I have this code snippet:
userjobs=$(grep -rw "$USER" /my/job/dir/|awk '{print $1}'|sort|uniq|rev|cut -c 2-|rev)
for job in "${userjobs[#]}"; do
cat "$job"
done
exit 0
When I run it as is, I get the following output:
cat: /my/job/dir/45
/my/job/dir/46: No such file or directory
However, if I unquote $job, I no longer receive this behavior, and it cats each of the files as expected.
I've done some reading up on globbingand splitting to see if this is occurring, but it seems like double-quoting should prevent that from happening. Can anyone explain why the behavior is different between "$job" and $job?
This happens because your variable looks like:
userjobs='/my/job/dir/45
/my/job/dir/46'
If you expand it as an array, with "${userjobs[#]}", that it acts as an array with exactly one element -- that string. Thus, behavior is identical to:
userjobs=( [0]='/my/job/dir/45
/my/job/dir/46' )
...still exactly one string with a literal newline in it.
Thus, cat "$job" looks for a file with a literal newline in its name.
To load your result into a real array you can iterate over with "${userjobs[#]}" expanding to a distinct element per line, use:
readarray -t userjobs < <(grep ...)
userjobs needs to be an array. Put parentheses around the value when assigning it:
userjobs=($(grep -rw "$USER" /my/job/dir/|awk '{print $1}'|sort|uniq|rev|cut -c 2-|rev))
Related
I am a newbie in bash script.
Here is my environment:
Mac OS X Catalina
/bin/bash
I found here a mix of several commands to remove the duplicate string in a string.
I needed for my program which updates the .zhrc profile file.
Here is my code:
#!/bin/bash
a='export PATH="/Library/Frameworks/Python.framework/Versions/3.8/bin:/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/opt/local/bin:"'
myvariable=$(echo "$a" | tr ':' '\n' | sort | uniq | xargs)
echo "myvariable : $myvariable"
Here is the output:
xargs: unterminated quote
myvariable :
After some test, I know that the source of the issue is due to some quotes "" inside my variable '$a'.
Why am I so sure?
Because when I execute this code for example:
#!/bin/bash
a="/Library/Java/JavaVirtualMachines/jdk1.8.0_271.jdk/Contents/Home:/Library/Java/JavaVirtualMachines/jdk1.8.0_271.jdk/Contents/Home"
myvariable=$(echo "$a" | tr ':' '\n' | sort | uniq | xargs)
echo "myvariable : $myvariable"
where $a doesn't contain any quotes, I get the correct output:
myvariable : /Library/Java/JavaVirtualMachines/jdk1.8.0_271.jdk/Contents/Home
I tried to search for a solution for "xargs: unterminated quote" but each answer found on the web is for a particular case which doesn't correspond to my problem.
As I am a newbie and this line command is using several complex commands, I was wondering if anyone know the magic trick to make it work.
Basically, you want to remove duplicates from a colon-separated list.
I don't know if this is considered cheating, but I would do this in another language and invoke it from bash. First I would write a script for this purpose in zsh: It accepts as parameter a string with colon separtors and outputs a colon-separated list with duplicates removed:
#!/bin/zsh
original=${1?Parameter missing} # Original string
# Auxiliary array, which is set up to act like a Set, i.e. without
# duplicates
typeset -aU nodups_array
# Split the original strings on the colons and store the pieces
# into the array, thereby removing duplicates. The core idea for
# this is stolen from:
# https://stackoverflow.com/questions/2930238/split-string-with-zsh-as-in-python
nodups_array=("${(#s/:/)original}")
# Join the array back with colons and write the resulting string
# to stdout.
echo ${(j':')nodups_array}
If we call this script nodups_string, you can invoke it in your bash-setting as:
#!/bin/bash
a_path="/Library/Frameworks/Python.framework/Versions/3.8/bin:/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/opt/local/bin:"
nodups_a_path=$(nodups_string "$a_path")
my_variable="export PATH=$nodups_a_path"
echo "myvariable : $myvariable"
The overall effect would be literally what you asked for. However, there is still an open problem I should point out: If one of the PATH components happens to contain a space, the resulting export statement can not validly be executed. This problem is also inherent into your original problem; you just didn't mention it. You could do something like
my_variable=export\ PATH='"'$nodups_a_path"'"'
to avoid this. Of course, I wonder why you take such an effort to generat a syntactically valid export command, instead of simply building the PATH by directly where it is needed.
Side note: If you would use zsh as your shell instead of bash, and only want to keep your PATH free of duplicates, a simple
typeset -iU path
would suffice, and zsh takes care of the rest.
With awk:
awk -v RS=[:\"] 'NR > 1 { pth[$0]="" } END { for (i in pth) { if (i !~ /[[:space:]]+/ && i != "" ) { printf "%s:",i } } }' <<< "$a"
Set the record separator to : and double quotes. Then when the number record is greater than one, set up an array called pth with the path as the index. At the end, loop through the array, re printing the paths separated with :
Given the following script:
#!/bin/bash
asteriskFiles=("sip.conf" "extensions.conf")
for asteriskFile in $asteriskFiles
do
# backup current configuration file
cp somepath/${asteriskFile} test/
echo "test"
done
This gives me the output "test" only once, so the loop runs only once instead of two times (two entries in asteriskFiles array). What am I doing wrong? Thanks for any hint!
An illustration:
$ asteriskFiles=("sip.conf" "extensions.conf")
$ echo $asteriskFiles # is equivalent to echo ${asteriskFiles[0]}
sip.conf
$ echo "${asteriskFiles[#]}"
sip.conf extensions.conf
Note that the quotes are important. echo ${asteriskFiles[#]} might seem to work, but bash would wordsplit on whitespace if any of your files had whitespace in them.
