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Script parameters in Bash
(5 answers)
Closed 2 years ago.
At the moment I am manually searching for three characters which can be anything in dir1 dir2 dir3 etc
By going grep -i -r abc dir1
then
grep -i -r abc dir2
grep -i -r abc dir3
etc
Trying to automate this somewhat and thought about writing a shell script, something like
search.sh
and then when I want to search for something in the above directories I can put the three letters that I'm searching for
For example: run search.sh $Mid = abc
The shell script would do something like this:
$mid = Mid;
grep -i -r $mid nab-prep1001 | grep -i -r $mid nab-prep1002 | grep -i -r $mid multi-account-bpay-report | grep -i -r $mid nab-prep1004 | grep -i -r $mid nab-prep100 | grep -i -r $mid nab-prep1006 | grep -i -r $mid nab-prep1007
Very simple script and straightforward approach. Arguments are passed with $n, here n is number of the arguments 1,2,3 etc.
#!/bin/bash
echo "Simple Script"
echo "$1" "$2
Output:
$ ./simple.sh hello world
Simple Script
hello world
You can pass in arguments to a script and retrieve them as positional parameters inside your script.
So running:
./search.sh abc
You can access the argument "abc" with $1 inside the script (assuming you are passing in a single parameter).
I would recommend simply reading up on Linux Script Arguments online.
Related
I am trying to make a script so that I can give command with variable number of arguments myfind one two three and it finds all files in the folder, then applies grep -i one then grep -i two, and grep -i three and so on.
I tried following code:
#! /bin/bash
FULLARG="find . | "
for arg in "$#"
do
FULLARG=$FULLARG" grep -i "$arg" | "
done
echo $FULLARG
$FULLARG
However, though the command is created but it is not working and giving following error:
$ ./myfind one two three
find . | grep -i one | grep -i two | grep -i three |
find: unknown predicate `-i'
Where is the problem and how can it be solved?
You could store the result of find . and keep filtering that out till you have command line arguments:
#!/bin/bash
result="$(find .)"
for arg in "$#"
do
result=$(echo "$result" | grep -i "${arg}")
done
echo "$result"
am trying to execute the following command:
$ ssh root#10.10.10.50 "tail -F -n 1 $(ls -t /var/log/alert_ARCDB.log | head -n1 )"
ls: cannot access /var/log/alert_ARCDB.log: No such file or directory
tail: cannot follow `-' by name
notice the error returned, when i login to ssh separately and then execute
tail -F -n 1 $(ls -t /var/log/alert_ARCDB.log | head -n1 )"
see the below:
# ls -t /var/log/alert_ARCDB.log | head -n1
/var/log/alert_ARCDB.log
why is that happening and how to fix it. am trying to do this in one line as i don't want to create a script file.
Thanks a lot
Shell parameter expansion happens before command execution.
Here's a simple example. If I type...
ls "$HOME"
...the shell replaces $HOME with the path to my home directory first, then runs something like ls /home/larsks. The ls command has no idea that the command line originally had $HOME.
If we look at your command...
$ ssh root#10.10.10.50 "tail -F -n 1 $(ls -t /var/log/alert_ARCDB.log | head -n1 )"
...we see that you're in exactly the same situation. The $(ls -t ...) expression is expanded before ssh is executed. In other words, that command is running your local system.
You can inhibit the shell expansion on your local system by using single quotes. For example, running:
echo '$HOME'
Will produce:
$HOME
So you can run:
ssh root#10.10.10.50 'tail -F -n 1 $(ls -t /var/log/alert_ARCDB.log | head -n1 )'
But there's another problem here. If /var/log/alert_ARCDB.log is a file, your command makes no sense: calling ls -t on a single file gets you nothing.
If alert-ARCDB.log is a directory, you have a different problem. The result of ls /some/directory is a list of filenames without any directory prefix. If I run something like:
ls -t /tmp
I will get output like
file1
file2
If I do this:
tail $(ls -t /tmp | head -1)
I end up with a command that looks like:
tail file1
And that will fail, because there is no file1 in my current directory.
One approach would be to pipe the commands you want to perform to ssh. One simple way to achieve that is to first create a function that will echo the commands you want executed :
remote_commands()
{
echo 'cd /var/log/alert_ARCDB.log'
echo 'tail -F -n 1 "$(ls -t | head -n1 )"'
}
The cd will allow you to use the relative path listed by ls. The single quotes make sure that everything will be sent as-is to the remote shell, with no local expansion occurring.
Then you can do
ssh root#10.10.10.50 bash < <(remote_commands)
This assumes alert_ARCDB.log is a directory (or else I am not sure why you would want to add head -n1 after that).
