I'm trying to store the result of this command that is written in a script
ls -l /etc|wc -l
in a variable on another file.
To summarize, I have a script with that command and when I execute it, I want the result to be stored in a variable in another file.
Can someone help me with this please?
You may try to use temporary file (if possible).
This command:
ls -l /etc|wc -l > /tmp/myvar.txt
Another file:
myvar="$(cat /tmp/myvar.txt)"
You just need to use '> path/to/file' at the end of your command to redirect the output to a file (this will override the file content).
If you need another behavior, like append the content, you should use '>>' instead of '>'.
Take a look here for more details.
I'm not sure I understand what you're trying to do so I'll give you two solutions.
If the command you mention is in some file script_A.sh and you want the results of that script stored in some variable $var when running some other script script_B.sh, randomir's solution is good. In script_B:
var=$(bash path/to/script_A.sh)
If what you're asking is to run script_A.sh and then have it write a new line to a file that would store the results to a value when you run script_B.sh, I suppose you could run something like:
result=$(ls -l /etc|wc -l)
echo "var=\"$result\"" > path/to/script_B.sh
or even replace a line in a script_B.sh that already exists:
result= $(ls -l /etc|wc -l)
sed -i "s|var=SOMEPLACEHOLDER|var='$result'|" path/to/script_B.sh
If the latter is what you want, though, can you tell us more about what you're trying to accomplish? There's probably a better way than what you propose.
Related
I am trying to run script on output files that can be further used as input files for gaussian.
I wanted to know what are the commands used in Linux to run the script on .log files and .HSCP1 files.
Many thanks,
Regards,
The generic syntax of passing an argument to a script in linux, assuming your script is named script.sh and your target file is named arg.log, would be
script.sh arg.log
This assigns the name arg.log to $1 inside the environment of the executing copy of script.sh. If you don't then do something with that, it won't matter.
You might also have your script read its stdin like this:
script.sh < arg.log
which will put the contents os arg.log on script.sh's stdin, but unless it reads them accordingly, it won't matter.
Of course, both these assume script.sh is in your $PATH; otherwise you will need to apply a path for the OS, such as /path/to/dir/with/script.sh or (if you are in the same directory) ./script.sh.
If what you are asking is how to get a lorge number of files assigned as arguments, you could pass wildcards - for the first example above, that could be done as
./script.sh /path/to/*.log /also/to/other.*
or you could use find, maybe with xargs like so -
find /path/to/files/ -name *.log | xargs /path/of/script.sh
which will call the script over and over.
I hope one of these helps, but you really must provide more context for what you are doing and how.
From terminal i need execute a command like this :
test /home/root/myimg.jpg
where test is an executable file and myimg.jpg is an image that i pass to exe file.
The image is, in may case, the last file added to the /home/root directory and i can see it from terminal by typing ls /home/root | tail -n 1.
i need to write an alias that allows me to pass the latest image saved in the directory. How can i do it ? Because is ever the last image added that i need to pass and its name changes everytime !!!
thanks a lot for every answer :)
alias name='test $(ls /home/root | tail -n 1.)'
should do the job.
Explanation:
$(command) replaces itself with the output of command
I'm struggling with passing a shell command. Specifically, I have written a shell file that the user will run. In it, another shell file is written based on the user inputs. This file will then be passed to a command that submits the job.
Now in this internal shell file I have a variable containing a function. However, when I run the user shell script I can't get this function to pass in a way that the internal shell file can use it.
I can't share my work but I'll try to make an example
#User shell script
cat >test.txt <<EOF
#a bunch of lines that are not relevant
var=`grep examples input.txt`
/bin/localfoo -${var}
EOF
# pass test.txt to localfoo2
/bin/localfoo2 /test.txt
When I run the 'User Shell Script' it prints that grep can't find the file, but I don't want grep to be evaluated. I need it to be written, as is, so that when the line '/bin/localfoo2 /test.txt' is read, grep is evaluated.
I've tried a number of things. I've tried double back ticks, i've tried using 'echo $(eval $var)'. But none of the methods I've found through googling have managed to pass this var in a way that will accomplish what I want.
Your help is much appreciated.
You can try with single quote (').
You have to put the single quote in before the grep command and end of the grep command like below.
#User shell script
cat >test.txt <<EOF
#a bunch of lines that are not relevant
var='`grep examples input.txt`'
/bin/localfoo -${var}
EOF
# pass test.txt to localfoo2
/bin/localfoo2 /test.txt
I did not understand where you have to execute that grep command.
If you want to execute the grep command inside the localfoo script, I hope this method will help.
I am writing a shell script to sync to a github repo, kick off the build, then take the output file, rename it, and move it to a location where it can be seen by Apache.
It's the renaming of the file that I've got not the faintest how to do within a shell script (I have virtually no experience with shell scripts - my understanding
Compiler will create /var/espbuild/firstpart_1vXX_secondpart.bin
I need to move this file to:
/var/www/html/builds/espbuild/firstpart_1vXX_DATE_secondpart_postfix.bin
1vXX is the version number
DATE is the output of date +%m-%d
postfix is just a string.
I'm not really certain where to start for something like this - I'm sure there's a graceful way, since this is the kind of thing shell scripts are made for, but I know just about nothing about shell scripts.
Thanks in advance
You can get the result of a command into a variable by using $():
DATE=$(date +%m-%d)
Then just use it in the new filename:
INPUT=/var/espbuild/firstpart_1vXX_secondpart.bin
OUTPUT=/var/www/html/builds/espbuild/firstpart_1vXX_${DATE}_secondpart_postfix.bin
mv ${INPUT} ${OUTPUT}
Edit: To get out the version part, here's a quick example:
VERSION=$(grep -o 1v.. <<< ${INPUT})
Then OUTPUT should be set like:
OUTPUT=/var/www/html/builds/espbuild/firstpart_${VERSION}_${DATE}_secondpart_postfix.bin
You can use this in BASH:
f='/var/espbuild/firstpart_1vXX_secondpart.bin'
s="${f##*/}"
s2=${s##*_}
dest="/var/www/html/builds/espbuild/${s%_*}_$(date '+%m-%d')_${s2%.*}_postfix.bin"
echo "$dest"
/var/www/html/builds/espbuild/firstpart_1vXX_07-14_secondpart_postfix.bin
cp "$f" "$dest"
Can I execute command within another command in UNIX shells?
If impossible, can I use the output of the previous command as the input of next command, as in:
command x then command y,
where in command y I want use the output of command x?
You can use the backquotes for this.
For example this will cat the file.txt
cat `echo file.txt`
And this will print the date
echo the date is `date`
The code between back-quotes will be executed and be replaced by its result.
You can do something like;
x=$(grep $(dirname "$path") file)
here dirname "$path" will run first and its result will be substituted and then grep will run, searching for the result of dirname in the file
What exactly are you trying to do? It's not clear from the commands you are executing. Perhaps if you describe what you're looking for we can point you in the right direction. If you want to execute a command over a range of file (or directory) names returned by the "find" command, Colin is correct, you need to look at the "-exec" option of "find". If you're looking to execute a command over a bunch of arguments listed in a file or coming from stdin, you need to check out the "xargs" commands. If you want to put the output of a single command on to the command line of another command, then using "$(command)" (or 'command' [replace the ' with a backquote]) will do the job. There's a lot of ways to do this, but without knowing what it is you're trying it's hard to be more helpful.
Here is an example where I have used nested system commands.
I had run "ls -ltr" on top of find command. And it executes
it serially on the find output.
ls -ltr $(find . -name "srvm.jar")