Write the beginning of your loop like this
for asteriskFile in "${asteriskFiles[#]}"
The Probem
The asteriskFiles variable holds an array. If you dereference it like a scalar, you only get the first element of the array.
The Solution
You want to use the correct shell parameter expansion to access all the subscript elements. For example:
$ echo "${asteriskFiles[#]}"
sip.conf extensions.conf
The # subscript (when correctly quoted) will expand to the properly-tokenized elements of your array, which your for-loop will then be able to iterate over.
I need to take specific variables from one script and use them in a different script.
Example:
Original script:
VARA=4 # Some description of VARA
VARB=6 # Some description of VARB
SOMEOTHERVAR="Foo"
/call/to/some/program
I want to write a second script that needs VARA and VARB, but not SOMEOTHERVAR or the call to the program.
I can already do:
eval $(grep 'VARA=' origscript.sh)
eval $(grep 'VARB=' origscript.sh)
This seems to work, but when I want to do both, like this, it only sets the first:
eval $(grep 'VAR[AB]=' origscript.sh)
because it seems to concatenate the two lines that grep returns. (Which probably means that the comments save the first assignments.)
Put quotes around it, so that the newlines in the output of grep will not be turned into spaces.
eval "$(grep 'VAR[AB]=' origscript.sh)"
Long story short, I'm trying to grep a value contained in the first column of a text file by using a variable.
Here's a sample of the script, with the grep command that doesn't work:
for ii in `cat list.txt`
do
grep '^$ii' >outfile.txt
done
Contents of list.txt :
123,"first product",description,20.456789
456,"second product",description,30.123456
789,"third product",description,40.123456
If I perform grep '^123' list.txt, it produces the correct output... Just the first line of list.txt.
If I try to use the variable (ie grep '^ii' list.txt) I get a "^ii command not found" error. I tried to combine text with the variable to get it to work:
VAR1= "'^"$ii"'"
but the VAR1 variable contained a carriage return after the $ii variable:
'^123
'
I've tried a laundry list of things to remove the cr/lr (ie sed & awk), but to no avail. There has to be an easier way to perform the grep command using the variable. I would prefer to stay with the grep command because it works perfectly when performing it manually.
You have things mixed in the command grep '^ii' list.txt. The character ^ is for the beginning of the line and a $ is for the value of a variable.
When you want to grep for 123 in the variable ii at the beginning of the line, use
ii="123"
grep "^$ii" list.txt
(You should use double quotes here)
Good moment for learning good habits: Continue in variable names in lowercase (well done) and use curly braces (don't harm and are needed in other cases) :
ii="123"
grep "^${ii}" list.txt
Now we both are forgetting something: Our grep will also match
1234,"4-digit product",description,11.1111. Include a , in the grep:
ii="123"
grep "^${ii}," list.txt
And how did you get the "^ii command not found" error ? I think you used backquotes (old way for nesting a command, better is echo "example: $(date)") and you wrote
grep `^ii` list.txt # wrong !
#!/bin/sh
# Read every character before the first comma into the variable ii.
while IFS=, read ii rest; do
# Echo the value of ii. If these values are what you want, you're done; no
# need for grep.
echo "ii = $ii"
# If you want to find something associated with these values in another
# file, however, you can grep the file for the values. Use double quotes so
# that the value of $ii is substituted in the argument to grep.
grep "^$ii" some_other_file.txt >outfile.txt
done <list.txt
Suppose I've got a list of files
file1
"file 1"
file2
a for...in loop breaks it up between whitespace, not newlines:
for x in $( ls ); do
echo $x
done
results:
file
1
file1
file2
I want to execute a command on each file. "file" and "1" above are not actual files. How can I do that if the filenames contains things like spaces or commas?
It's a little trickier than I think find -print0 | xargs -0 could handle, because I actually want the command to be something like "convert input/file1.jpg .... output/file1.jpg" so I need to permutate the filename in the process.
Actually, Mark's suggestion works fine without even doing anything to the internal field separator. The problem is running ls in a subshell, whether by backticks or $( ) causes the for loop to be unable to distinguish between spaces in names. Simply using
for f in *
instead of the ls solves the problem.
#!/bin/bash
for f in *
do
echo "$f"
done
UPDATE BY OP: this answer sucks and shouldn't be on top ... #Jordan's post below should be the accepted answer.
one possible way:
ls -1 | while read x; do
echo $x
done
I know this one is LONG past "answered", and with all due respect to eduffy, I came up with a better way and I thought I'd share it.
What's "wrong" with eduffy's answer isn't that it's wrong, but that it imposes what for me is a painful limitation: there's an implied creation of a subshell when the output of the ls is piped and this means that variables set inside the loop are lost after the loop exits. Thus, if you want to write some more sophisticated code, you have a pain in the buttocks to deal with.
My solution was to take the "readline" function and write a program out of it in which you can specify any specific line number that you may want that results from any given function call. ... As a simple example, starting with eduffy's:
ls_output=$(ls -1)
# The cut at the end of the following line removes any trailing new line character
declare -i line_count=$(echo "$ls_output" | wc -l | cut -d ' ' -f 1)
declare -i cur_line=1
while [ $cur_line -le $line_count ] ;
do
# NONE of the values in the variables inside this do loop are trapped here.
filename=$(echo "$ls_output" | readline -n $cur_line)
# Now line contains a filename from the preceeding ls command
cur_line=cur_line+1
done
Now you have wrapped up all the subshell activity into neat little contained packages and can go about your shell coding without having to worry about the scope of your variable values getting trapped in subshells.
I wrote my version of readline in gnuc if anyone wants a copy, it's a little big to post here, but maybe we can find a way...
Hope this helps,
RT