blocks|key|2874589|text|使用Apache+Commons+IO中的FileUtils#readFileToByteArray(File)，然后使用ByteArrayInputStream(byte[])构造函数创建ByteArrayInputStream。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2874590|public+static+ByteArrayInputStream+retrieveByteArrayInputStream(File+file)+{
++++return+new+ByteArrayInputStream(FileUtils.readFileToByteArray(file));
}|code-block|syntax|javascript|2874591|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/io|1|http://commons.apache.org/io/api-release/org/apache/commons/io/FileUtils.html#readFileToByteArray%2528java.io.File%2529|2|http://docs.oracle.com/javase/1.4.2/docs/api/java/io/ByteArrayInputStream.html#ByteArrayInputStream%2528byte[]%2529^0|L|Z|1P|S|2N|K|2|H|0|L|Z|1|1P|S|2|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]|$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|G|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]|U|$5|N|O|P|E|$Q|V]]]]

Use the <a href="http://commons.apache.org/io/api-release/org/apache/commons/io/FileUtils.html#readFileToByteArray%28java.io.File%29" rel="noreferrer"><code>FileUtils#readFileToByteArray(File)</code></a> from <a href="http://commons.apache.org/io" rel="noreferrer">Apache Commons IO</a>, and then create the <code>ByteArrayInputStream</code> using the <a href="http://docs.oracle.com/javase/1.4.2/docs/api/java/io/ByteArrayInputStream.html#ByteArrayInputStream%28byte[]%29" rel="noreferrer"><code>ByteArrayInputStream(byte[])</code></a> constructor.

<pre><code>public static ByteArrayInputStream retrieveByteArrayInputStream(File file) {
 return new ByteArrayInputStream(FileUtils.readFileToByteArray(file));
}
</code></pre>

blocks|key|2874629|text|一般的想法是，文件会产生一个FileInputStream，而byte[]会产生一个ByteArrayInputStream。两者都实现了InputStream，因此它们应该与任何使用InputStream作为参数的方法兼容。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2874630|当然，可以将所有文件内容放入ByteArrayInputStream中：|2874631|2874632|将完整文件读取到byte[]中；Java版本>=+7包含从文件中读取所有数据的convenience+method+called+readAllBytes；|ordered-list-item|2874633|围绕文件内容创建一个ByteArrayInputStream，该内容现在位于内存中。|2874634|2874635|请注意，对于非常大的文件，这可能不是最佳解决方案-所有文件都将在同一时间点存储在内存中。为工作使用正确的流是很重要的。|2874636|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/nio/file/Files.html#readAllBytes%2528java.nio.file.Path%2529^0|E|F|V|6|16|K|1X|B|2K|B|0|E|K|0|0|8|6|1T|C|13|12|0|0|A|K|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@$9|1I|A|1J|1|1K]]|E|$]]|$1|L|3|M|5|K|7|1L|8|@$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|1P|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1Q|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

The general idea is that a File would yield a <code>FileInputStream</code> and a <code>byte[]</code> a <code>ByteArrayInputStream</code>. Both implement <code>InputStream</code> so they should be compatible with any method that uses <code>InputStream</code> as a parameter.

Putting all of the file contents in a <code>ByteArrayInputStream</code> can be done of course:

<ol>
<li>read in the full file into a <code>byte[]</code>; Java version >= 7 contains a <a href="http://docs.oracle.com/javase/7/docs/api/java/nio/file/Files.html#readAllBytes%28java.nio.file.Path%29" rel="noreferrer">convenience method called <code>readAllBytes</code></a> to read all data from a file;</li>
<li>create a <code>ByteArrayInputStream</code> around the file content, which is now in memory.</li>
</ol>

Note that this may not be optimal solution for very large files - all the file will stored in memory at the same point in time. Using the right stream for the job is important.

blocks|key|2874670|text|这并不完全是您所要求的，但它是读取字节文件的一种快速方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2874671|File+file+=+new+File(yourFileName);
RandomAccessFile+ra+=+new+RandomAccessFile(yourFileName,+"rw"):
byte[]+b+=+new+byte[(int)file.length()];
try+{
++++ra.read(b);
}+catch(Exception+e)+{
++++e.printStackTrace();
}

//Then+iterate+through+b|code-block|syntax|javascript|2874672|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This isn't exactly what you are asking, but is a fast way of reading files in bytes. 

