I want to print the text at a specified line number from a file.
Here is my bash script
line=12
sed -n "$line{p;q;}"
My line number comes in a variable. But the above code is not working. What should I do?
Using sed
line=12
sed -n "${line}p" my_file
# Multiple lines
line1=10
line2=15
sed -n "${line1},${line2}p" my_file
In awk:
awk "NR==${line}" my_file
# Multiple lines
awk "NR >= ${line1} && NR <= ${line2}" my_file
Or using head and tail but probably not as efficient:
head -${line} my_file | tail -1
# Multiple lines
head -${line2} my_file | tail -$(($line2-$line1+1))
You have to give the file name as an argument to sed.
line=12
sed -n "$line{p;q;}" filename
If you are passing the filename as an argument to a bash script, you need to use:
line=12
sed -n "$line{p;q;}" "$1"
Fast sed command (useful for bigger files) is:
n=12; sed $n'q;d' file
Related
I have shell script variable var="7,8,9"
These are the line number use to delete to file using sed.
Here I tried:
sed -i "$var"'d' test_file.txt
But i got error `sed: -e expression #1, char 4: unknown command: ,'
Is there any other way to remove the line?
sed command doesn't accept comma delimited line numbers.
You can use this awk command that uses a bit if BASH string manipulation to form a regex with the given comma separated line numbers:
awk -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt
This will set awk variable var as this regex:
^(7|8|9)$
And then condition NR !~ var ensures that we print only those lines that don't match above regex.
For inline editing, if you gnu-awk with version > 4.0 then use:
awk -i inplace -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt
Or for older awk use:
awk -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt > $$.tmp && mv $$.tmp test_file.txt
I like sed, you were close to it. You just need to split each line number into a separate command. How about this:
sed -e "$(echo 1,3,4 | tr ',' '\n' | while read N; do printf '%dd;' $N; done)"
do like this:
sed -i "`echo $var|sed 's/,/d;/g'`d;" file
Another option to consider would be ed, with printf '%s\n' to put commands onto separate lines:
lines=( 9 8 7 )
printf '%s\n' "${lines[#]/%/d}" w | ed -s file
The array lines contains the line numbers to be deleted; it's important to put these in descending order! The expansion ${lines[#]/%/d} adds a d (delete) command to each line number and w writes to the file at the end. You can change this to ,p instead, to check the output before overwriting your file.
As an aside, for this example, you could also just use 7,9 as a single entry in the array.
I would like to remove a file name only from the following configuration file.
Configuration File -- test.conf
knowledgebase/arun/test.rf
knowledgebase/arunraj/tester/test.drl
knowledgebase/arunraj2/arun/test/tester.drl
The above file should be read. And removed contents should went to another file called output.txt
Following are my try. It is not working to me at all. I am getting empty files only.
#!/bin/bash
file=test.conf
while IFS= read -r line
do
# grep --exclude=*.drl line
# awk 'BEGIN {getline line ; gsub("*.drl","", line) ; print line}'
# awk '{ gsub("/",".drl",$NF); print line }' arun.conf
# awk 'NF{NF--};1' line arun.conf
echo $line | rev | cut -d'/' -f 1 | rev >> output.txt
done < "$file"
Expected Output :
knowledgebase/arun
knowledgebase/arunraj/tester
knowledgebase/arunraj2/arun/test
There's the dirname command to make it easy and reliable:
#!/bin/bash
file=test.conf
while IFS= read -r line
do
dirname "$line"
done < "$file" > output.txt
There are Bash shell parameter expansions that will work OK with the list of names given but won't work reliably for some names:
file=test.conf
while IFS= read -r line
do
echo "${line%/*}"
done < "$file" > output.txt
There's sed to do the job — easily with the given set of names:
sed 's%/[^/]*$%%' test.conf > output.txt
It's harder if you have to deal with names like /plain.file (or plain.file — the same sorts of edge cases that trip up the shell expansion).
You could add Perl, Python, Awk variants to the list of ways of doing the job.
