I have a script which takes in only one positional parameter which is a list of values, and I'm trying to get the parameter from stdin with xargs.
However by default, xargs passes all the lists to my script as positional parameters, e.g. when doing:
echo 1 2 3 | xargs myScript, it will essentially be myScript 1 2 3, and what I'm looking for is myScript "1 2 3". What is the best way to achieve this?
Change the delimiter.
$ echo 1 2 3 | xargs -d '\n' printf '%s\n'
1 2 3
Not all xargs implementations have -d though.
And not sure if there is an actual use case for this but you can also resort to spawning another shell instance if you have to. Like
$ echo -e '1 2\n3' | xargs sh -c 'printf '\''%s\n'\'' "$*"' sh
1 2 3
If the input can be altered, you can do this. But not sure if this is what you wanted.
echo \"1 2 3\"|xargs ./myScript
Here is the example.
$ cat myScript
#!/bin/bash
echo $1; shift
echo $1; shift
echo $1;
$ echo \"1 2 3\"|xargs ./myScript
1 2 3
$ echo 1 2 3|xargs ./myScript
1
2
3
Related
I have simple bash script which only outputs the filenames that are given to the script as positional arguments:
#!/usr/bin/env bash
for file; do
echo "$file"
done
Say I have files with spaces (say "f 1" and "f 2"). I can call the script with a wildcard and get the expected output:
$ ./script f*
> f 1
> f 2
But if I use command substitution it doesn't work:
$ ./script $(echo f*)
> f
> 1
> f
> 2
How can I get the quoting right when my command substition outputs multiple filenames with spaces?
Edit: What I ultimatively want is to pass filenames to a script (that is slightly more elaborate than just echoing their names) in a random order, e.g. something like that:
./script $(ls f* | shuf)
With GNU shuf and Bash 4.3+:
readarray -d '' files < <(shuf --zero-terminated --echo f*)
./script "${files[#]}"
where the --zero-terminated can handle any filenames, and readarray also uses the null byte as the delimiter.
With older Bash where readarray doesn't support the -d option:
while IFS= read -r -d '' f; do
files+=("$f")
done < <(shuf --zero-terminated --echo f*)
./script "${files[#]}"
In extreme cases with many files, this might run into command line length limitations; in that case,
shuf --zero-terminated --echo f*
could be replaced by
printf '%s\0' f* | shuf --zero-terminated
Hat tip to Socowi for pointing out --echo.
It's very difficult to get this completely correct. A simple attempt would be to use %q specifier to printf, but I believe that is a bashism. You still need to use eval, though. eg:
$ cat a.sh
#!/bin/sh
for x; do echo $((i++)): "$x"; done
$ ./a.sh *
0: a.sh
1: name
with
newlines
2: name with spaces
$ eval ./a.sh $(printf "%q " *)
0: a.sh
1: name
with
newlines
2: name with spaces
This feels like an XY Problem. Maybe you should explain the real problem, someone might have a much better solution.
Nonetheless, working with what you posted, I'd say read this page on why you shouldn't try to parse ls as it has relevant points; then I suggest an array.
lst=(f*)
./script "${lst[#]}"
This will still fail if you reparse it as the output of a subshell, though -
./script $( echo "${lst[#]}" ) # still fails same way
./script "$( echo "${lst[#]}" )" # *still* fails same way
Thinking about how we could make it work...
You can use xargs:
$ ls -l
total 4
-rw-r--r-- 1 root root 0 2021-08-13 00:23 ' file 1'
-rw-r--r-- 1 root root 0 2021-08-13 00:23 ' file 2'
-rw-r--r-- 1 root root 0 2021-08-13 00:23 ' file 3'
-rw-r--r-- 1 root root 0 2021-08-13 00:23 ' file 4'
-rwxr-xr-x 1 root root 35 2021-08-13 00:25 script
$ ./script *file*
file 1
file 2
file 3
file 4
$ ls *file* | shuf | xargs -d '\n' ./script
file 4
file 2
file 1
file 3
If your xargs does not support -d:
$ ls *file* | shuf | tr '\n' '\0' | xargs -0 ./script
file 3
file 1
file 4
file 2
I am new to programming, so plz bear with the way I try to explain my problem (also any help regarding how to elegantly phrase the tile is welcome).
