pipe then hyphen (stdin) as an alternative to for loop - linux

I wrote a few sed an awk commands to extract a set of IDs that are associated with file names. I would like to run a set of commands using these filenames from id.txt
cat id.txt
14235.gz
41231.gz
41234.gz
I usually write for loops as follows:
for i in $(cat id.txt);
do
command <options> $i
done
I thought I could also do cat id.txt | command <options> -
Is there a way to pipe the output of cat, awk, sed, etc, line by line into a command?

Use a while read loop see Don't read lines wit for
while IFS= read -r line_in_text_file; do
echo "$line_in_text_file"
done < id.txt

Commands don't usually get their filename arguments on standard input. Using - as an argument means to read the file contents from standard input instead of a named file, it doesn't mean to get the filename from stdin.
You can use command substitution to use the contents of the file as all the filename arguments to the command:
command <options> $(cat id.txt)
or you can use xargs
xargs command <options> < id.txt

Is there a way to pipe the output of cat, awk, sed, etc, line by line into a command?
Compound commands can be placed in a pipe, the syntax is not very strict. The usual:
awk 'some awk script' |
while IFS= read -r line; do
echo "$line"
done |
sed 'some sed script'
I avoid reading input line by line using a while read - it's very slow. It's way faster to use awk scripts and other commands.
Command groups can be used to:
awk 'some awk script' |
{ # or '(', but there is no need for a subshell
echo "header1,header2"
# remove first line
IFS= read -r first_line
# ignore last line
sed '$d'
} |
sed 'some sed script'
Remember that pipe command are run in a subshell, so variable changes will not affect parent shell.
Bash has process substitution extension that let's you run a while loop inside parent shell:
var=1
while IFS= read -r line; do
if [[ "$line" == 2 ]]; then
var=2
fi
done <(
seq 10 |
sed '$d'
)
echo "$var" # will output 2

xargs can do this
cat id.txt | xargs command
From xargs help
$ xargs --help
Usage: xargs [OPTION]... COMMAND [INITIAL-ARGS]...
Run COMMAND with arguments INITIAL-ARGS and more arguments read from input.
Mandatory and optional arguments to long options are also
mandatory or optional for the corresponding short option.
-0, --null items are separated by a null, not whitespace;
disables quote and backslash processing and
logical EOF processing
-a, --arg-file=FILE read arguments from FILE, not standard input
-d, --delimiter=CHARACTER items in input stream are separated by CHARACTER,
not by whitespace; disables quote and backslash
...

Related

how to not escape space and backslashes in echo and while in bash?

I'm passing two positional args to a script to run, both args are a path, and while in the scenario analyzing the paths, the problem is sometimes there is some path like: m i sc . . . . .. . . it has dots and spaces, and sometimes even we have a backslash in dir names.
It is so tried to get arguments via two procedures, directly and via at sign.
SOURCE_ARG=$1
DESTINATION_ARG=$2
and
ARG_COUNT=0
for POSITIONAL_ARGUMENTS in "${#}"
do
((ARG_COUNT++))
ARGUMENT_ARRAY[$ARG_COUNT]=$POSITIONAL_ARGUMENTS
done
In the loop, I iterate through the result of commands that have forwarded to them.
while IFS= read -r dir
do
echo "${ARGUMENT_ARRAY[1]}"
echo "${dir}"
while IFS= read -r item
do
# do some stuff
done < <(ls -A "$dir"/)
done < <(du -hP "$SOURCE_ARG" | awk '{$1=""; print $0}' | grep -v "^.$" | sed "s/^ //g")
when i use echo "${ARGUMENT_ARRAY[1]}" i get the same path as i need to check but when using loop iteration varible as dir in here ->echo "${dir}" i got all the spaces escaped, since other commands for that path could not do their jobs.
What I'm Asking for is that how can I get the output of $dir within the loop and as like as echo "${ARGUMENT_ARRAY[1]}" that i mentioned above(input with all spaces and backslashes)
Thanks to #Barmar in comments.
The only reason that filenames are without escapes (i.e. you see directories with no special character or special characters have been escaped) is because du is printing the filenames with escapes, so $dir variable would have escaped once and special characters are no longer available for the other loop iteration in my problem.
Now that we know the problem was raised by using du in my script:
while IFS= read -r dir
# do sth
done < <(du -hP "$SOURCE_ARG" | awk '{$1=""; print $0}' | grep -v "^.$" | sed "s/^ //g")
We can change the du to find and the problem is solved:
while IFS= read -r dir
# do sth
done < <(find "$SOURCE_ARG" -type d –)
PS 1:
Another problem raised as I wanted to print the lines to check them if they are ok or not (i.e. while debugging application) was with echo.
So be sure to try printf "%s\n" "$dir" instead of echo, as some versions of echo process escape sequences.
echo "${dir}"
printf "%s\n" "$dir"
PS 2:
Also If a filename has more than one space in a row, The way I used awk, was collapsing them into a single space.
awk '{$1=""; print $0}' | grep -v "^.$" | sed "s/^ //g"

How to move files where the first line contains a string?