You can get the path like this:
path=${fullpath%/*}
It cuts away the string after the last /
Using awk one liner you can do this:
awk 'BEGIN{FS=OFS="/"} {NF--} 1' test.conf
Output:
knowledgebase/arun
knowledgebase/arunraj/tester
knowledgebase/arunraj2/arun/test
Currently, I have a command in a bash script that greps for a given string in a text file and prints the line numbers only using sed ...
grep -n "<string>" file.txt | sed -n 's/^\([0-9]*\).*/\1/p'
The grep could find multiple matches, and thus, print multiple line numbers. From this command's output, I would like to extract the minimum and maximum values, and assign those to respective bash variables. How could I best modify my existing command or add new commands to accomplish this? If using awk or sed will be necessary, I have a preference of using sed. Thanks!
You can get the minimum and maximum with this:
grep -n "<string>" input | sed -n -e 's/^\([0-9]*\).*/\1/' -e '1p;$p'
You can also read them into an array:
F=($(grep -n "<string>" input | sed -n -e 's/^\([0-9]*\).*/\1/' -e '1p;$p'))
echo ${F[0]} # min
echo ${F[1]} # max
grep -n "<string>" file.txt | sed -n -e '1s/^\([0-9]*\).*/\1/p' -e '$s/^\([0-9]*\).*/\1/p'
grep .... |awk -F: '!f{print $1;f=1} END{print $1}'
Here's how I'd do it, since grep -n 'pattern' file prints output in the format line number:line contents ...
minval=$(grep -n '<string>' input | cut -d':' -f1 | sort -n | head -1)
maxval=$(grep -n '<string>' input | cut -d':' -f1 | sort -n | tail -1)
the cut -d':' -f1 command splits the grep output around the colon and pulls out only the first field (the line numbers), sort -n sorts the numeric line numbers in ascending order (which they would already be in, but it's good practice to ensure it), then head -1 and tail -1 remove the first, and last value in the sorted list respectively, i.e. the minimum and maximum values and assign them to variables $minval and $maxval respectively.
Hope this helps!
Edit: Turns out you can't do it the way I had it originally, since echoing out a list of newline-separated values apparently concatenates them into one line.
It can be done with one process. Like this:
awk '/expression/{if(!n)print NR;n=NR} END {print n}' file.txt
Then You can assign to an array (as perreal suggested). Or You can modify this script and assign to varables using eval
eval $(awk '/expression/{if(!n)print "A="NR;n=NR} END {print "B="n}' file.txt)
echo $A
echo $B
Output (file.txt contains three lines of expression)
1
3
I have number of files which have similar names like
DWH_Export_AUSTA_20120701_20120731_v1_1.csv.397.dat.2012-10-02 04-01-46.out
DWH_Export_AUSTA_20120701_20120731_v1_2.csv.397.dat.2012-10-02 04-03-12.out
DWH_Export_AUSTA_20120801_20120831_v1_1.csv.397.dat.2012-10-02 04-04-16.out
etc.
I need to get number before .csv(1 or 2) from the file name and put it into end of every line in file with TAB separator.
I have written this code, it finds number that I need, but i do not know how to put this number into file. There is space in the filename, my script breaks because of it.
Also I am not sure, how to send to script list of files. Now I am working only with one file.
My code:
#!/bin/sh
string="DWH_Export_AUSTA_20120701_20120731_v1_1.csv.397.dat.2012-10-02 04-01-46.out"
out=$(echo $string | awk 'BEGIN {FS="_"};{print substr ($7,0,1)}')
awk ' { print $0"\t$out" } ' $string
for file in *
do
sfx=$(echo "$file" | sed 's/.*_\(.*\).csv.*/\1/')
sed -i "s/$/\t$sfx/" "$file"
done
Using sed:
$ sed 's/.*_\(.*\).csv.*/&\t\1/' file
DWH_Export_AUSTA_20120701_20120731_v1_1.csv.397.dat.2012-10-02 04-01-46.out 1
DWH_Export_AUSTA_20120701_20120731_v1_2.csv.397.dat.2012-10-02 04-03-12.out 2
DWH_Export_AUSTA_20120801_20120831_v1_1.csv.397.dat.2012-10-02 04-04-16.out 1
To make this for many files:
sed 's/.*_\(.*\).csv.*/&\t\1/' file1 file2 file3
OR
sed 's/.*_\(.*\).csv.*/&\t\1/' file*
To make this changed get saved in the same file(If you have GNU sed):
sed -i 's/.*\(.\).csv.*/&\t\1/' file
Untested, but this should do what you want (extract the number before .csv and append that number to the end of every line in the .out file)
awk 'FNR==1 { split(FILENAME, field, /[_.]/) }
{ print $0"\t"field[7] > FILENAME"_aaaa" }' *.out
for file in *_aaaa; do mv "$file" "${file/_aaaa}"; done
If I understood correctly, you want to append the number from the filename to every line in that file - this should do it:
#!/bin/bash
while [[ 0 < $# ]]; do
num=$(echo "$1" | sed -r 's/.*_([0-9]+).csv.*/\t\1/' )
#awk -e "{ print \$0\"\t${num}\"; }" < "$1" > "$1.new"
#sed -r "s/$/\t$num/" < "$1" > "$1.mew"
#sed -ri "s/$/\t$num/" "$1"
shift
done
Run the script and give it names of the files you want to process. $# is the number of command line arguments for the script which is decremented at the end of the loop by shift, which drops the first argument, and shifts the other ones. Extract the number from the filename and pick one of the three commented lines to do the appending: awk gives you more flexibility, first sed creates new files, second sed processes them in-place (in case you are running GNU sed, that is).
