I have a source input, input.txt
a.txt
b.txt
c.txt
I want to feed these input into a program as the following:
my-program --file=a.txt --file=b.txt --file=c.txt
So I try to use xargs, but with no luck.
cat input.txt | xargs -i echo "my-program --file"{}
It gives
my-program --file=a.txt
my-program --file=b.txt
my-program --file=c.txt
But I want
my-program --file=a.txt --file=b.txt --file=c.txt
Any idea?
Don't listen to all of them. :) Just look at this example:
echo argument1 argument2 argument3 | xargs -l bash -c 'echo this is first:$0 second:$1 third:$2'
Output will be:
this is first:argument1 second:argument2 third:argument3
None of the solutions given so far deals correctly with file names containing space. Some even fail if the file names contain ' or ". If your input files are generated by users, you should be prepared for surprising file names.
GNU Parallel deals nicely with these file names and gives you (at least) 3 different solutions. If your program takes 3 and only 3 arguments then this will work:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo a2.txt; echo b2.txt; echo c2.txt;) |
parallel -N 3 my-program --file={1} --file={2} --file={3}
Or:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo a2.txt; echo b2.txt; echo c2.txt;) |
parallel -X -N 3 my-program --file={}
If, however, your program takes as many arguments as will fit on the command line:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo d1.txt; echo e1.txt; echo f1.txt;) |
parallel -X my-program --file={}
Watch the intro video to learn more: http://www.youtube.com/watch?v=OpaiGYxkSuQ
How about:
echo $'a.txt\nb.txt\nc.txt' | xargs -n 3 sh -c '
echo my-program --file="$1" --file="$2" --file="$3"
' argv0
It's simpler if you use two xargs invocations: 1st to transform each line into --file=..., 2nd to actually do the xargs thing ->
$ cat input.txt | xargs -I# echo --file=# | xargs echo my-program
my-program --file=a.txt --file=b.txt --file=c.txt
You can use sed to prefix --file= to each line and then call xargs:
sed -e 's/^/--file=/' input.txt | xargs my-program
Here is a solution using sed for three arguments, but is limited in that it applies the same transform to each argument:
cat input.txt | sed 's/^/--file=/g' | xargs -n3 my-program
Here's a method that will work for two args, but allows more flexibility:
cat input.txt | xargs -n 2 | xargs -I{} sh -c 'V="{}"; my-program -file=${V% *} -file=${V#* }'
I stumbled on a similar problem and found a solution which I think is nicer and cleaner than those presented so far.
The syntax for xargs that I have ended with would be (for your example):
xargs -I X echo --file=X
with a full command line being:
my-program $(cat input.txt | xargs -I X echo --file=X)
which will work as if
my-program --file=a.txt --file=b.txt --file=c.txt
was done (providing input.txt contains data from your example).
Actually, in my case I needed to first find the files and also needed them sorted so my command line looks like this:
my-program $(find base/path -name "some*pattern" -print0 | sort -z | xargs -0 -I X echo --files=X)
Few details that might not be clear (they were not for me):
some*pattern must be quoted since otherwise shell would expand it before passing to find.
-print0, then -z and finally -0 use null-separation to ensure proper handling of files with spaces or other wired names.
Note however that I didn't test it deeply yet. Though it seems to be working.
xargs doesn't work that way. Try:
myprogram $(sed -e 's/^/--file=/' input.txt)
It's because echo prints a newline. Try something like
echo my-program `xargs --arg-file input.txt -i echo -n " --file "{}`
I was looking for a solution for this exact problem and came to the conclution of coding a script in the midle.
to transform the standard output for the next example use the -n '\n' delimeter
example:
user#mybox:~$ echo "file1.txt file2.txt" | xargs -n1 ScriptInTheMiddle.sh
inside the ScriptInTheMidle.sh:
!#/bin/bash
var1=`echo $1 | cut -d ' ' -f1 `
var2=`echo $1 | cut -d ' ' -f2 `
myprogram "--file1="$var1 "--file2="$var2
For this solution to work you need to have a space between those arguments file1.txt and file2.txt, or whatever delimeter you choose, one more thing, inside the script make sure you check -f1 and -f2 as they mean "take the first word and take the second word" depending on the first delimeter's position found (delimeters could be ' ' ';' '.' whatever you wish between single quotes .
Add as many parameters as you wish.
Problem solved using xargs, cut , and some bash scripting.
