io.FileNotFoundException:打开失败:ENOENT [英] Reading File from Uri gives java.io.FileNotFoundException: open failed: ENOENT

问题描述

当使用ACTION_GET_CONTENT意图拾取图像时，我得到一个无法从中打开文件的URI.如果我尝试打开文件，如下所示:

When picking an image using an ACTION_GET_CONTENT intent, I get a URI that I can't open the file from. If I try to open the file, like this:

InputStream in = new FileInputStream(new File(uri.getPath()));

它具有以下异常:

03-11 15:14:36.132  20557-20557/my.package W/System.err﹕ java.io.FileNotFoundException: /document/image:9537: open failed: ENOENT (No such file or directory)
03-11 15:14:36.138  20557-20557/my.package W/System.err﹕ at libcore.io.IoBridge.open(IoBridge.java:456)
03-11 15:14:36.138  20557-20557/my.package W/System.err﹕ at java.io.FileInputStream.<init>(FileInputStream.java:76)

/document/image:9537似乎确实是错误的路径，但是如何获得正确的路径?

/document/image:9537 seems to indeed be an incorrect path, but how do I get the correct path?

我使用此逻辑打开图像选择器:

I use this logic to open the image picker:

Intent photoPickerIntent = new Intent(Intent.ACTION_GET_CONTENT);
photoPickerIntent.setType("image/*");
photoPickerIntent.putExtra("return-data", false);
startActivityForResult(Intent.createChooser(photoPickerIntent, "Complete action using"), PICK_FROM_FILE);

并像这样在onActivityResult中检索Uri:

And retrieve the Uri in the onActivityResult like this:

public void onActivityResult(int requestCode, int resultCode, Intent data)
{
    ....
    Uri uri = data.getData();

我需要获取文件进行解码以使其更小.

I need to get the file to do decoding to make it smaller.

推荐答案

如果我尝试打开文件，如下所示:

If I try to open the file, like this:

这不适用于大多数现代Android设备.您很可能收到了content: Uri.这在较新版本的Android上是很正常的.未来的Android版本可能会阻止file: Uri值.

That will not work for most modern Android devices. Most likely, you received a content: Uri. This is fairly normal on newer versions of Android. Future versions of Android might block file: Uri values.

我需要获取文件进行解码以使其更小.

I need to get the file to do decoding to make it smaller.

不必与给定的Uri关联的文件. Uri可能指向:

There does not have to be a file associated with a given Uri. That Uri might point to:

• 外部存储中的本地文件
• 内部存储中另一个应用程序的本地文件
• 可移动存储中的本地文件
• 已加密且需要动态解密的本地文件
• 保存在数据库的BLOB列中的字节流
• 需要先由其他应用下载的内容
• ...等等

• A local file on external storage
• A local file on internal storage for the other app
• A local file on removable storage
• A local file that is encrypted and needs to be decrypted on the fly
• A stream of bytes held in a BLOB column in a database
• A piece of content that needs to be downloaded by the other app first
• ...and so on

使用ContentResolver和openInputStream()在Uri指向的内容上获取InputStream.然后，将其传递给您的解码逻辑，例如BitmapFactory及其decodeStream()方法.

Use a ContentResolver and openInputStream() to get an InputStream on the content pointed to by the Uri. Then, pass that to your decoding logic, such as BitmapFactory and its decodeStream() method.

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