blocks|key|1020587|text|使用openRawResource()读取字节流|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1020588|像这样的事情应该是可行的|1020589|InputStream+is+=+context.getResources().openRawResource(R.raw.urfilename);|code-block|syntax|javascript|1020590|检查此链接|1020591|http://developer.android.com/guide/topics/resources/accessing-resources.html#ResourcesFromCode|offset|length|1020592|它清楚地说明了以下内容|1020593|虽然不常见，但您可能需要访问原始文件和目录。如果这样做，那么将文件保存在res/中将不会对您起作用，因为从res/读取资源的惟一方法是使用资源ID|style|BOLD|1020594|如果你想给出一个类似于你的代码中提到的文件名，你可能需要把它保存在assets文件夹中。|1020595|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|0|2M|0|0|0|0|21|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|14|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|15|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|16|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|17|8|@]|9|@$M|18|N|19|1|1A]]|A|$]]|$1|O|3|P|5|6|7|1B|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1C|8|@$M|1D|N|1E|S|T]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1F|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|L]]]]

read a stream of bytes using openRawResource()

some thing like this should work

<pre><code>InputStream is = context.getResources().openRawResource(R.raw.urfilename);
</code></pre>

Check this link

<a href="http://developer.android.com/guide/topics/resources/accessing-resources.html#ResourcesFromCode" rel="nofollow noreferrer">http://developer.android.com/guide/topics/resources/accessing-resources.html#ResourcesFromCode</a>

It clearly says the following

While uncommon, you might need access your original files and directories. If you do, then saving your files in res/ won't work for you, because the only way to read a resource from res/ is with the resource ID

If you want to give a file name like the one mentioned in ur code probably you need to save it on assets folder.

blocks|key|1764204|text|您也许能够对原始文件使用Resources.getIdentifier(name,+type,+package)。这将为您获取id，然后您可以继续使用setImageResource(id)或其他任何东西。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1764205|int+id+=+getResources().getIdentifier("3150-MCM",+"raw",+getPackageName());
if+(id+!=+0)+//if+it's+zero+then+its+not+valid
+++image.setImageResource(id);|code-block|syntax|javascript|1764206|是你想要的吗？它可能不喜欢多个文件夹，但值得一试。|1764207|entityMap^0|C|18|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You might be able to use <code>Resources.getIdentifier(name, type, package)</code> with raw files. This'll get the id for you and then you can just continue with setImageResource(id) or whatever.

<pre><code>int id = getResources().getIdentifier("3150-MCM", "raw", getPackageName());
if (id != 0) //if it's zero then its not valid
 image.setImageResource(id);
</code></pre>

is what you want? It might not like the multiple folders though, but worth a try.

blocks|key|3003598|text|尝试{+//获取对AssetManager的引用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3003599|+++++++++AssetManager+mngr+=+getAssets();|code-block|syntax|javascript|3003600|++++++++//+Create+an+input+stream+to+read+from+the+asset+folder
++++++++InputStream+ins+=+mngr.open(imdir);

++++++++//+Convert+the+input+stream+into+a+bitmap
++++++++img+=+BitmapFactory.decodeStream(ins);

++}+catch+(final+IOException+e)+{
++++++++e.printStackTrace();
++}+|3003601|其中图像目录是资产的路径|3003602|喜欢|3003603|assest+->+image+->+somefolder+->+some.jpg+(评估某个文件夹映像文件)|3003604|则路径将为|3003605|image/某文件夹/某.jpg|3003606|现在不需要图像的资源id，您可以在运行时使用以下内容填充图像|3003607|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

try {
 // Get reference to AssetManager 
 AssetManager mngr = getAssets();

<pre><code> // Create an input stream to read from the asset folder
 InputStream ins = mngr.open(imdir);

 // Convert the input stream into a bitmap
 img = BitmapFactory.decodeStream(ins);

 } catch (final IOException e) {
 e.printStackTrace();
 } 
</code></pre>

here image directory is path of assets 

like

assest -> image -> somefolder -> some.jpg

then path will be

image/somefolder/some.jpg

now no need of resource id for image , you can populate image on runtime using this

i want to show image in imageview without using id.

i will place all images in raw folder and open 

<pre><code> try {
 String ss = "res/raw/images/inrax/3150-MCM.jpg";
 in = new FileInputStream(ss);
 buf = new BufferedInputStream(in);
 Bitmap bMap = BitmapFactory.decodeStream(buf);
 image.setImageBitmap(bMap);
 if (in != null) {
 in.close();
 }
 if (buf != null) {
 buf.close();
 }
 } catch (Exception e) {
 Log.e("Error reading file", e.toString());
 }
</code></pre>

but this is not working i want to access image using its path not by name

Android Show image by path

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想不使用id在imageview中显示图像。我会把所有的图片放在raw文件夹中并打开     try {            String ss = "res/raw/images/inrax/3150-MCM.jpg";             in = new FileInputStream(ss);     ...

问Android按路径显示图片
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