myFile.renameTo(new File(“/the/new/place/newName.file”)); File#renameTo does that (it can not only rename, but also move between directories, at least on the same file system). Renames the file denoted by this abstract pathname. Many aspects of the behavior of this method are inherently platform-dependent: The rename operation might not be able to move a file from one filesystem … Read more
using System.IO; … foreach (string file in Directory.EnumerateFiles(folderPath, “*.xml”)) { string contents = File.ReadAllText(file); } Note the above uses a .NET 4.0 feature; in previous versions replace EnumerateFiles with GetFiles). Also, replace File.ReadAllText with your preferred way of reading xml files – perhaps XDocument, XmlDocument or an XmlReader.
Usually when a method accepts a file, there’s another method nearby that accepts a stream. If this isn’t the case, the API is badly coded. Otherwise, you can use temporary files, where permission is usually granted in many cases. If it’s applet, you can request write permission. An example: try { // Create temp file. … Read more
You do it the same way as you would a directory, but you specify the full path to the filename as the source. In your example: rsync -avz –progress /var/www/public_html/.htaccess root@<remote-ip>:/var/www/public_html/ As mentioned in the comments: since -a includes recurse, one little typo can make it kick off a full directory tree transfer, so a … Read more
You can use os.listdir() to get the files in the source directory, os.path.isfile() to see if they are regular files (including symbolic links on *nix systems), and shutil.copy to do the copying. The following code copies only the regular files from the source directory into the destination directory (I’m assuming you don’t want any sub-directories … Read more
The issue is that the file isn’t being copied to your app bundle. To fix it: Click your project Click your target Select Build Phases Expand Copy Bundle Resources Click ‘+’ and select your file.
You can call source(“abc.R”) followed by source(“xyz.R”) (assuming that both these files are in your current working directory. If abc.R is: fooABC <- function(x) { k <- x+1 return(k) } and xyz.R is: fooXYZ <- function(x) { k <- fooABC(x)+1 return(k) } then this will work: > source(“abc.R”) > source(“xyz.R”) > fooXYZ(3) [1] 5 > … Read more
For Python 2.x, use the StringIO module. For example: >>> from cStringIO import StringIO >>> f = StringIO(‘foo’) >>> f.read() ‘foo’ I use cStringIO (which is faster), but note that it doesn’t accept Unicode strings that cannot be encoded as plain ASCII strings. (You can switch to StringIO by changing “from cStringIO” to “from StringIO”.) … Read more
Yet another method based on file(1) behavior: >>> textchars = bytearray({7,8,9,10,12,13,27} | set(range(0x20, 0x100)) – {0x7f}) >>> is_binary_string = lambda bytes: bool(bytes.translate(None, textchars)) Example: >>> is_binary_string(open(‘/usr/bin/python’, ‘rb’).read(1024)) True >>> is_binary_string(open(‘/usr/bin/dh_python3’, ‘rb’).read(1024)) False
Have you tried mode ‘a+’? with open(filename, ‘a+’) as f: f.write(…) Note however that f.tell() will return 0 in Python 2.x. See https://bugs.python.org/issue22651 for details.