blocks|key|649863|text|要复制文件并将其保存到目标路径，您可以使用以下方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|649864|public+static+void+copy(File+src,+File+dst)+throws+IOException+{
++++InputStream+in+=+new+FileInputStream(src);
++++try+{
++++++++OutputStream+out+=+new+FileOutputStream(dst);
++++++++try+{
++++++++++++//+Transfer+bytes+from+in+to+out
++++++++++++byte[]+buf+=+new+byte[1024];
++++++++++++int+len;
++++++++++++while+((len+=+in.read(buf))+>+0)+{
++++++++++++++++out.write(buf,+0,+len);
++++++++++++}
++++++++}+finally+{
++++++++++++out.close();
++++++++}
++++}+finally+{
++++++++in.close();
++++}
}|code-block|syntax|javascript|649865|在19%2B接口上，您可以使用Java自动资源管理：|649866|public+static+void+copy(File+src,+File+dst)+throws+IOException+{
++++try+(InputStream+in+=+new+FileInputStream(src))+{
++++++++try+(OutputStream+out+=+new+FileOutputStream(dst))+{
++++++++++++//+Transfer+bytes+from+in+to+out
++++++++++++byte[]+buf+=+new+byte[1024];
++++++++++++int+len;
++++++++++++while+((len+=+in.read(buf))+>+0)+{
++++++++++++++++out.write(buf,+0,+len);
++++++++++++}
++++++++}
++++}
}|649867|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

To copy a file and save it to your destination path you can use the method below.

<pre><code>public static void copy(File src, File dst) throws IOException {
 InputStream in = new FileInputStream(src);
 try {
 OutputStream out = new FileOutputStream(dst);
 try {
 // Transfer bytes from in to out
 byte[] buf = new byte[1024];
 int len;
 while ((len = in.read(buf)) &gt; 0) {
 out.write(buf, 0, len);
 }
 } finally {
 out.close();
 }
 } finally {
 in.close();
 }
}
</code></pre>

On API 19+ you can use Java Automatic Resource Management:

<pre><code>public static void copy(File src, File dst) throws IOException {
 try (InputStream in = new FileInputStream(src)) {
 try (OutputStream out = new FileOutputStream(dst)) {
 // Transfer bytes from in to out
 byte[] buf = new byte[1024];
 int len;
 while ((len = in.read(buf)) &gt; 0) {
 out.write(buf, 0, len);
 }
 }
 }
}
</code></pre>

blocks|key|435618|text|它的Kotlin扩展|type|unstyled|depth|inlineStyleRanges|entityRanges|data|435619|fun+File.copyTo(file:+File)+{
++++inputStream().use+{+input+->
++++++++file.outputStream().use+{+output+->
++++++++++++input.copyTo(output)
++++++++}
++++}
}|code-block|syntax|javascript|435620|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|$]]

Kotlin extension for it

<pre><code>fun File.copyTo(file: File) {
 inputStream().use { input -&gt;
 file.outputStream().use { output -&gt;
 input.copyTo(output)
 }
 }
}
</code></pre>

blocks|key|650069|text|这在Android+O+(API+26)上很简单，如您所见：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|650070|++@RequiresApi(api+=+Build.VERSION_CODES.O)
++public+static+void+copy(File+origin,+File+dest)+throws+IOException+{
++++Files.copy(origin.toPath(),+dest.toPath());
++}|code-block|syntax|javascript|650071|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 is simple on Android O (API 26), As you see:


<pre><code> @RequiresApi(api = Build.VERSION_CODES.O)
 public static void copy(File origin, File dest) throws IOException {
 Files.copy(origin.toPath(), dest.toPath());
 }
</code></pre>

blocks|key|649936|text|这些对我来说很好用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|649937|public+static+void+copyFileOrDirectory(String+srcDir,+String+dstDir)+{

++++try+{
++++++++File+src+=+new+File(srcDir);
++++++++File+dst+=+new+File(dstDir,+src.getName());

