I had name file in JAVA I/O:
f1 = new File("C:\\Program Files\directory\FileDir\file.txt");
I want to recover the filename only: FileDir(type String).
.getName() should help you.
new File("C:\Program Files\directory\FileDir\file.txt").getName();
There are other methods for File as well. Please go through the documentation here to find out how to get absolute path, canonical path, etc. and what are the differences between them.
Related
Oracle's sqlldr defaults to a .dat extension. That I want to override. I don't like to rename the file. When googled get to know few answers to use . like data='fileName.' which is not working. Share your ideas, please.
Error message is fileName.dat is not found.
Sqlloder has default extension for all input files data,log,control...
data= .dat
log= .log
control = .ctl
bad =.bad
PARFILE = .par
But you have to pass filename without apostrophe and dot
sqlloder pass/user#db control=control data=data
sqloader will add extension. control.ctl data.dat
Nevertheless i do not understand why you do not want to specify extension?
You can't, at least in Unix/Linux environments. In Windows you can use the trailing period trick, specifying either INFILE 'filename.' in the control file or DATA=filename. on the command line. WIndows file name handling allows that; you can for instance do DIR filename. at a command prompt and it will list the file with no extension (as will DIR filename). But you can't do that with *nix, from a shell prompt or anywhere else.
You said you don't want to copy or rename the file. Temporarily renaming it might be the simplest solution, but as you may have a reason not to do that even briefly you could instead create a hard or soft link to the file which does have an extension, and use that link as the target instead. You could wrap that in a shell script that takes the file name argument:
# set variable from correct positional parameter; if you pass in the control
# file name or other options, this might not be $1 so adjust as needed
# if the tmeproary file won't be int he same directory, need to be full path
filename=$1
# optionally check file exists, is readable, etc. but overkill for demo
# can also check temporary file does not already exist - stop or remove
# create soft link somewhere it won't impact any other processes
ln -s ${filename} /tmp/${filename##*/}.dat
# run SQL*Loader with soft link as target
sqlldr user/password#db control=file.ctl data=/tmp/${filename##*/}.dat
# clean up
rm -f /tmp/${filename##*/}.dat
You can then call that as:
./scriptfile.sh /path/to/filename
If you can create the link in the same directory then you only need to pass the file, but if it's somewhere else - which may be necessary depending on why renaming isn't an option, and desirable either way - then you need to pass the full path of the data file so the link works. (If the temporary file will be int he same filesystem you could use a hard link, and you wouldn't have to pass the full path then either, but it's still cleaner to do so).
As you haven't shown your current command line options you may have to adjust that to take into account anything else you currently specify there rather than in the control file, particularly which positional argument is actually the data file path.
I have the same issue. I get a monthly download of reference data used in medical application and the 485 downloaded files don't have file extensions (#2gb). Unless I can load without file extensions I have to copy the files with .dat and load from there.
I am creating files from within Ruby scripts and adding stuff to them. But where are these files stored that I am creating?
I'm very new to this, sorry!
The files are created at whatever location you specified. For instance:
f = File.new("another_test.txt","w+")
that will create the file in the current working directory. You specify the path along with the file name. For example:
f = File.new("~/Desktop/another_test.txt","w+") # will create the file on the desktop.
For more details, check the File documentation.
Updated:
Included mu is too short correction.
How to make from
f6f6c3408e67bf6473d65de172f0e5da.jpg
file name
That structure of folder f6/f6/c3/40/8e/67/bf/64/73/d6/5d/e1/72/f0/e5/da.jpg to escape file system overload
This should do it:
'f6f6c3408e67bf6473d65de172f0e5da.jpg'.sub /(.*)(\..*)/ do
filename, extension = $1, $2
filename.scan(/../).join('/') + extension
end
Why don't you follow what systems like git do and just extract the first two characters to make a directory and put the files in it. Creating so many directories for this seems pointless (and might affect performance too ) and if it works for systems like git to avoid inefficiencies of the file system, it should work for you too. And of course, if you follow this approach, the implementation is going to pretty simple as well.
directory,filename = filename[0..1],filename[2..-1]
On a Linux system I have a binary (bin.exe) which needs to read an input file (input.cfg), where the names of other data files (data.txt) are specified. Usually both binary, input file and data files were in the same directory. Now and for organization reasons I need binary file to be in $SOMEPATH/bin and input and data files in $SOMEPATH/input.
I do not know how to do this. If I try
$SOMEPATH/bin/bin.exe $SOMEPATH/input/input.cfg
I get
error, "data.txt" not found
One solution would be to include absolute of relative path of "data.txt" in input.cfg, but the binary does not accept this.
I thought about somehow fooling the binary so that it thinks it is in $SOMEPATH/input, so that I just do
$SOMEPATH/bin/bin.exe input.cfg
and it works, but I do not know whether this is possible. any hints?
(cd $SOMEPATH/input && $SOMEPATH/bin/bin.exe input.cfg)
This is assuming that the program is relying on the current working directory to find the files. If the program is trying hard to find them in the same location as the executable, by consulting /proc/<pid>/exe for example, then you may be out of luck.
I would like to scan text of textfiles in Matlab with the textscan function. Before I can open the textfile with fid = fopen('C:\path'), I need to unzip the files first. The files have the extension: *.gz
There are thousands of files which I need to analyze and high performance is important.
I have two ideas:
(1) Use an external program an call it from the command line in Matlab
(2) Use a Matlab 'zip'toolbox. I have heard of gunzip, but don't know about its performance.
Does anyone knows a way to unzip these files as quick as possible from within Matlab?
Thanks!
You could always try the Matlab unzip() function:
unzip
Extract contents of zip file
Syntax
unzip(zipfilename)
unzip(zipfilename, outputdir)
unzip(url, ...)
filenames = unzip(...)
Description
unzip(zipfilename) extracts the archived contents of zipfilename into the current folder and sets the files' attributes, preserving the timestamps. It overwrites any existing files with the same names as those in the archive if the existing files' attributes and ownerships permit it. For example, files from rerunning unzip on the same zip filename do not overwrite any of those files that have a read-only attribute; instead, unzip issues a warning for such files.
Internally, this uses Java's zip library org.apache.tools.zip. If your zip archives each contain many text files it might be faster to drop down into Java and extract them entry by entry, without explicitly unzipped files. look at the source of unzip.m to get some ideas, and also the Java documentation.
I've found 7zip-commandline(Windows) / p7zip(Unix) to be somewhat speedier for this.
[edit]From some quick testing, it seems making a system call to gunzip is faster than using MATLAB's native gunzip. You could give that a try as well.
Just write a new function that imitates basic MATLAB gunzip functionality:
function [] = sunzip(fullfilename,output_dir)
if ~exist('output_dir','var'), output_dir = fileparts(fullfilename); end
app_path = '/usr/bin/7za';
switches = ' e'; %extract files ignoring directory structure
options = [' -o' output_dir];
system([app_path switches options '_' fullfilename]);
Then use it as you would use gunzip:
sunzip('/data/time_1000.out.gz',tmp_dir);
With MATLAB's toc timer, I get the following extraction times with 6 uncompressed 114MB ASCII files:
gunzip: 10.15s
sunzip: 7.84s
worked well, just needed a minor change to Max's syntax calling the executable.
system([app_path switches ' ' fullfilename options ]);