I want to write a command with removes an item from a file .todolist like that:
path="$HOME/.todolist"
if [ "$task" == "- " ];
then
exit 1
fi
cat $path | grep -v $1 > $path
When I cat $path | grep -v $1 all works perfect but when I try writing back to the file it leaves the file empty. What am I doing wrong?
As simple as that-
path="$HOME/.todolist"
grep -v $1 $path | sponge $path
This may be simpler:
sed -ie "/$1/d" $path
That will delete any line in the file containing the value in $1. The -i option tells sed to edit the file in place (not really, but it works as if it did). If $1 contains /s you'll get an error, though, but you can use other delimiters if you need to:
sed -ie "\%$1%d" $path
Note that with an alternate delimiter you must escape the first one with "\".
path="$HOME/.todolist"
if [ "$task" == "- " ];
then
exit 1
fi
a=$1
export a
perl -pi -e 'undef $_ if(/$ENV{a}/)' $path
Related
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"
I'm studying about how to use 'cut'.
#!/bin/bash
for file in *.c; do
name = `$file | cut -d'.' -f1`
gcc $file -o $name
done
What's wrong with the following code?
There are a number of problems on this line:
name = `$file | cut -d'.' -f1`
First, to assign a shell variable, there should be no whitespace around the assignment operator:
name=`$file | cut -d'.' -f1`
Secondly, you want to pass $file to cut as a string, but what you're actually doing is trying to run $file as if it were an executable program (which it may very well be, but that's beside the point). Instead, use echo or shell redirection to pass it:
name=`echo $file | cut -d. -f1`
Or:
name=`cut -d. -f1 <<< $file`
I would actually recommend that you approach it slightly differently. Your solution will break if you get a file like foo.bar.c. Instead, you can use shell expansion to strip off the trailing extension:
name=${file%.c}
Or you can use the basename utility:
name=`basename $file .c`
You should use the command substitution (https://www.gnu.org/software/bash/manual/bashref.html#Command-Substitution) to execute a command in a script.
With this the code will look like this
#!/bin/bash
for file in *.c; do
name=$(echo "$file" | cut -f 1 -d '.')
gcc $file -o $name
done
With the echo will send the $file to the standard output.
Then the pipe will trigger after the command.
The cut command with the . delimiter will split the file name and will keep the first part.
This is assigned to the name variable.
Hope this answer helps
I have the following grep:
grep -Po '(?<=PROGRAM\()[^\)]+(?=\))' /home/programs/hello_word.sh
Wich displays the string between PROGRAM( and ):
RECTONTER
Then, I need to know if these string extracted is contained in a file, so:
grep -Po '(?<=PROGRAM\()[^\)]+(?=\))' /home/programs/hello_word.sh | xargs -I % grep -e % /home/leherad/pgm_currentdate
File content:
RECTONTER
CORASFE
RENTOASD
UBICARP
If its found, returns the line of /home/leherad/pgm_currentdate, but I want to print the line extracted in the first grep (RECTONTER). If not found, then wouldn't return nothing.
There is a simple way to do this, or I should not complicate and would be better build a script and save the first grep in a variable?
You can store it on a variable first:
read -r FIRST < <(exec grep -Po '(?<=PROGRAM\()[^\)]+(?=\))' /home/programs/hello_word.sh) && grep -e "$FIRST" /home/leherad/pgm_currentdate
Update 01
#!/bin/bash
shopt -s nullglob
for FILE in /home/programs/*; do
read -r FIRST < <(exec grep -Po '(?<=PROGRAM\()[^\)]+(?=\))' "$FILE") && grep -e "$FIRST" /home/leherad/pgm_currentdate && echo "$FIRST"
done
I think a straightforward way to solve this is to use a function.
Also, your grep pattern will match shell comments, which could cause unexpected behavior in your xargs command when there are more than one matches; you might want to take steps to only grab the first match. It's hard to say without actually seeing the input files, so I'm guessing this is either ok or comments are actually the expected place for your target pattern.
Anyway, here's my best guess at a function that would work for you.
get_program() {
local filename="$1"
local program="$( grep -m1 -Po '(?<=PROGRAM\()[^\)]+(?=\))' "$filename" )"
if grep -q -e "$program" /home/leherad/pgm_currentdate; then
echo $program
grep -e "$program" /home/leherad/pgm_currentdate
fi
}
get_program /home/programs/hello_word.sh
For ex the file is this:
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
I want to rename this file to:
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using ${parameter%word} (Remove matching suffix pattern):
$ echo "$fn"
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
$ echo "${fn%:*}"
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using cut
$ echo $fn
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
$ echo $fn |cut -d: -f1
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using awk
echo $fn |awk -F : '{print $1}'
more ways...
According to the link here:
This should work:
awk '{old=$0;gsub(/...$/,"",$0);system("mv \""old"\" "$0)}'
provided the file name is given as input.
For eg:
ls -1 NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00|nawk '{old=$0;gsub(/...$/,"",$0);system("mv \""old"\" "$0)}'
Rename file using bash string manipulations:
# Filename needs to be in a variable
file=NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
# Rename file
mv "$file" "${file%???}"
This removes the last three characters from filename.
Using just bash:
fn='NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00'
mv "$fn" "${fn::-3}"
if you have Ruby
echo NBDG6_CD* | ruby -e 'f=gets.chomp;File.rename(f, f[0..-4])'
I have the following bash code which loops through a text file, line by line .. im trying to prefix the work 'prefix' to each line but instead am getting this error:
rob#laptop:~/Desktop$ ./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"