++++++++if+(src.isDirectory())+{

++++++++++++String+files[]+=+src.list();
++++++++++++int+filesLength+=+files.length;
++++++++++++for+(int+i+=+0;+i+<+filesLength;+i%2B%2B)+{
++++++++++++++++String+src1+=+(new+File(src,+files[i]).getPath());
++++++++++++++++String+dst1+=+dst.getPath();
++++++++++++++++copyFileOrDirectory(src1,+dst1);

++++++++++++}
++++++++}+else+{
++++++++++++copyFile(src,+dst);
++++++++}
++++}+catch+(Exception+e)+{
++++++++e.printStackTrace();
++++}
}

public+static+void+copyFile(File+sourceFile,+File+destFile)+throws+IOException+{
++++if+(!destFile.getParentFile().exists())
++++++++destFile.getParentFile().mkdirs();

++++if+(!destFile.exists())+{
++++++++destFile.createNewFile();
++++}

++++FileChannel+source+=+null;
++++FileChannel+destination+=+null;

++++try+{
++++++++source+=+new+FileInputStream(sourceFile).getChannel();
++++++++destination+=+new+FileOutputStream(destFile).getChannel();
++++++++destination.transferFrom(source,+0,+source.size());
++++}+finally+{
++++++++if+(source+!=+null)+{
++++++++++++source.close();
++++++++}
++++++++if+(destination+!=+null)+{
++++++++++++destination.close();
++++++++}
++++}
}|code-block|syntax|javascript|649938|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|$]]

These worked nice for me

<pre><code>public static void copyFileOrDirectory(String srcDir, String dstDir) {

 try {
 File src = new File(srcDir);
 File dst = new File(dstDir, src.getName());

 if (src.isDirectory()) {

 String files[] = src.list();
 int filesLength = files.length;
 for (int i = 0; i &lt; filesLength; i++) {
 String src1 = (new File(src, files[i]).getPath());
 String dst1 = dst.getPath();
 copyFileOrDirectory(src1, dst1);

 }
 } else {
 copyFile(src, dst);
 }
 } catch (Exception e) {
 e.printStackTrace();
 }
}

public static void copyFile(File sourceFile, File destFile) throws IOException {
 if (!destFile.getParentFile().exists())
 destFile.getParentFile().mkdirs();

 if (!destFile.exists()) {
 destFile.createNewFile();
 }

 FileChannel source = null;
 FileChannel destination = null;

 try {
 source = new FileInputStream(sourceFile).getChannel();
 destination = new FileOutputStream(destFile).getChannel();
 destination.transferFrom(source, 0, source.size());
 } finally {
 if (source != null) {
 source.close();
 }
 if (destination != null) {
 destination.close();
 }
 }
}
</code></pre>

blocks|key|650120|text|现在使用Kotlin要简单得多：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|650121|+File("originalFileDir",+"originalFile.name")
++++++++++++.copyTo(File("newFileDir",+"newFile.name"),+true)|code-block|syntax|javascript|650122|true或false用于覆盖目标文件|offset|length|style|CODE|650123|https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.io/java.io.-file/copy-to.html|650124|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|4|5|5|0|0|29|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|X|8|@$I|Y|J|Z|K|L]|$I|10|J|11|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|12|8|@]|9|@$I|13|J|14|1|15]]|A|$]]|$1|O|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|N]]]]

Much simpler now with Kotlin:

<pre><code> File("originalFileDir", "originalFile.name")
 .copyTo(File("newFileDir", "newFile.name"), true)
</code></pre>

