blocks|key|269202|text|此方法在应用程序的私有数据区域中打开一个文件。您不能使用此方法打开此区域的子目录中的任何文件，也不能从整个其他区域打开任何文件。因此，直接使用FileInputStream的构造函数来传递其中包含目录的路径。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|269203|entityMap^0|1Z|F|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

This method opens a file in the private data area of the application. You cannot open any files in subdirectories in this area or from entirely other areas using this method. So use the constructor of the <code>FileInputStream</code> directly to pass the path with a directory in it.

blocks|key|2967573|text|解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2967574|FileInputStream+fis+=+new+FileInputStream+(new+File(NAME_OF_FILE));++//+2nd+line|code-block|syntax|javascript|2967575|openFileInput方法不接受路径分隔符。|2967576|别忘了|2967577|fis.close();|2967578|在最后。|2967579|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The solution is:

<pre><code>FileInputStream fis = new FileInputStream (new File(NAME_OF_FILE)); // 2nd line
</code></pre>

The openFileInput method doesn't accept path separators.

Don't forget to

<pre><code>fis.close();
</code></pre>

at the end.

blocks|key|272713|text|openFileInput()不接受路径，只接受文件名如果您想访问路径，请使用File+file+=+new+File(path)和相应的FileInputStream|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|272714|entityMap^0|0|F|13|Q|1X|F|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

<code>openFileInput()</code> doesn't accept paths, only a file name
if you want to access a path, use <code>File file = new File(path)</code> and corresponding <code>FileInputStream</code>

blocks|key|269261|text|在对子目录Dir使用openFileInput("/Dir/data.txt")方法从内部存储器访问文件时，我收到了上面的错误消息。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|269262|您无法使用上述方式访问子目录。|269263|尝试如下所示：|269264|FileInputStream+fIS+=+new+FileInputStream+(new+File("/Dir/data.txt"));|code-block|syntax|javascript|269265|entityMap^0|5|3|A|U|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

I got the above error message while trying to access a file from Internal Storage using <code>openFileInput("/Dir/data.txt")</code> method with subdirectory <code>Dir</code>. 

You cannot access sub-directories using the above method.

Try something like:

<pre><code>FileInputStream fIS = new FileInputStream (new File("/Dir/data.txt"));
</code></pre>

blocks|key|269208|text|您不能直接使用带有目录分隔符的路径，但您必须为每个目录创建一个文件对象。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|269209|注意:此代码创建目录，您可能不需要该目录...|269210|File+file=+context.getFilesDir();
file.mkdir();

String[]+array=filePath.split("/");
for(int+t=0;+t<+array.length+-1+;t%2B%2B)
{
++++file=new+File(file,array[t]);
++++file.mkdir();
}

File+f=new+File(file,array[array.length-1]);

RandomAccessFileOutputStream+rvalue+=+new+RandomAccessFileOutputStream(f,append);|code-block|syntax|javascript|269211|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You cannot use path with directory separators directly, but you will
have to make a file object for every directory.

NOTE: This code makes directories, yours may not need that...

<pre><code>File file= context.getFilesDir();
file.mkdir();

String[] array=filePath.split("/");
for(int t=0; t&lt; array.length -1 ;t++)
{
 file=new File(file,array[t]);
 file.mkdir();
}

File f=new File(file,array[array.length-1]);

RandomAccessFileOutputStream rvalue = new RandomAccessFileOutputStream(f,append);
</code></pre>

blocks|key|269267|text|String+all+=+"";
++++++++try+{
++++++++++++BufferedReader+br+=+new+BufferedReader(new+FileReader(filePath));
++++++++++++String+strLine;
++++++++++++while+((strLine+=+br.readLine())+!=+null){
++++++++++++++++all+=+all+%2B+strLine;
++++++++++++}
++++++++}+catch+(IOException+e)+{
++++++++++++Log.e("notes_err",+e.getLocalizedMessage());
++++++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|269268|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>String all = "";
 try {
 BufferedReader br = new BufferedReader(new FileReader(filePath));
 String strLine;
 while ((strLine = br.readLine()) != null){
 all = all + strLine;
 }
 } catch (IOException e) {
 Log.e("notes_err", e.getLocalizedMessage());
 }
</code></pre>