I am currently using the following command:
grep -l -Z -E '.*?FindMyRegex' /home/user/folder/*.csv | xargs -0 -I{} mv {} /home/destination/folder
This works fine. The problem is it uses grep on the entire file.
I would like to use the grep command on the FIRST line of the file only.
I have tried to use head -1 file | at the beginning, but it did not work.
A change I would add to your script is -
for file in *.csv; do
head -1 "$file" | grep -l -Z -E '.*?FindMyRegex' | xargs -0 -I{} mv {} /home/destination/folder;
done
you can maybe try sed '1q' file.csv | grep ... to search the regexp only in the first line.
You don't need grep or find, as long as your files don't have embedded newlines.
I don't know an easy way off the top of my head to get sed to delimit with nulls.
mv $( for f in /home/user/folder/*.csv;
do sed -ns '1 { /yourPattern/F; q; }' $f;
done ) /home/destination/folder/
EDIT
Rewrote with a loop. This will run a separate instance of sed to check each file, but at least it shouldn't read beyond the first line. It will fail syntactically if there are no hits.
You might need -E depending on your regex.
-n says don't print records from the files.
-s says treat each file as a distinct input - this is so the filenames aren't always the first one.
This does require GNU sed for the F.
gawk 'FNR==1{if($0~/PATTERN/)
printf "mv %s %s\n",FILENAME, "/target";nextfile}' /path/*.csv
First of all, in your regex: .*?FindMyRegex the .*? doesn't make any sense, they could be removed.
The above awk (gawk) one-liner will build up mv file target command lines for you. You can check them, if you are satisfied with them, pipe the output to |sh , the commands are gonna be executed.
replace PATTERN by your regex pattern, and /target by the real target dir.
The one-liner is assuming that the filenames don't contain special chars (space i.e.), if it is the case, add "s to the mv cmd.
using GNU awk to find the filenames, pipe the filenames into xargs
gawk -v pattern="myRegex" '
FNR == 1 {if ($0 ~ pattern) printf "%s\0", FILENAME; nextfile}
' *.csv | xargs -0 echo mv -t destination
If it looks OK, remove "echo"
Try this Shellcheck-clean Bash code:
#! /bin/bash
shopt -s nullglob # Globs that match nothing expand to nothing
shopt -s dotglob # Globs match files whose names start with '.'
dest=/home/destination/folder
for file in *.csv ; do
head -n 1 -- "$file" | grep -qE '.*?FindMyRegex' && mv -- "$file" "$dest"
done
shopt -s nullglob prevents an error if there are no .csv files in the directory.
shopt -s dotglob ensures that files whose name starts with '.' are handled.
The -- in the options for head and mv ensures that files whose names begin with - are handled correctly.
The quotes in "$file" and "$dest" ensure that names that contain whitespace (actually $IFS) characters (including newlines) or glob metacharacters are handled correctly.
Note that the .*? in the reqular expression is probably redundant, and may not do what you think it does (grep -E doesn't do non-greedy matching).

Bash command using xargs and xargs -0

I just found difference between two commands:
echo sum.txt| xargs cat
This will output content of sum.txt
echo sum.txt| xargs -0 cat
This shows error:
cat: sum.txt
: No such file or directory
I know -0 will treat null bytes as delimiter. And i think the new line starts with : is because echo command produce new lines. And doesn't produce output like:
cat: sum.txt: No such file or directory
But if echo produce new lines why the first command can succeed? since xargs use white spaces as delimiter by default.
I think what is happening will become more clear if you replace cat with echo -n as an experiment.
echo sum.txt| xargs echo -n
echo sum.txt | xargs -0 echo -n
In the first example, xargs breaks on the newline and discards the newline, leaving just sum.txt.
In the second example, xargs breaks on the end of the EOF at the end of the output of the first echo, resulting in the string 'sum.txt\n'.

