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.
Related
I have one variable, which is coming from some where like:
VAR1='hhgfhfghhgf"";2Ddgfsaj!!!$#^$\'&%*%~*)_)(_{}||\\/'
Now i have command like this
./myscript.sh '$VAR1'
I am getting that $VAR1 from some diff process and when I display it look exactly as its above.
Now that command is failing as there is already single quote inside variable. In the process where I use it it is expanded at that point, which causes that error.
I have control over myscript.sh but not above command.
Is there any way I can get variable inside my script?
What you are saying is not possible to failing when passing to your script. Might your script has processing issue (or a command where this argument will passing into it) which cannot expand the variable correctly. You can either use printf with %q modifier to escape all special characters then pass it to your script:
./myscript.sh "$(printf '%q\n' "$VAR1")"
... or do the same within your script before you wanted to pass to some other commands:
VAR2="$(printf '%q\n' "$VAR1")"
I'm trying to store the result of this command that is written in a script
ls -l /etc|wc -l
in a variable on another file.
To summarize, I have a script with that command and when I execute it, I want the result to be stored in a variable in another file.
Can someone help me with this please?
You may try to use temporary file (if possible).
This command:
ls -l /etc|wc -l > /tmp/myvar.txt
Another file:
myvar="$(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 trying to grep the user command into a variable.
If the command begins with specific words I'm redirecting the command into another shell (not BASH), if not it will run regularly on bash.
When I'm using "read" I lose the BASH prompt then I'm unable to auto-complete, backspace etc.
My goal is:
if I type for example ls -la the command will not run and will be assigned into variable:
user#machine:~$ ls -la
user_command = 'ls -la'
if the command is not other shell command (not BASH)
eval $user_command
Is there any way to achieve this?
I guess you want to : invoke some specific commands through another shell instead of bash
You could define a function for each of those commands
exemple:
ls () { # function that, instead of bash, launch 'ls' through ksh
ksh -c "ls $*"
}
that way, with each function's name matching the command you want to run with another shell (edit the inside of the function accordingly if it needs to be run with another shell than ksh)
Place those function in specific user's "~/.bashrc" files.
You could write your implementation of ls and place it in ~/bin/ls.
Give full permissions to this file
Can I execute command within another command in UNIX shells?
If impossible, can I use the output of the previous command as the input of next command, as in:
command x then command y,
where in command y I want use the output of command x?
You can use the backquotes for this.
For example this will cat the file.txt
cat `echo file.txt`
And this will print the date
echo the date is `date`
The code between back-quotes will be executed and be replaced by its result.
You can do something like;
x=$(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")
I want to write a very simple script , which takes a process name , and return the tail of the last file name which contains the process name.
I wrote something like that :
#!/bin/sh
tail $(ls -t *"$1"*| head -1) -f
My question:
Do I need the first line?
Why isn't ls -t *"$1"*| head -1 | tail -f working?
Is there a better way to do it?
1: The first line is a so called she-bang, read the description here:
In computing, a shebang (also called a
hashbang, hashpling, pound bang, or
crunchbang) refers to the characters
"#!" when they are the first two
characters in an interpreter directive
as the first line of a text file. In a
Unix-like operating system, the
program loader takes the presence of
these two characters as an indication
that the file is a script, and tries
to execute that script using the
interpreter specified by the rest of
the first line in the file
2: tail can't take the filename from the stdin: It can either take the text on the stdin or a file as parameter. See the man page for this.
3: No better solution comes to my mind: Pay attention to filenames containing spaces: This does not work with your current solution, you need to add quotes around the $() block.
$1 contains the first argument, the process name is actually in $0. This however can contain the path, so you should use:
#!/bin/sh
tail $(ls -rt *"`basename $0`"*| head -1) -f
You also have to use ls -rt to get the oldest file first.
You can omit the shebang if you run the script from a shell, in that case the contents will be executed by your current shell instance. In many cases this will cause no problems, but it is still a bad practice.
Following on from #theomega's answer and #Idan's question in the comments, the she-bang is needed, among other things, because some UNIX / Linux systems have more than one command shell.
Each command shell has a different syntax, so the she-bang provides a way to specify which shell should be used to execute the script, even if you don't specify it in your run command by typing (for example)
./myscript.sh
instead of
/bin/sh ./myscript.sh
Note that the she-bang can also be used in scripts written in non-shell languages such as Perl; in the case you'd put
#!/usr/bin/perl
at the top of your script.