Instead of awk, you may want to go with sed or coreutils.
Grab number from filename, with grep for variety:
num=$(<<<filename grep -Eo '[^_]+\.csv' | cut -d. -f1)
<<<filename is equivalent to echo filename.
With sed
Append num to each line with GNU sed:
sed "s/\$/\t$num" filename
Use the -i switch to modify filename in-place.
With paste
You also need to know the length of the file for this method:
len=$(<filename wc -l)
Combine filename and num with paste:
paste filename <(seq $len | while read; do echo $num; done)
Complete example
for filename in DWH_Export*; do
num=$(echo $filename | grep -Eo '[^_]+\.csv' | cut -d. -f1)
sed -i "s/\$/\t$num" $filename
done
How can I delete the first (!) line of a text file if it's empty, using e.g. sed or other standard UNIX tools. I tried this command:
sed '/^$/d' < somefile
But this will delete the first empty line, not the first line of the file, if it's empty. Can I give sed some condition, concerning the line number?
With Levon's answer I built this small script based on awk:
#!/bin/bash
for FILE in $(find some_directory -name "*.csv")
do
echo Processing ${FILE}
awk '{if (NR==1 && NF==0) next};1' < ${FILE} > ${FILE}.killfirstline
mv ${FILE}.killfirstline ${FILE}
done
The simplest thing in sed is:
sed '1{/^$/d}'
Note that this does not delete a line that contains all blanks, but only a line that contains nothing but a single newline. To get rid of blanks:
sed '1{/^ *$/d}'
and to eliminate all whitespace:
sed '1{/^[[:space:]]*$/d}'
Note that some versions of sed require a terminator inside the block, so you might need to add a semi-colon. eg sed '1{/^$/d;}'
Using sed, try this:
sed -e '2,$b' -e '/^$/d' < somefile
or to make the change in place:
sed -i~ -e '2,$b' -e '/^$/d' somefile
If you don't have to do this in-place, you can use awk and redirect the output into a different file.
awk '{if (NR==1 && NF==0) next};1' somefile
This will print the contents of the file except if it's the first line (NR == 1) and it doesn't contain any data (NF == 0).
NR the current line number,NF the number of fields on a given line separated by blanks/tabs
E.g.,
$ cat -n data.txt
1
2 this is some text
3 and here
4 too
5
6 blank above
7 the end
$ awk '{if (NR==1 && NF==0) next};1' data.txt | cat -n
1 this is some text
2 and here
3 too
4
5 blank above
6 the end
and
cat -n data2.txt
1 this is some text
2 and here
3 too
4
5 blank above
6 the end
$ awk '{if (NR==1 && NF==0) next};1' data2.txt | cat -n
1 this is some text
2 and here
3 too
4
5 blank above
6 the end
Update:
This sed solution should also work for in-place replacement:
sed -i.bak '1{/^$/d}' somefile
The original file will be saved with a .bak extension
Delete the first line of all files under the actual directory if the first line is empty :
find -type f | xargs sed -i -e '2,$b' -e '/^$/d'
This might work for you:
sed '1!b;/^$/d' file