Pass a list of files to perl script via pipe

I am having a problem where my perl script will fail upon having an input piped, but works fine when I just list all the file names individually.
For reference, input of the perl script is read with while(<>).
Example:
script.pl file1.tag file2.tag file3.tag
runs fine.
But the following all fail.
find ./*.tag | chomp | script.pl
ls -l *.tag | perl -pe 's/\n/ /g' | script.pl
find ./*.tag | perl -pe 's/\n/ /g' | script.pl
I also tested dumping it into a text file and catting that into the perl:
cat files.text | script.pl
All of them fail the same way. It is like the script is passed no input arguments and the program just finishes.
From perldoc perlop:
The null filehandle <> is special [...] Input from <> comes either from standard input, or from each file listed on the command line. Here's how it works: the first time <> is evaluated, the #ARGV array is checked, and if it is empty, $ARGV[0] is set to -, which when opened gives you standard input. The #ARGV array is then processed as a list of filenames.
You're not passing any command line arguments to your Perl scripts, so everything you pipe into them is read into STDIN instead of being treated as filenames:
$ echo foo > foo.txt
$ echo bar > bar.txt
$ ls | perl -e 'print "<$_>\n" while <>'
<bar.txt
>
<foo.txt
>
Notice that the files foo.txt and bar.txt are not actually read; all we get is the file names. If you want the files to be opened and read, you have to pass them as command line arguments or explicitly set #ARGV:
$ perl -e 'print "<$_>\n" while <>' *
<bar
>
<foo
>
If you have a large number of files, like you're likely to get from find, you should use xargs as Dyno Hongjun Fu suggested.
However, you don't need find, ls, cat, or your Perl one-liner to run your script on all the .tag files in the current directory. Simply do:
script.pl *.tag
you need xargs, e.g.
find ./ -type f -name "*.tag" | xargs -i script.pl {}
what is chomp?

How can I prepend a string to the beginning of each line in a file?

I have the following bash code which loops through a text file, line by line .. im trying to prefix the work 'prefix' to each line but instead am getting this error:
rob#laptop:~/Desktop$ ./appendToFile.sh stusers.txt kp
stusers.txt
kp
./appendToFile.sh: line 11: /bin/sed: Argument list too long
115000_210org#house.com,passw0rd
This is the bash script ..
#!/bin/bash
file=$1
string=$2
echo "$file"
echo "$string"
for line in `cat $file`
do
sed -e 's/^/prefix/' $line
echo "$line"
done < $file
What am i doing wrong here?
Update:
Performing head on file dumps all the lines onto a single line of the terminal, probably related?
rob#laptop:~/Desktop$ head stusers.txt
rob#laptop:~/Desktop$ ouse.com,passw0rd
a one-line awk command should do the trick also:
awk '{print "prefix" $0}' file
Concerning your original error:
./appendToFile.sh: line 11: /bin/sed: Argument list too long
The problem is with this line of code:
sed -e 's/^/prefix/' $line
$line in this context is file name that sed is running against. To correct your code you should fix this line as such:
echo $line | sed -e 's/^/prefix/'
(Also note that your original code should not have the < $file at the end.)
William Pursell addresses this issue correctly in both of his suggestions.
However, I believe you have correctly identified that there is an issue with your original text file. dos2unix will not correct this issue, as it only strips the carriage returns Windows sticks on the end of lines. (However, if you are attempting to read a Linux file in Windows, you would get a mammoth line with no returns.)
Assuming that it is not an issue with the end of line characters in your text file, William Pursell's, Andy Lester's, or nullrevolution's answers will work.
A variation on the while read... suggestion:
while read -r line; do echo "PREFIX " $line; done < $file
This could be run directly from the shell (no need for a batch / script file):
while read -r line; do echo "kp" $line; done < stusers.txt
The entire loop can be replaced by a single sed command that operates on the entire file:
sed -e 's/^/prefix/' $file
A Perl way to do it would be:
perl -p -e's/^/prefix' filename
or
perl -p -e'$_ = "prefix $_"' filename
In either case, that reads from filename and prints the prefixed lines to STDOUT.
If you add a -i flag, then Perl will modify the file in place. You can also specify multiple filenames and Perl will magically do all of them.
Instead of the for loop, it is more appropriate to use while read...:
while read -r line; do
do
echo "$line" | sed -e 's/^/prefix/'
done < $file
But you would be much better off with the simpler:
sed -e 's/^/prefix/' $file
Use sed. Just change the word prefix.
sed -e 's/^/prefix/' file.ext
If you want to save the output in another file
sed -e 's/^/prefix/' file.ext > file_new.ext
You don't need sed, just concatenate the strings in the echo command
while IFS= read -r line; do
echo "prefix$line"
done < filename
Your loop iterates over each word in the file:
for line in `cat file`; ...
sed -i '1a\
Your Text' file1 file2 file3
A solution without sed/awk and while loops:
xargs -n1 printf "$prefix%s\n" < "$file"

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