my program takes input from the user of a file directory via command line in windows , and then runs diruse on specified path to gather information about disk space available. For some reason the error keeps saying "diruse.exe is not recognized as an internal or external command". Below is a section of my code, to view it in its entirety please go here. I took the code out of its sub to see if it would work, but it did not.
&argument_checking; #calling sub checking user arguements.
#&reading_directory; # reads user given directory.
chdir($user_directory), or die " Directory $user_directory does not exist"; #change to user directory or exit.
open(DIRUSE, "diruse.exe /* . |"); #opening user given directory path.
foreach my $directory_lines(<DIRUSE>)
{
chomp $directory_lines;
push(#directory_lines, $directory_lines); #pushing directory lines from file into an array
}
close(DIRUSE); #closing
shift(#directory_lines);
shift(#directory_lines);
pop(#directory_lines);
&chop_and_save;
&gui_creator;
That message comes from the command shell, and it means it can't find diruse.exe in the current directory or in any directory in that process's PATH.
Related
I want to schedule my Perl code to be run every day at a specific time. so I put the below code in bash file:
Automate.sh
#!/bin/sh
perl /tmp/Taps/perl.pl
The schedule has been specified in below path:
10 17 * * * sh /tmp/Taps/Automate.sh > /tmp/Taps/result.log
When the time arrived to 17:10 the .sh file hasn't been running. however, when I run ./Automate.sh (manually) it is running and I see the result. I don't know what is the problem.
Perl Code
#!/usr/bin/perl -w
use strict;
use warnings;
use Data::Dumper;
use XML::Dumper;
use TAP3::Tap3edit;
$Data::Dumper::Indent=1;
$Data::Dumper::Useqq=1;
my $dump = new XML::Dumper;
use File::Basename;
my $perl='';
my $xml='';
my $tap3 = TAP3::Tap3edit->new();
foreach my $file(glob '/tmp/Taps/X*')
{
$files= basename($file);
$tap3->decode($files) || die $tap3->error;
}
my $filename=$files.".xml\n";
$perl = $tap3->structure;
$dump->pl2xml($perl, $filename);
print "Done \n";
error:
No such file or directory for file X94 at /tmp/Taps/perl.pl line 22.
X94.xml
foreach my $file(glob 'Taps/X*') -- when you're running from cron, your current directory is /. You'll want to provide the full path to that Taps directory. Also specify the output directory for Out.xml
Cron uses a minimal environment and a short $PATH, which may not necessarily include the expected path to perl. Try specifying this path fully. Or source your shell settings before running the script.
There are a lot of things that can go wrong here. The most obvious and certain one is that if you use a glob to find the file in directory "Taps", then remove the directory from the file name by using basename, then Perl cannot find the file. Not quite sure what you are trying to achieve there. The file names from the glob will be for example Taps/Xfoo, a relative path to the working directory. If you try to access Xfoo from the working directory, that file will not be found (or the wrong file will be found).
This should also (probably) lead to a fatal error, which should be reported in your error log. (Assuming that the decode function returns a false value upon error, which is not certain.) If no errors are reported in your error log, that is a sign the program does not run at all. Or it could be that decode does not return false on missing file, and the file is considered to be empty.
I assume that when you test the program, you cd to /tmp and run it, or your "Taps" directory is in your home directory. So you are making assumptions about where your program looks for the files. You should be certain where it looks for files, probably by using only absolute paths.
Another simple error might be that crontab does not have permission to execute the file, or no read access to "Taps".
Edit:
Other complications in your code:
You include Data::Dumper, but never actually use that module.
$xml variable is not used.
$files variable not declared (this code would never run with use strict)
Your $files variable is outside your foreach loop, which means it will only run once. Since you use glob I assumed you were reading more than one file, in which case this solution will probably not do what you want. It is also possible that you are using a glob because the file name can change, e.g. X93, X94, etc. In that case you will read the last file name returned by the glob. But this looks like a weak link in your logic.
You add a newline \n to a file name, which is strange.
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.
In our inbound directory we have the following file structure.
abc.csv.timestamp
abc.csv.timestamp
abc.csv.timestamp
I want to find the most recent file in the directory and MOVE to another directory.
However i want to pass the file pattern abc.csv.* as a string to the command.
Any pointers ?
I have a script written using ActivePerl that creates files where the user specifies. If the directory doesn't already exist, it uses mkpath to attempt to create it and trap any error conditions (such as not having permission to create the directory there). This seems fine. What I'm running into trouble with is determine the permissions of a directory that already exists. I don't want a user to be able to specify a protected folder that they can read from (c:\windows\system32 comes to mind) and the script silently fail attempting to create its files there.
From other perl examples on the web I've tried using the following, but I'm always given 0777 as the result for any existing directory:
use File::stat;
#
#...
#
my $info = stat($candiDirectory);
my $retMode = $info->mode;
my $mymode = sprintf("0%o, $retMode & 07777);
print "retMode for $candiDirectory is $mymode \n";
While this seems reasonable for unix/Linux, I'd be surprised if it didn't require something different under Win32 or 64.
from perldoc perlfunc:
-w $filename
unless (-w $filename) {
say "i can't write this file";
}
I have a NSIS script that is attempting to include a .NSI file that sits in a different folder.
My Problem: When I go to compile my script I get the compile error !include: could not find: "../Utilities.nsi". This file exists and is in the correct location I am specifying(in the parent directory - one step back).
How can I include a file that sits in another directory? I hope its possible.
!include "../Utilities.nsi" # include error: '!include: could not find: "../Utilities.nsi"'
InstallDir "abc"
Name "def"
OutFile "def.exe"
Section
DetailPrint "Hello World"
SectionEnd
The manual says this about !include:
This command will include 'file' as if it was part of the original
script. Note that if a file is included in another directory, the
current directory is still where the script was compiled from (not
where the included file resides). If the compiler can't find the file
it will look for it in every include directory. See !addincludedir for
more information. If the /nonfatal switch is used and no files are
found, a warning will be issued instead of an error."
Also, the examples in the manual do not use quotation marks -- did you try
removing them? Also, "/" => "\".