linux join command for multiple files

ntubski · 03-15-2013, 01:26 PM

Quote:

Originally Posted by danielbmartin

When I enter /home/daniel/Desktop/LQfiles/dbm684.bin the result is bash: /home/daniel/Desktop/LQfiles/dbm684.bin: Permission denied

You need give execute permission for the file (chmod +x /home/daniel/Desktop/LQfiles/dbm684.bin).

Quote:

It's unclear what benefit is obtained by the #!/bin/bash

The benefit is that the person/program executing the file doesn't need to know which interpreter to use because it's encoded in the file itself.

whizje · 03-15-2013, 03:22 PM

Quote:

... so your Path= and mine are not equivalent.

I was on the assumption that you start the program from the directory where the datafiles are.
While you are on the assumption that the data files are in the same directory as the script.
I assumed the OP's shell doesn't support the direction operator and tried to find substitutes for it.
offtopic I just installed Centos for RHCE qualification. What a terrible installer it has if you don't want a standard install.

danielbmartin · 03-15-2013, 03:40 PM

Quote:

Originally Posted by whizje

While you are on the assumption that the data files are in the same directory as the script.

Not an assumption, it is a requirement (unless you change the file definitions in the code).

Daniel B. Martin

whizje · 03-15-2013, 04:08 PM

Quote:

Not an assumption, it is a requirement (unless you change the file definitions in the code).

Ok do you have a alternative for the redirection operator.

danielbmartin · 03-16-2013, 10:10 AM

Quote:

Originally Posted by whizje

Ok do you have a alternative for the redirection operator.

The file identifier for my program is /home/daniel/Desktop/LQfiles/dbm678.bin
The input files, work files, and the output file are all located in the same folder and with similar file names.
All of the file names start with /home/daniel/Desktop/LQfiles/dbm678
This code ...

Code:

Path=$(cut -d'.' -f1 <<< ${0})

... produces this desirable result ...

Code:

/home/daniel/Desktop/LQfiles/dbm678

... but you could hard code it thusly ...

Code:

Path="/home/daniel/Desktop/LQfiles/dbm678"

The output file is /home/daniel/Desktop/LQfiles/dbm678out.txt

The input files are ...

Code:

/home/daniel/Desktop/LQfiles/dbm678inp1.txt
/home/daniel/Desktop/LQfiles/dbm678inp2.txt
/home/daniel/Desktop/LQfiles/dbm678inp3.txt
/home/daniel/Desktop/LQfiles/dbm678inp4.txt
/home/daniel/Desktop/LQfiles/dbm678inp5.txt

... and the corresponding work files are ...

Code:

/home/daniel/Desktop/LQfiles/dbm678w1.txt
/home/daniel/Desktop/LQfiles/dbm678w2.txt
/home/daniel/Desktop/LQfiles/dbm678w3.txt
/home/daniel/Desktop/LQfiles/dbm678w4.txt
/home/daniel/Desktop/LQfiles/dbm678w5.txt

The OP would have to substitute appropriate file identifiers for his computer.
They might be of the form ...

Code:

/home/cheah/What/Ever/dbm678.bin
 ... and ...
/home/cheah/What/Ever/dbm678inp1.txt
 ... and ...
/home/cheah/What/Ever/dbm678w1.txt
 ... and ...
/home/cheah/What/Ever/dbm678out.txt

If there are 12 input files he would need input file identifiers numbered 1 through 12, and corresponding work files numbered 1 through 12.

Daniel B. Martin

whizje · 03-16-2013, 01:28 PM

Code:

Path=$(cut -d'.' -f1 <<< ${0})
alternative
Path=$(readlink -f $0)

danielbmartin · 03-16-2013, 01:35 PM

Quote:

Originally Posted by whizje

Code:

Path=$(cut -d'.' -f1 <<< ${0})
alternative
Path=$(readlink -f $0)

Your suggested alternative is not equivalent.
However, this is ...

Code:

Path=$(readlink -f $0 | cut -d'.' -f1)

... and I will use it henceforth.

Perhaps this change is the only obstacle preventing OP from using my program.

Daniel B. Martin