I have a a Shell script that contain a Perl script and R script.
my Shell script R.sh:-
#!/bin/bash
./R.pl #calling Perl script
`perl -lane 'print $F[0]' /media/data/abc.cnv > /media/data/abc1.txt`;
#Shell script
Rscript R.r #calling R script
This is my R.pl (head):-
`export path=$PATH:/media/exe_folder/bin`;
print "Enter the path to your input file:";
$base_dir ="/media/exe_folder";
chomp($CEL_dir = <STDIN>);
opendir (DIR, "$CEL_dir") or die "Couldn't open directory $CEL_dir";
$cel_files = "$CEL_dir"."/cel_files.txt";
open(CEL,">$cel_files")|| die "cannot open $file to write";
print CEL "cel_files
";
for ( grep { /^[wd]/ } readdir DIR ){
print CEL "$CEL_dir"."/$_
";
}close (CEL);
The output of Perl script is input for Shell script and Shell's output is input for R script.
I want to run the Shell script by providing the input file name and output file name like :-
./R.sh home/folder/inputfile.txt home/folder2/output.txt
If folder contain many files then it will take only user define file and process it.
Is There is a way to do this?
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