<pre><code>File file = new File(yourFileName);
RandomAccessFile ra = new RandomAccessFile(yourFileName, "rw"):
byte[] b = new byte[(int)file.length()];
try {
 ra.read(b);
} catch(Exception e) {
 e.printStackTrace();
}

//Then iterate through b
</code></pre>

blocks|key|223314|text|ByteArrayInputStream是一个围绕字节数组的InputStream包装器。这意味着您必须将文件完全读取到byte[]中，然后使用其中一个ByteArrayInputStream构造函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|223315|你能提供更多关于你正在用ByteArrayInputStream做什么的细节吗？很可能有更好的方法来实现你想要实现的目标。|223316|编辑：|BOLD|223317|如果你正在使用Apache+FTPClient上传，你只需要一个InputStream。你可以做到这一点；|223318|String+remote+=+"whatever";
InputStream+is+=+new+FileInputStream(new+File("your+file"));
ftpClient.storeFile(remote,+is);|code-block|syntax|javascript|223319|当然，您应该记得在使用完输入流之后关闭它。|223320|entityMap^0|0|K|U|B|1O|6|24|K|0|C|K|0|0|3|0|W|B|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|17|8|@$9|18|A|19|B|J]]|D|@]|E|$]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|M|3|N|5|O|7|1D|8|@]|D|@]|E|$P|Q]]|$1|R|3|S|5|6|7|1E|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|U|$]]

A <code>ByteArrayInputStream</code> is an <code>InputStream</code> wrapper around a byte array. This means you'll have to fully read the file into a <code>byte[]</code>, and then use one of the <code>ByteArrayInputStream</code> constructors.

Can you give any more details of what you are doing with the <code>ByteArrayInputStream</code>? Its likely there are better ways around what you are trying to achieve.

Edit: 
If you are using Apache FTPClient to upload, you just need an <code>InputStream</code>. You can do this;

<pre><code>String remote = "whatever";
InputStream is = new FileInputStream(new File("your file"));
ftpClient.storeFile(remote, is);
</code></pre>

You should of course remember to close the input stream once you have finished with it.

blocks|key|2874709|text|下面这段代码很方便：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2874710|private+static+byte[]+readContentIntoByteArray(File+file)
{
++FileInputStream+fileInputStream+=+null;
++byte[]+bFile+=+new+byte[(int)+file.length()];
++try
++{
+++++//convert+file+into+array+of+bytes
+++++fileInputStream+=+new+FileInputStream(file);
+++++fileInputStream.read(bFile);
+++++fileInputStream.close();
++}
++catch+(Exception+e)
++{
+++++e.printStackTrace();
++}
++return+bFile;
}|code-block|syntax|javascript|2874711|参考：http://howtodoinjava.com/2014/11/04/how-to-read-file-content-into-byte-array-in-java/|offset|length|2874712|entityMap|0|LINK|mutability|MUTABLE|url|http://howtodoinjava.com/2014/11/04/how-to-read-file-content-into-byte-array-in-java/^0|0|0|3|2D|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

This piece of code comes handy:

<pre><code>private static byte[] readContentIntoByteArray(File file)
{
 FileInputStream fileInputStream = null;
 byte[] bFile = new byte[(int) file.length()];
 try
 {
 //convert file into array of bytes
 fileInputStream = new FileInputStream(file);
 fileInputStream.read(bFile);
 fileInputStream.close();
 }
 catch (Exception e)
 {
 e.printStackTrace();
 }
 return bFile;
}
</code></pre>

Reference: <a href="http://howtodoinjava.com/2014/11/04/how-to-read-file-content-into-byte-array-in-java/" rel="nofollow">http://howtodoinjava.com/2014/11/04/how-to-read-file-content-into-byte-array-in-java/</a>

I have a file that can be any thing like ZIP, RAR, txt, CSV, doc etc. I would like to create a ByteArrayInputStream from it. 
I'm using it to upload a file to FTP through FTPClient from Apache Commons Net.

Does anybody know how to do it?

For example:

<pre><code>String data = "hdfhdfhdfhd";
ByteArrayInputStream in = new ByteArrayInputStream(data.getBytes());
</code></pre>

My code:

<pre><code>public static ByteArrayInputStream retrieveByteArrayInputStream(File file) {
 ByteArrayInputStream in;

 return in; 
}
</code></pre>

How to create ByteArrayInputStream from a file in Java?

Java

我有一个文件，可以是任何东西像ZIP，RAR，txt，CSV，文件等。我想创建一个ByteArrayInputStream从它。我正在使用它通过FTPClient从Apache Commons Net上传一个文件到FTP。有人知道怎么做吗？例如：String data = "hdfhdfhdfhd";ByteArray...

问如何在Java中从文件创建ByteArrayInputStream？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在Java中从文件创建ByteArrayInputStream？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在Java中从文件创建ByteArrayInputStream？
EN