Bash script to take file input and print content with a certain condition

Question

Right. Basically what I want to achieve is a bash script which takes the name of a file as the first argument and displays lines from the input file that are:

Not directories
Executable And these lines should be sorted in an ascending file size order.

The input file example that I'm using has the following content:

-rw-r--r-- 1 root     software       36 Dec  3 14:27 config
-rwxr-xr-x 1 root     software       72 Dec  3 14:27 config2
-rw-rw-r-- 1 jonathan software  7410294 Dec  3 14:28 even larger file
-rwxrwxr-x 1 jonathan software 17290686 Dec  3 14:29 even larger file2
-rwxrwxr-x 1 jonathan software  2470098 Dec  3 14:28 large file
-rwxr-xr-x 1 andrew   software   823366 Feb 11 16:25 myprogram
-rwxrwxr-x 1 jonathan software    29568 Dec  3 14:25 script
drwxrwsr-x 2 andrew   software     4096 Dec  3 19:24 testdir1
drwxrwsr-x 2 bob      software     4096 Dec  3 10:21 testdir2
drwxrwsr-x 2 david    software     4096 Aug 23 14:24 testdir3
drwxrwsr-x 2 david    software     4096 Dec  3 13:32 testdir4
drwxrwsr-x 2 james    software     4096 Jan  5 14:24 testdir5
-rw-rw-r-- 1 james    software       85 Dec  3 14:24 testfile1

I have written the script to ask for the file name and print the content of the file:

#!/bin/bash
echo "Please enter your filename:"
read filename
while read line;
do echo -e "$line\n";
done < $filename

But I don't know what's the command to look for the content line by line and search for lines which don't contain "drw" and also how to sort the lines by the size given in the example input file.

EDIT: This is the 2nd part of my problem. Now I want to modify the script to take a second argument, which if supplied is assumed to be the user name. Only lines matching the user name will be printed.

Eg: "jonathan" would print the following:

-rwxrwxr-x 1 jonathan software    29568 Dec  3 14:25 script
-rwxrwxr-x 1 jonathan software  2470098 Dec  3 14:28 large file
-rwxrwxr-x 1 jonathan software 17290686 Dec  3 14:29 even larger file2

How are you getting input data like this? from ls's output?

anubhava
– anubhava

2014-01-17 16:54:55 +00:00
Commented Jan 17, 2014 at 16:54 — anubhava
– anubhava, Commented Jan 17, 2014 at 16:54

cnicutar · Accepted Answer · 2014-01-17 17:34:13Z

2

Here is a solution with awk:

$ awk '!/^d/ && $1 ~ /x/' text.txt | sort -k5,5n
-rwxr-xr-x 1 root     software       72 Dec  3 14:27 config2
-rwxrwxr-x 1 jonathan software    29568 Dec  3 14:25 script
-rwxr-xr-x 1 andrew   software   823366 Feb 11 16:25 myprogram
-rwxrwxr-x 1 jonathan software  2470098 Dec  3 14:28 large file
-rwxrwxr-x 1 jonathan software 17290686 Dec  3 14:29 even larger file2

You filter out things that start with a d
You check for things whose first field contains an x
You sort based on the 5 field, numerically

Here is a version you can drop in a shell script:

file=$1
user=$2

awk '!/^d/ && $1 ~ /x/ && $3 ~/'"${user}/" "$file" | sort -k5,5n

edited Jan 17, 2014 at 17:34

answered Jan 17, 2014 at 16:52

cnicutar

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10 Comments

Bhanu Chawla Over a year ago

Not much familiar with awk tool mate. How do I implement that awk command in my bash script?

cnicutar Over a year ago

@BhanuChawla You might get away with just calling that but with $1 instead of text.txt.

Bhanu Chawla Over a year ago

I have edited the question and added another issue I need to solve. I guess awk will solve that. Can you help mate? :)

cnicutar Over a year ago

@BhanuChawla Add another && $3 ~ /$2/ maybe ?

Bhanu Chawla Over a year ago

Sorry to be a noob but what will be the final script? Can you update your answer? I tried awk '!/^d/ && $1 ~ /x/' text.txt | sort -k5,5n && $3 ~ /$2/ in my script but it's not working. pastebin.com/dERMnRzM

|

BMW · Accepted Answer · 2014-01-17 21:16:54Z

0

using shell.

ls -l |grep -v "^d" |grep "^-rwx"|sort -k5n

answered Jan 17, 2014 at 21:16

BMW

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