Cheers!
if you wanna pass by I have some useful tips http://hongouru.blogspot.com
Actually, it's relatively easy:
... | sed 's/^/--prefix=/g' | xargs echo | xargs -I PARAMS your_cmd PARAMS
The sed 's/^/--prefix=/g' is optional, in case you need to prefix each param with some --prefix=.
The xargs echo turns the list of param lines (one param in each line) into a list of params in a single line and the xargs -I PARAMS your_cmd PARAMS allows you to run a command, placing the params where ever you want.
So cat input.txt | sed 's/^/--file=/g' | xargs echo | xargs -I PARAMS my-program PARAMS does what you need (assuming all lines within input.txt are simple and qualify as a single param value each).
There is another nice way of doing this, if you do not know the number of files upront:
my-program $(find . -name '*.txt' -printf "--file=%p ")
Nobody has mentioned echoing out from a loop yet, so I'll put that in for completeness sake (it would be my second approach, the sed one being the first):
for line in $(< input.txt) ; do echo --file=$line ; done | xargs echo my-program
Old but this is a better answer:
cat input.txt | gsed "s/\(.*\)/\-\-file=\1/g" | tr '\n' ' ' | xargs my_program
# i like clean one liners
gsed is just gnu sed to ensure syntax matches version brew install gsed or just sed if your on gnu linux already...
test it:
cat input.txt | gsed "s/\(.*\)/\-\-file=\1/g" | tr '\n' ' ' | xargs echo my_program
Related
I've got a command to perform a series of commands that produce a variable output string such as 123456. I want to pipe that to a sed command replacing a known string in a csv file that looks like this:
Fred,Wilma,Betty,Barney
However, the command below does not work and I haven't found any other references to using pipe values as the variable for a replace.
How does this code change if the values in the csv are in a random order and I always want to change the second value?
Example code:
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 > test.txt |
sed -i -e '/Wilma/ r test.txt' -e 's/Wilma//' input.csv
Contents of input.csv should become: Fred,0d522cd316,Betty,Barney
Okay, in
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 > test.txt | sed -i -e '/Wilma/ r test.txt' -e 's/Wilma//' input.csv
you have a bug. That "> test.txt" after cut is going to eat your stdin on sed, so things go weird with that pipe afterwards taking stdin. You don't want a pipe there, or you don't want to redirect to a file.
The way to take piped stdin and use it as a parameter in a command is through xargs.
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 | xargs --replace=INSERTED -- sed -i -e 's/Wilma/INSERTED/' input.csv
(...though that find|shasum is suspect too, in that the order of files is random(ish) and it matters for a reliable sum. You prpobably mean to "|sort" after find.)
(Some would sed -i -e "s/Wilma/$(find|sort|shasum|cut)" f, but I ain't among them. Animals.)
For replacing a fixed string like "Wilma", try:
sed -i 's/Wilma/'"$(find / -iname awk 2>/dev/null |
sha256sum | cut -c1-10)"'/' input.csv
To replace the 2nd field no matter what's in it, try:
sed -i 's/[^,]*/'"$(find / -iname awk 2>/dev/null |
sha256sum | cut -c1-10)"'/2' input.csv
I want to do the equivalent of:
for i in `ls -1`; do echo $i | mv $i `sed 's/profile/account/'`; done
with xargs
ls -1 | xargs -I{} mv {} `echo {} | sed 's/profile/account/'`
But the sed after the pipe within backticks is ignored.
Anyone know why that's the case?
Edit: More info
The root problem is simply to rename files in a directly given a pattern replacement (here replace profile with account, however that is solved with for in loop over the files.
The question I'm posing is why is the following not working.
ls -1 | xargs -I{} mv {} `echo {} | sed 's/profile/account/'`
Why does the
`echo {} | sed 's/profile/account/'`
portion not return the replaced filename but the original filename. It's as if the | doesn't work inside the backticks.
To write the problem differently:
If I have a list like in a file called list.txt
profile1.txt
profile2.txt
And want to generate
account1.txt
account2.txt
While operating on each individual line separately so I can run a command on it.
cat list.txt | xargs -I{} echo `echo {} | sed 's/profile/account/'`
Why does command return:
profile1.txt
profile2.txt
Instead of changing profile to account?
This might work for you (GNU parallel):
parallel 'old={}; new=${old/profile/account}; echo mv $old $new' ::: *
Remove echo once checked.
Agree with the comments above regarding using shell glob expansion rather than ls -1, or the straightforward solution using prename, nevertheless the exercise is interesting also in case you wanted to feed the pipe other outputs (e.g. find or alike).