blocks|key|2967581|text|File+file+=+context.getFilesDir();+
file.mkdir();
String[]+array+=+filePath.split("/");+
for(int+t+=+0;+t+<+array.length+-+1;+t%2B%2B)+{
++++file+=+new+File(file,+array[t]);+
++++file.mkdir();
}
File+f+=+new+File(file,array[array.length-+1]);+
RandomAccessFileOutputStream+rvalue+=+
++++new+RandomAccessFileOutputStream(f,+append);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2967582|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>File file = context.getFilesDir(); 
file.mkdir();
String[] array = filePath.split("/"); 
for(int t = 0; t &lt; array.length - 1; t++) {
 file = new File(file, array[t]); 
 file.mkdir();
}
File f = new File(file,array[array.length- 1]); 
RandomAccessFileOutputStream rvalue = 
 new RandomAccessFileOutputStream(f, append);
</code></pre>

blocks|key|2967612|text|我解决了这种类型的错误，在onCreate事件中创建了一个目录，然后通过在方法中创建一个新的文件对象来访问该目录，该方法需要执行某些操作，如在该目录中保存或检索文件，希望这会有所帮助！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2967613|+public+class+MyClass+{++++

+private+String+state;
+public+File+myFilename;

+@Override
+protected+void+onCreate(Bundle+savedInstanceState)+{//create+your+directory+the+user+will+be+able+to+find
++++super.onCreate(savedInstanceState);
++++if+(Environment.MEDIA_MOUNTED.equals(state))+{
++++++++myFilename+=+new+File(Environment.getExternalStorageDirectory().toString()+%2B+"/My+Directory");
++++++++if+(!myFilename.exists())+{
++++++++++++myFilename.mkdirs();
++++++++}
++++}
+}

+public+void+myMethod+{

+File+fileTo+=+new+File(myFilename.toString()+%2B+"/myPic.png");++
+//+use+fileTo+object+to+save+your+file+in+your+new+directory+that+was+created+in+the+onCreate+method
+}
}|code-block|syntax|javascript|2967614|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|$]]

I solved this type of error by making a directory in the onCreate event, then accessing the directory by creating a new file object in a method that needs to do something such as save or retrieve a file in that directory, hope this helps!

<pre><code> public class MyClass { 

 private String state;
 public File myFilename;

 @Override
 protected void onCreate(Bundle savedInstanceState) {//create your directory the user will be able to find
 super.onCreate(savedInstanceState);
 if (Environment.MEDIA_MOUNTED.equals(state)) {
 myFilename = new File(Environment.getExternalStorageDirectory().toString() + "/My Directory");
 if (!myFilename.exists()) {
 myFilename.mkdirs();
 }
 }
 }

 public void myMethod {

 File fileTo = new File(myFilename.toString() + "/myPic.png"); 
 // use fileTo object to save your file in your new directory that was created in the onCreate method
 }
}
</code></pre>

I have a filename in my code as :

<pre><code>String NAME_OF_FILE="//sdcard//imageq.png";
FileInputStream fis =this.openFileInput(NAME_OF_FILE); // 2nd line
</code></pre>

I get an error on 2nd line :

<blockquote>
 05-11 16:49:06.355: ERROR/AndroidRuntime(4570): Caused by: java.lang.IllegalArgumentException: File //sdcard//imageq.png contains a path separator
</blockquote>

I tried this format also:

<pre><code>String NAME_OF_FILE="/sdcard/imageq.png";
</code></pre>

java.lang.IllegalArgumentException: contains a path separator

我的代码中有一个文件名，如下所示：String NAME_OF_FILE="//sdcard//imageq.png";FileInputStream fis =this.openFileInput(NAME_OF_FILE); // 2nd line我在第二行得到一个错误：镜像05-11 16:49:06.355: ERROR/AndroidRuntime(4570)：由: java.lang.

问java.lang.IllegalArgumentException:包含路径分隔符
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