<code>true</code>or<code>false</code> is for overwriting the destination file

<a href="https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.io/java.io.-file/copy-to.html" rel="noreferrer">https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.io/java.io.-file/copy-to.html</a>

blocks|key|649985|text|现在回答可能太晚了，但最方便的方法是使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|649986|FileUtils的|offset|length|style|CODE|649987|static+void+copyFile(File+srcFile,+File+destFile)|649988|这就是我所做的|649989|`|649990|private+String+copy(String+original,+int+copyNumber){
++++String+copy_path+=+path+%2B+"_copy"+%2B+copyNumber;
++++++++try+{
++++++++++++FileUtils.copyFile(new+File(path),+new+File(copy_path));
++++++++++++return+copy_path;
++++++++}+catch+(IOException+e)+{
++++++++++++e.printStackTrace();
++++++++}
++++++++return+null;
++++}|code-block|syntax|javascript|649991|649992|entityMap^0|0|0|9|0|0|1D|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|12|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|13|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|14|8|@]|9|@]|A|$Q|R]]|$1|S|3|M|5|6|7|15|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|U|$]]

It might be too late for an answer but the most convenient way is using 

<code>FileUtils</code>'s

<code>static void copyFile(File srcFile, File destFile)</code>

e.g. this is what I did

`

<pre><code>private String copy(String original, int copyNumber){
 String copy_path = path + "_copy" + copyNumber;
 try {
 FileUtils.copyFile(new File(path), new File(copy_path));
 return copy_path;
 } catch (IOException e) {
 e.printStackTrace();
 }
 return null;
 }
</code></pre>

`

blocks|key|650167|text|在kotlin中，只需：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|650168|val+fileSrc+:+File+=+File("srcPath")
val+fileDest+:+File+=+File("destPath")

fileSrc.copyTo(fileDest)|code-block|syntax|javascript|650169|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|$]]

in kotlin , just :

<pre><code>val fileSrc : File = File("srcPath")
val fileDest : File = File("destPath")

fileSrc.copyTo(fileDest)
</code></pre>

blocks|key|435477|text|这是一个解决方案，如果在复制过程中发生错误，则实际关闭输入/输出流。这个解决方案利用apache+Commons+IO+IOUtils方法来复制和处理流的关闭。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|435478|++++public+void+copyFile(File+src,+File+dst)++{
++++++++InputStream+in+=+null;
++++++++OutputStream+out+=+null;
++++++++try+{
++++++++++++in+=+new+FileInputStream(src);
++++++++++++out+=+new+FileOutputStream(dst);
++++++++++++IOUtils.copy(in,+out);
++++++++}+catch+(IOException+ioe)+{
++++++++++++Log.e(LOGTAG,+"IOException+occurred.",+ioe);
++++++++}+finally+{
++++++++++++IOUtils.closeQuietly(out);
++++++++++++IOUtils.closeQuietly(in);
++++++++}
++++}|code-block|syntax|javascript|435479|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|$]]

Here is a solution that actually closes the input/output streams if an error occurs while copying. This solution utilizes apache Commons IO IOUtils methods for both copying and handling the closing of streams.

<pre><code> public void copyFile(File src, File dst) {
 InputStream in = null;
 OutputStream out = null;
 try {
 in = new FileInputStream(src);
 out = new FileOutputStream(dst);
 IOUtils.copy(in, out);
 } catch (IOException ioe) {
 Log.e(LOGTAG, "IOException occurred.", ioe);
 } finally {
 IOUtils.closeQuietly(out);
 IOUtils.closeQuietly(in);
 }
 }
</code></pre>

blocks|key|435819|text|现在在kotlin你可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|435820|file1.copyTo(file2)|code-block|syntax|javascript|435821|其中，file1是原始文件的对象，file2是要复制到的新文件的对象|435822|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

now in kotlin you could just use
<pre><code>file1.copyTo(file2)
</code></pre>
where file1 is an object of the original file and file2 is an object of the new file you want to copy to

In my app I want to save a copy of a certain file with a different name (which I get from user)

Do I really need to open the contents of the file and write it to another file?

What is the best way to do so?

How to make a copy of a file in android?

Android

Kotlin

Java

在我的应用程序中，我想用不同的名称保存某个文件的副本(这是我从用户处获得的)我真的需要打开文件的内容并将其写入另一个文件吗？最好的方法是什么？

问如何在android中制作文件的副本？
EN

回答 9

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问如何在android中制作文件的副本？EN

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问如何在android中制作文件的副本？
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