Below does it via calling a mini scriptlet where its argument is each xargs fed line, using var replacement bash-isms (note the narrowed ls ... to be able to re-run these without errors):
ls -1 *profile* | xargs -l1 bash -xc 'echo mv "$1" "${1/profile/account}"' --
Note above is a dry-run, remove the echo to actually do it.
In the example:
cat list.txt | xargs -I{} echo `echo {} | sed 's/profile/account/'`
xargs does not interpolate the {} inside of backticks.
So:
`echo {} | sed 's/profile/account/'`
Echos the literal {}. The subsequent sed does not change the {}.
`echo {} | sed 's/profile/account/'`
Will return: {}
Which gives:
cat list.txt | xargs -I{} echo {}
I want to move the output of the command:
ls -1 /${TMP_DIR}/*0000000221*.dbf | xargs | sed 's/ /,/g'
In the end of a command that come after it, like that:
ls -1 /${TMP_DIR}/*0000000221*.dbf | xargs | sed 's/ /,/g' | impdp sim/sim files=$1
For example:
execute ls -1 /${TMP_DIR}/*0000000221*.dbf | xargs | sed 's/ /,/g' will give me:
/tmp/a_0000000221.dbf,/tmp/a_00000002212.dbf,/tmp/b_0000000221.dbf
So I want the final command will look like:
impdp sim/sim files=/tmp/a_0000000221.dbf,/tmp/a_00000002212.dbf,/tmp/b_0000000221.dbf
EDIT:
Sorry I didnt write this from the beginning - I've variable in the command ${TMP_DIR}
You probably don't need that many pipes. You can use it like this:
printf "impdp sim/sim files=" && printf "%s," /tmp/*0000000221*.dbf
impdp sim/sim files=/tmp/a_0000000221.dbf,/tmp/a_00000002212.dbf,/tmp/b_0000000221.dbf,
ls is a bit redundant if you just want to get the file names.
You can get the shell to glob those and then use printf to put them one per line.
To separate those items with ',' rather than '\n', you can use paste
Finally, putting all that within $() will execute that in a subshell,
and output the result for the command in the current shell.
impdp sim/sim files=$(printf '%s\n' /${TMP_DIR}/*0000000221*.dbf | paste -d, -s)
You can try other order of commands:
impdp sim/sim files=$(ls -1 /tmp/*0000000221*.dbf | xargs | sed 's/ /,/g')
You can use globbing, an array and IFS to construct the parameter string:
$ ls -1
1.txt
2.txt
3.txt
$ echo impdp sim/sim files="$(a=(*.txt);IFS=',';echo "${a[*]}")"
impdp sim/sim files=1.txt,2.txt,3.txt
Obviously this will break on filenames with spaces or newlines.
To run, just remove the echo.
(all solutions including mine assumes your filenames do not contain spaces)
sed is a little overkill, you can use tr and avoid xargs too:
impdp sim/sim files=$(ls /tmp/*0000000221*.dbf | tr "\n" ",")
I need help using xargs(1) and bc(1) in the same line. I can do it multiple lines, but I really want to find a solution in one line.
Here is the problem: The following line will print the size of a file.txt
ls -l file.txt | cut -d" " -f5
And, the following line will print 1450 (which is obviously 1500 - 50)
echo '1500-50' | bc
Trying to add those two together, I do this:
ls -l file.txt | cut -d" " -f5 | xargs -0 -I {} echo '{}-50' | bc
The problem is, it's not working! :)
I know that xargs is probably not the right command to use, but it's the only command I can find who can let me decide where to put the argument I get from the pipe.
This is not the first time I'm having issues with this kind of problem. It will be much of a help..
Thanks
If you do
ls -l file.txt | cut -d" " -f5 | xargs -0 -I {} echo '{}-50'
you will see this output:
23
-50
This means, that bc does not see a complete expression.
Just use -n 1 instead of -0:
ls -l file.txt | cut -d" " -f5 | xargs -n 1 -I {} echo '{}-50'
and you get
23-50
which bc will process happily:
ls -l file.txt | cut -d" " -f5 | xargs -n 1 -I {} echo '{}-50' | bc
-27
So your basic problem is, that -0 expects not lines but \0 terminated strings. And hence the newline(s) of the previous commands in the pipe garble the expression of bc.
