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I'm new to bash scripting, can someone tell me what is the meaning of the below command.
grep -v _1_ <file_name> > <new_file>
The -v switch inverts the criteria, so it shows only the lines which does NOT contain the string _1_ and logs the output into the file <new_file>.
The question is wrong, you have written the command as:
grep -v _1_ <file_name> > <new_file>
While it should be:
grep -v _1_ <file1> > <file2>
You can use grep on more file than one, which means that you are looking for something in more than one file.
The -v part is already explained by Antonio.
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I know what the commands ls and wc do, but I can not find out what ls > wc and ls | wc will show. Can someone please help me flush out the meaning of this commands?
ls | wc The output from the ls command is piped into the wc command. So it will count the words which are in the output of ls. So you see simply the number of files read by ls.
ls > wc This creates a new file in your current working directory with the name wc with the output of your ls command. The program wc is not used here, simply a new file with the same name is created. You can simply look into this new file with your favorite editor or simply use cat for it.
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How can I print the total number of file descriptors with index 24 for all the running processes?
I tried something: $
ls /proc/*/fd 2> errors.txt > stdout.txt | grep "^24" stdout.txt | wc -l
This solution returns 0 everytime.
I mention that my task ask me to write an one liner in order to solve it.
ls /proc/*/fd 2>/dev/null | grep -c '^24$'
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My friend just passed me this command to count the number of files in a directory:
$ ls -lt | wc -l
Can someone please help me flush out the meaning of this command? I know that ls is to list all the files. But what does -lt mean?
Also, I get a different count if I use ls | wc -l with no -lt option. Why is that the case?
You'll want to get familiar with the "man (manual) pages":
$ man ls
In this case you'll see:
-l (The lowercase letter ``ell''.) List in long format. (See below.) If
the output is to a terminal, a total sum for all the file sizes is
output on a line before the long listing.
-t Sort by time modified (most recently modified first) before sorting the
operands by lexicographical order.
Another way you can see the effect of the options is to run ls without piping to the wc command. Compare
$ ls
with
$ ls -l
and
$ ls -lt
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I am building a AIX bash shell utility whereby i get a dynamic variable with value like status(running).
I just need the string within brakets that is running.
Right now i am able to get the whole word with status along with the brackets using awk print.
Can anyone suggest me how to just extract the running out of it. Thanks.
Let's say:
s='(running)'
Using pure BASH:
echo "${s//[()]/}"
running
Using sed:
echo "$s" | sed 's/[()]//g'
running
Using tr:
tr -d '()' <<< "$s"
running
UPDATE: As per comments by OP:
s='status(running)'
Using sed:
echo "$s" | sed 's/^.*(\(.*\)).*$/\1/g'
running
Using pure BASH:
t="${s#*\(}"
echo "${t%)*}"
running
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Is there a way to have each result of grep followed by newline
for example if the result of grep is:
1
2
3
I need it to be:
1
2
3
grep "pattern" /path/to/file | awk '{print $0,"\n"}'