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2014/10/16 ¤W¤È 08:12:05

android Programming Glossary: data.xml

How to store large blobs in an android content provider?

http://stackoverflow.com/questions/3883211/how-to-store-large-blobs-in-an-android-content-provider

cache root.mkdirs File path new File root uri.getEncodedPath So if the uri was content com.example.myapp some data.xml we'll end up accessing Android data com.example.myapp cache some data.xml int imode 0 if mode.contains w imode ParcelFileDescriptor.MODE_WRITE_ONLY.. the uri was content com.example.myapp some data.xml we'll end up accessing Android data com.example.myapp cache some data.xml int imode 0 if mode.contains w imode ParcelFileDescriptor.MODE_WRITE_ONLY if path.exists try path.createNewFile catch..

Open XML file from res/xml in Android

http://stackoverflow.com/questions/4329308/open-xml-file-from-res-xml-in-android

canine mammal color blue color animal ... AnimalTree And I can open it using File fXmlFile getResources.getXml res xml data.xml DocumentBuilderFactory dbFactory DocumentBuilderFactory.newInstance DocumentBuilder dBuilder dbFactory.newDocumentBuilder.. getTagValue mammal eElement ... Plain and simple Now only in Android I cannot simply read the file res xml data.xml because File requires a String not an integer id . This is where I am lost. Is there some way I can make File open the file..

R.raw.anything cannot be resolved [duplicate]

http://stackoverflow.com/questions/5884268/r-raw-anything-cannot-be-resolved

resolved Android error 78 answers I'm developing an android apps with Eclipse. In my app I try to read a file data.xml. I put it in res raw and to access it i'm supposed to use getRessources .openRawResource R.raw.data But Eclipse show me..

Check if a file exists before calling openFileInput

http://stackoverflow.com/questions/8867334/check-if-a-file-exists-before-calling-openfileinput

FILE_NAME if file.exists ... Hope this helps you out EDIT Also here is another way. tring fileUrl appname data.xml String file android.os.Environment.getExternalStorageDirectory .getPath fileUrl File f new File file if f.exists return..