This might work for you:
ls -l file.txt | cut -d" " -f5 | sed 's/.*/&-50/' | bc
Infact you could remove the cut:
ls -l file.txt | sed -r 's/^(\S+\s+){4}(\S+).*/\2-50/' | bc
Or use awk:
ls -l file.txt | awk '{print $5-50}'
Parsing output from the ls command is not the best idea. (really).
you can use many other solutions, like:
find . -name file.txt -printf "%s\n"
or
stat -c %s file.txt
or
wc -c <file.txt
and can use bash arithmetics, for avoid unnecessary slow process forks, like:
find . -type f -print0 | while IFS= read -r -d '' name
do
size=$(wc -c <$name)
s50=$(( $size - 50 ))
echo "the file=$name= size:$size minus 50 is: $s50"
done
Here is another solution, which only use one external command: stat:
file_size=$(stat -c "%s" file.txt) # Get the file size
let file_size=file_size-50 # Subtract 50
If you really want to combine them into one line:
let file_size=$(stat -c "%s" file.txt)-50
The stat command gets you the file size in bytes. The syntax above is for Linux (I tested against Ubuntu). On the Mac the syntax is a little different:
let file_size=$(stat -f "%z" mini.csv)-50
I'm having some rather unusual problems using grep in a bash script. Below is an example of the bash script code that I'm using that exhibits the behaviour:
UNIQ_SCAN_INIT_POINT=1
cat "$FILE_BASENAME_LIST" | uniq -d >> $UNIQ_LIST
sed '/^$/d' $UNIQ_LIST >> $UNIQ_LIST_FINAL
UNIQ_LINE_COUNT=`wc -l $UNIQ_LIST_FINAL | cut -d \ -f 1`
while [ -n "`cat $UNIQ_LIST_FINAL | sed "$UNIQ_SCAN_INIT_POINT"'q;d'`" ]; do
CURRENT_LINE=`cat $UNIQ_LIST_FINAL | sed "$UNIQ_SCAN_INIT_POINT"'q;d'`
CURRENT_DUPECHK_FILE=$FILE_DUPEMATCH-$CURRENT_LINE
grep $CURRENT_LINE $FILE_LOCTN_LIST >> $CURRENT_DUPECHK_FILE
MATCH=`grep -c $CURRENT_LINE $FILE_BASENAME_LIST`
CMD_ECHO="$CURRENT_LINE matched $MATCH times," cmd_line_echo
echo "$CURRENT_DUPECHK_FILE" >> $FILE_DUPEMATCH_FILELIST
let UNIQ_SCAN_INIT_POINT=UNIQ_SCAN_INIT_POINT+1
done
On numerous occasions, when grepping for the current line in the file location list, it has put no output to the current dupechk file even though there have definitely been matches to the current line in the file location list (I ran the command in terminal with no issues).
I've rummaged around the internet to see if anyone else has had similar behaviour, and thus far all I have found is that it is something to do with buffered and unbuffered outputs from other commands operating before the grep command in the Bash script....
However no one seems to have found a solution, so basically I'm asking you guys if you have ever come across this, and any idea/tips/solutions to this problem...
Regards
Paul
The `problem' is the standard I/O library. When it is writing to a terminal
it is unbuffered, but if it is writing to a pipe then it sets up buffering.
try changing
CURRENT_LINE=`cat $UNIQ_LIST_FINAL | sed "$UNIQ_SCAN_INIT_POINT"'q;d'`
to
CURRENT LINE=`sed "$UNIQ_SCAN_INIT_POINT"'q;d' $UNIQ_LIST_FINAL`
Are there any directories with spaces in their names in $FILE_LOCTN_LIST? Because if they are, those spaces will need escaped somehow. Some combination of find and xargs can usually deal with that for you, especially xargs -0
A small bash script using md5sum and sort that detects duplicate files in the current directory:
CURRENT="" md5sum * |
sort |
while read md5sum filename;
do
[[ $CURRENT == $md5sum ]] && echo $filename is duplicate;
CURRENT=$md5sum;
done
you tagged linux, some i assume you have tools like GNU find,md5sum,uniq, sort etc. here's a simple example to find duplicate files
$ echo "hello world">file
$ md5sum file
6f5902ac237024bdd0c176cb93063dc4 file
$ cp file file1
$ md5sum file1
6f5902ac237024bdd0c176cb93063dc4 file1
$ echo "blah" > file2
$ md5sum file2
0d599f0ec05c3bda8c3b8a68c32a1b47 file2
$ find . -type f -exec md5sum "{}" \; |sort -n | uniq -w32 -D
6f5902ac237024bdd0c176cb93063dc4 ./file
6f5902ac237024bdd0c176cb93063dc4 ./file1