I am trying to get input stream from something like this.
我试图从这样的东西获取输入流。
InputSource myInputSource = new InputSource(activity.getResources().openRawResource(com.MYCLass.R.xml.programs));
myXMLReader.parse(myInputSource);
and then call parse on the parser instance i Created. Some how i get nothing.
Works fine if I use a server XML....
然后在我创建的解析器实例上调用解析。有些我什么都没得到。如果我使用服务器XML,工作正常....
2 个解决方案
#1
11
Put the xml file into /res/raw folder
. It looks like openRawResource opens resources from that folder only. You can also try getResources().getXml(com.MYCLass.R.xml.programs);
which will return you an instance of XML parser.
将xml文件放入/ res / raw文件夹。看起来openRawResource只打开该文件夹中的资源。您也可以尝试getResources()。getXml(com.MYCLass.R.xml.programs);这将返回一个XML解析器的实例。
A piece of code taken from @Krupa
一段代码取自@Krupa
InputStream object = this.getResources()
.openRawResource(R.raw.fileName);
#2
4
You can read the file by the following code :
您可以通过以下代码读取该文件:
InputStream object = this.getResources()
.openRawResource(R.rawFolderName.fileName);
#1
11
Put the xml file into /res/raw folder
. It looks like openRawResource opens resources from that folder only. You can also try getResources().getXml(com.MYCLass.R.xml.programs);
which will return you an instance of XML parser.
将xml文件放入/ res / raw文件夹。看起来openRawResource只打开该文件夹中的资源。您也可以尝试getResources()。getXml(com.MYCLass.R.xml.programs);这将返回一个XML解析器的实例。
A piece of code taken from @Krupa
一段代码取自@Krupa
InputStream object = this.getResources()
.openRawResource(R.raw.fileName);
#2
4
You can read the file by the following code :
您可以通过以下代码读取该文件:
InputStream object = this.getResources()
.openRawResource(R.rawFolderName.fileName);