ByteArrayOutputStream到FileBody [英] ByteArrayOutputStream to a FileBody

问题描述

我有一个URI来要么采取或我想加载和COM preSS为JPEG，75％的质量的画廊选择的图像。我相信我已经实现了与以下code：

I have a Uri to an image that was either taken or selected from the Gallery that I want to load up and compress as a JPEG with 75% quality. I believe I have achieved that with the following code:

ByteArrayOutputStream bos = new ByteArrayOutputStream();
Bitmap bm = BitmapFactory.decodeFile(imageUri.getPath());
bm.compress(CompressFormat.JPEG, 60, bos);

不，我已经藏到一个 ByteArrayOutputStream 名为 BOS 我需要把它添加到一个 MultipartEntity 为 HTTP POST 到网站。 我想不通的是如何给ByteArrayOutputStream转换为FileBody。

Not that I have tucked it into a ByteArrayOutputStream called bos I need to then add it to a MultipartEntity in order to HTTP POST it to a website. What I can't figure out is how to convert the ByteArrayOutputStream to a FileBody.

推荐答案

使用一个<一个href="http://hc.apache.org/httpcomponents-client-ga/httpmime/apidocs/org/apache/http/entity/mime/content/ByteArrayBody.html"><$c$c>ByteArrayBody而不是（自了HTTPClient 4.1），尽管它的名字它需要一个文件名，也：

Use a ByteArrayBody instead (available since HTTPClient 4.1), despite its name it takes a file name, too:

ContentBody mimePart = new ByteArrayBody(bos.toByteArray(), "filename");

如果你被卡住了HTTPClient 4.0，使用<一个href="http://hc.apache.org/httpcomponents-client-ga/httpmime/apidocs/org/apache/http/entity/mime/content/InputStreamBody.html"><$c$c>InputStreamBody而不是：

If you are stuck with HTTPClient 4.0, use InputStreamBody instead:

InputStream in = new ByteArrayInputStream(bos.toByteArray());
ContentBody mimePart = new InputStreamBody(in, "filename") 

（这两个类也有采取addtional MIME类型字符串构造）

(Both classes also have constructors that take an addtional MIME type string)

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