I have a bash script (say for example script1.sh ) that takes in arguments a, b and c(another script). Essentially, argument c for script1.sh is the name of another script (let's say script2.sh). However, script2.sh takes in arguments d,e and f. So my question is, how do I pass arguments to script1.sh ?? (example, ./script1.sh -a 1 -b 2 -c script2.sh -d 3 -e 4 -f 5)
Sorry in advance if the above does not make sense, not sure how else to phrase it...
You should use "" for that
./script1.sh -a 1 -b 2 -c "script2.sh -d 3 -e 4 -f 5"
Try script1.sh with this code
#!/bin/bash
for arg in "$#"; { # loop through all arguments passed to the script
echo $arg
}
The output will be
$ ./script1.sh -a 1 -b 2 -c "script2.sh -d 3 -e 4 -f 5"
-a
1
-b
2
-c
script2.sh -d 3 -e 4 -f 5
But if you run this
#!/bin/bash
for arg in $#; { # no double quotes around $#
echo $arg
}
The output will be
$ ./script1.sh -a 1 -b 2 -c "script2.sh -d 3 -e 4 -f 5"
-a
1
-b
2
-c
script2.sh
-d
3
4
-f
5
But there is no -e why? Coz echo supports argument -e and use it.
I have a while loop, where A=1~3
mysql -e "select A from xxx;" while read A;
do
whatever
done
The mySQL command will return only numbers, each number in each line. So the while loop here will have A=1, A=2, A=3
I would like to append the integer number in the loop (here is A=1~3) into a command line to run outside the while loop. Any bash way to do this?
parallel --joblog test.log --jobs 2 -k sh ::: 1.sh 2.sh 3.sh
You probably want something like this:
mysql -e "select A from xxx;" | while read A; do
whatever > standard_out 2>standard_error
echo "$A.sh"
done | xargs parallel --joblog test.log --jobs 2 -k sh :::
Thanks for enlightening me. xargs works perfectly here:
Assuming we have A.csv (mimic the mysql command)
1
2
3
4
We can simply do:
cat A.csv | while read A; do
echo "echo $A" > $A.sh
echo "$A.sh"
done | xargs -I {} parallel --joblog test.log --jobs 2 -k sh ::: {}
The above will print the following output as expected
1
2
3
4
Here -I {} & {} are the argument list markers:
https://www.cyberciti.biz/faq/linux-unix-bsd-xargs-construct-argument-lists-utility/
My script v.sh
select f in "$#" ; do
echo $f
done
v.sh 1 2 3 I can select options after the command executed.
echo 1 2 3 | v.sh Showing nothing.
echo 1 2 3 | xargs v.sh Showing the options, but I can't select them.
How to select the options? Thx in advance.
With bash:
#!/bin/bash
if [[ -z $1 ]]; then
# no arguments, read from stdin
mapfile -t input </dev/stdin
exec </dev/tty
else
# use arguments
input=("$#")
fi
select f in "${input[#]}" ; do
echo "$f"
done
Example with stdin:
cut -d : -f 1 /etc/passwd | ./script.sh
with arguments:
./script.sh 1 2 "3 3" 4 5
If you have a list in python, and you want the elements from 2 to n can do something nice like
list[2:]
I'd like to something similar with argv in Bash. I want to pass all the elements from $2 to argc to a command. I currently have
command $2 $3 $4 $5 $6 $7 $8 $9
but this is less than elegant. Would would be the "proper" way?
you can do "slicing" as well, $# gets all the arguments in bash.
echo "${#:2}"
gets 2nd argument onwards
eg
$ cat shell.sh
#!/bin/bash
echo "${#:2}"
$ ./shell.sh 1 2 3 4
2 3 4
Store $1 somewhere, then shift and use $#?
script1.sh:
#!/bin/bash
echo "$#"
script2.sh:
#!/bin/bash
shift
echo "$#"
$ sh script1.sh 1 2 3 4
1 2 3 4
$ sh script2.sh 1 2 3 4
2 3 4