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:
$(command) replaces itself with the output of command
Related
I'm trying to store the result of this command that is written in a script
ls -l /etc|wc -l
in a variable on another file.
To summarize, I have a script with that command and when I execute it, I want the result to be stored in a variable in another file.
Can someone help me with this please?
You may try to use temporary file (if possible).
This command:
ls -l /etc|wc -l > /tmp/myvar.txt
Another file:
myvar="$(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.
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'm pretty new to Bash and Linux in general. I've created a couple scripts that I would like to be able to use by typing the command, rather than the directory and the executable file. I'm using Debian Jessie if that makes a difference.
The path to one of my scripts is ~/Scripts/DIR_1/My_Script.sh while another is in ~Scripts/DIR_2/My_Other_Script.sh. I would like ALL of the scripts contained withing the Scripts directory to be indexed as commands regardless of directory/path depth.
I've appended this text to the end of my .bashrc file...
PATH=${PATH}:$(find ~/Scripts -type d | sed '/\/\\./d' | tr '\n' ':' | sed 's/:$//')
Since I'm pretty new to this kind of thing, I had to steal that line from here.
When I try to run My_Script from the command line withing a sub directory of my home folder (or anywhere else for that matter) I get My_Script: command not found
I will readily admit that I might have misunderstood the process of adding a bash script to the command line.
How do I recursively add bash scripts as commands? What is wrong with the process I'm currently using?
I think your issue is that you're not putting the .sh, that is part of your file name.
Normally, pressing tab after having typed only the first letter should complete the command up to the point where there is an ambiguity (or completely if there's none). In case of ambiguity, pressing tab a second time shows the options. So in your case, if you type My<tab><tab> you should have options My_Script.sh and My_Other_Script.sh displayed. And if you type My_Script<tab> it should complete by putting My_Script.sh
Edit
I forgot to precise that you can check the value of PATH by doing echo $PATH. This will allow you to check that the command you copied did what you wanted.
I know history will capture commands that I run, but it is shell specific. I work with multiple shells and multiple hosts and would like to write a small script which, after every command I run, dumps that command to some file along with the host name. This way, i can implement my own history command which reads from that file, and can take a host as an argument which would be handy for me. I'm not sure how to get the first part though..i.e., get every shell command I type to trigger a "dump that command into a file" part. Any ideas?
Thanks
In bash, the PROMPT_COMMAND environment variable contains a command that will be executed before the PS1 prompt is displayed. So yours could be something like history | tail -n1 | perl -npe 's/^\s+\d+\s+//' | yourcommand HOST
The script utility should solve your problem. It records everything you type and all that is printed on the terminal in a file (even including terminal control codes, so if you cat that file on the console, you even reproduce the original text colors).
Can I execute command within another command in UNIX shells?
If impossible, can I use the output of the previous command as the input of next command, as in:
command x then command y,
where in command y I want use the output of command x?
You can use the backquotes for this.
For example this will cat the file.txt
cat `echo file.txt`
And this will print the date
echo the date is `date`
The code between back-quotes will be executed and be replaced by its result.
You can do something like;
x=$(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")