So - I've got a third party library that needs a File
as input. I've got a byte array
.
所以 - 我有一个需要File作为输入的第三方库。我有一个字节数组。
I don't want to write the bytes to disk .. I'd like to keep this in memory. Any idea on how I can create a File
from the provided byte array
(without writing to disk)?
我不想把字节写入磁盘..我想把它保存在内存中。关于如何从提供的字节数组创建文件(无需写入磁盘)的任何想法?
3 个解决方案
#1
10
Sorry, not possible. A File is inherently an on-disk entity, unless you have a RAM disk - but that's not something you can create in Java.
对不起,不可能。文件本质上是一个磁盘上的实体,除非你有一个RAM磁盘 - 但这不是你可以用Java创建的东西。
That's exactly the reason why APIs should not be based on File objects (or be overloaded to accept an InputStream).
这就是为什么API不应该基于File对象(或者重载以接受InputStream)的原因。
#2
3
There's one possibility, but it's a real long-shot.
有一种可能性,但这是一个真正的长镜头。
If the API uses new FileReader(file)
or new FileInputStream(file)
then you're hosed, but...
如果API使用新的FileReader(文件)或新的FileInputStream(文件),那么你就被软管了,但......
If it converts the file to a URL or URI (using toURL()
or toURI()
) then, since File
is not final, you can pass in a subclass of File
in which you control the construction of the URL/URI and, more importantly, the handler.
如果它将文件转换为URL或URI(使用toURL()或toURI()),那么,由于File不是final,您可以传入File的子类,您可以在其中控制URL / URI的构造,以及更多重要的是,处理程序。
But the chances are VERY slim!
但机会非常渺茫!
#3
0
So I see there is an accepted answer (and this is old), but I found a way to do this. I was using the IDOL On Demand API and needed to convert a byte array
to a File
.
所以我看到有一个已接受的答案(这是旧的),但我找到了一种方法来做到这一点。我使用的是IDOL On Demand API,需要将字节数组转换为File。
Here is an example of taking a byte array
of an image and turning into a File
:
以下是获取图像的字节数组并转换为文件的示例:
//imageByte is the byte array that is already defined
BufferedImage image = null;
ByteArrayInputStream bis = new ByteArrayInputStream(imageByte);
image = ImageIO.read(bis);
bis.close();
// write the image to a file
File outputfile = new File("image.png");
ImageIO.write(image, "png", outputfile);
And so outputfile
is a File
that can be used later in your program.
因此outputfile是一个可以在程序中稍后使用的文件。
#1
10
Sorry, not possible. A File is inherently an on-disk entity, unless you have a RAM disk - but that's not something you can create in Java.
对不起,不可能。文件本质上是一个磁盘上的实体,除非你有一个RAM磁盘 - 但这不是你可以用Java创建的东西。
That's exactly the reason why APIs should not be based on File objects (or be overloaded to accept an InputStream).
这就是为什么API不应该基于File对象(或者重载以接受InputStream)的原因。
#2
3
There's one possibility, but it's a real long-shot.
有一种可能性,但这是一个真正的长镜头。
If the API uses new FileReader(file)
or new FileInputStream(file)
then you're hosed, but...
如果API使用新的FileReader(文件)或新的FileInputStream(文件),那么你就被软管了,但......
If it converts the file to a URL or URI (using toURL()
or toURI()
) then, since File
is not final, you can pass in a subclass of File
in which you control the construction of the URL/URI and, more importantly, the handler.
如果它将文件转换为URL或URI(使用toURL()或toURI()),那么,由于File不是final,您可以传入File的子类,您可以在其中控制URL / URI的构造,以及更多重要的是,处理程序。
But the chances are VERY slim!
但机会非常渺茫!
#3
0
So I see there is an accepted answer (and this is old), but I found a way to do this. I was using the IDOL On Demand API and needed to convert a byte array
to a File
.
所以我看到有一个已接受的答案(这是旧的),但我找到了一种方法来做到这一点。我使用的是IDOL On Demand API,需要将字节数组转换为File。
Here is an example of taking a byte array
of an image and turning into a File
:
以下是获取图像的字节数组并转换为文件的示例:
//imageByte is the byte array that is already defined
BufferedImage image = null;
ByteArrayInputStream bis = new ByteArrayInputStream(imageByte);
image = ImageIO.read(bis);
bis.close();
// write the image to a file
File outputfile = new File("image.png");
ImageIO.write(image, "png", outputfile);
And so outputfile
is a File
that can be used later in your program.
因此outputfile是一个可以在程序中稍后使用的文件。