I would like to create a perl or bash script that will read keyboard input and assign a variable, perform a fixed string grep recursively within the current directory filled with Snort logs, and then automatically tcpdump the matched files, grep its output, and print the specified lines to the terminal. Does anyone have a good idea of how this should work?
Here is an example of the methodology I want from the script:
step 1: Read keyboard input and assign it to variable named string.
step 2 command: grep -Fr "$string"
step 2 output: snort.log.1470609906 matches
step 3 command: tcpdump -r snort.log.1470609906 | grep -F "$string" C-10
step 3 output:
Snort log
Here's some bash code that does that:
s="google.com"
grep -Frl "$s" | \
while IFS= read -r x; do
tcpdump -r "$x" | grep -F "$s" -C10
done
idk about perl but you can do it easily enough just in shell:
str="google.com"
find . -type f -name 'snort.log.*' -exec grep -FlZ "$str" {} + |
xargs -0 -I {} sh -c 'tcpdump -r "{}" | grep -F '"$str"' -C10'
I'm building a little bash script to run another bash script that's found in multiple directories. Here's the code:
cd /home/mainuser/CaseStudies/
grep -R -o --include="Auto.sh" [\w] | wc -l
When I execute just that part, it finds the same file 5 times in each folder. So instead of getting 49 results, I get 245. I've written a recursive bash script before and I used it as a template for this problem:
grep -R -o --include=*.class [\w] | wc -l
This code has always worked perfectly, without any duplication. I've tried running the first code with and without the " ", I've tried -r as well. I've read through the bash documentation and I can't seem to find a way to prevent, or even why I'm getting, this duplication. Any thoughts on how to get around this?
As a separate, but related question, if I could launch Auto.sh inside of each directory so that the output of Auto.sh was dumped into that directory; without having to place Auto.sh in each folder. That would probably be much more efficient that what I'm currently doing and it would also probably fix my current duplication problem.
This is the code for Auto.sh:
#!/bin/bash
index=1
cd /home/mainuser/CaseStudies/
grep -R -o --include=*.class [\w] | wc -l
grep -R -o --include=*.class [\w] |awk '{print $3}' > out.txt
while read LINE; do
echo 'Path '$LINE > 'Outputs/ClassOut'$index'.txt'
javap -c $LINE >> 'Outputs/ClassOut'$index'.txt'
index=$((index+1))
done <out.txt
Preferably I would like to make it dump only the javap outputs for the application its currently looking at. Since those .class files could be in any number of sub-directories, I'm not sure how to make them all dump in the top folder, without executing a modified Auto.sh in the top directory of each application.
Ok, so to fix the multiple find:
grep -R -o --include="Auto.sh" [\w] | wc -l
Should be:
grep -R -l --include=Auto.sh '\w' | wc -l
The reason this was happening, was that it was looking for instances of the letter w in Auto.sh. Which occurred 5 times in the file.
However, the overall fix that doesn't require having to place Auto.sh in every directory, is something like this:
MAIN_DIR=/home/mainuser/CaseStudies/
cd $MAIN_DIR
ls -d */ > DirectoryList.txt
while read LINE; do
cd $LINE
mkdir ProjectOutputs
bash /home/mainuser/Auto.sh
cd $MAIN_DIR
done <DirectoryList.txt
That calls this Auto.sh code:
index=1
grep -R -o --include=*.class '\w' | wc -l
grep -R -o --include=*.class '\w' | awk '{print $3}' > ProjectOutputs.txt
while read LINE; do
echo 'Path '$LINE > 'ProjectOutputs/ClassOut'$index'.txt'
javap -c $LINE >> 'ProjectOutputs/ClassOut'$index'.txt'
index=$((index+1))
done <ProjectOutputs.txt
Thanks again for everyone's help!
For instance, if I'd like to reference the output of the previous command once, I can use the command below:
ls *.txt | xargs -I % ls -l %
But how to reference the output twice? Like how can I implement something like:
ls *.txt | xargs -I % 'some command' % > %
PS: I know how to do it in shell script, but I just want a simpler way to do it.
You can pass this argument to bash -c:
ls *.txt | xargs -I % bash -c 'ls -l "$1" > "out.$1"' - %
You can lookup up 'tpipe' on SO; it will also lead you to 'pee' (which is not a good search term elsewhere on the internet). Basically, they're variants of the tee command which write to multiple processes instead of writing to files like the tee command does.
However, with Bash, you can use Process Substitution:
ls *.txt | tee >(cmd1) >(cmd2)
This will write the input to tee to each of the commands cmd1 and cmd2.
You can arrange to lose standard output in at least two different ways:
ls *.txt | tee >(cmd1) >(cmd2) >/dev/null
ls *.txt | tee >(cmd1) | cmd2