This question already has answers here:
Delete line from file at specified line number in bourne shell [duplicate]
(3 answers)
Closed 6 years ago.
Hi i have a linenumber
i=10
Now I want to delete that line with sed
sed '$i d' file
But it looke like that this wont work..
any ideas?
In awk. First test material:
$ cat > foo
1
2
3
Set the i:
$ i=2
Awk it:
$ awk -v line="$i" 'NR!=line' foo
1
3
sed -i.bak "${i}d" data.txt
is what you're looking for.
Notes
The -i option with sed is used for inplace edit. A backup with extension .bak is created.
The double quotes with sed expands the shell variables
To delete second line and show result:
sed -e '2d' data.txt
So your answer is:
sed -e "$i d" file.txt > file.txt
Just add strong quotes around the quoted variable:
i=10
sed ''"$i"' d' file
Related
This question already has answers here:
Replace a string in shell script using a variable
(12 answers)
Closed 4 years ago.
I am trying to using sed -i command to insert a string variable in the 1st line of a text file.
This command work : sed -i '1st header' file.txt
But when i pass a variable this doesn't work.
example :
var=$(cat <<-END
This is line one.
This is line two.
This is line three.
END
)
sed -i '1i $var' file.txt # doesn't work
sed -i ’1i $var’ file.txt # doesn't work
Any help with this problem
Thank you
First, let's define your variable a simpler way:
$ var="This is line one.
This is line two.
This is line three."
Since sed is not good at working with variables, let's use awk. This will place your variable at the beginning of a file:
awk -v x="$var" 'NR==1{print x} 1' file.txt
How it works
-v x="$var"
This defines an awk variable x to have the value of shell variable $var.
NR==1{print x}
At the first line, this tells awk to insert the value of variable x.
1
This is awk's shorthand for print-the-line.
Example
Let's define your variable:
$ var="This is line one.
> This is line two.
> This is line three."
Let's work on this test file:
$ cat File
1
2
This is what the awk command produces:
$ awk -v x="$var" 'NR==1{print x} 1' File
This is line one.
This is line two.
This is line three.
1
2
Changing a file in-place
To change file.txt in place using a recent GNU awk:
awk -i inplace -v x="$var" 'NR==1{print x} 1' file.txt
On macOS, BSD or older GNU/Linux, use:
awk -v x="$var" 'NR==1{print x} 1' file.txt >tmp && mv tmp file.txt
Using printf...
$ var="This is line one.
This is line two.
This is line three.
"
Use cat - to read from stdin and then print into a new file. Move it to the original file if you want to modify it.
$ printf "$var" | cat - file > newfile && mv newfile file;
Not the best job for sed. What about a simple cat ?
cat - file.txt <<EOF > newfile.txt
This is line one.
This is line two.
This is line three.
EOF
# you can add mv, if you really want the original file gone
mv newfile.txt file.txt
And for the original problem - sed does not like newlines and spaces in it's 'program', you need to quote and escape the line breaks:
# this works
sed $'1i "abc\\\ncde"' file.txt
# this does not, executes the `c` command from the second line
sed $'1i "abc\ncde"' file.txt
This question already has answers here:
Replace first few lines with first few lines from other file
(3 answers)
Closed 4 years ago.
i have 2 txt files : file1 and file2
i would like to replace the first line of file2 with the first line of file1, with a bash command
file1:
aaaaaaaa
bbbbbbbb
cccccccc
file2:
zzzzzzzz
yyyyyyyy
wwwwwwww
expected result of file2:
aaaaaaaa
yyyyyyyy
wwwwwwww
this can't be done with sed as you don't know what to replace with what...i'm right? so how to do this ?
EDIT:
so in my particular case (i do it in my openwrt router), what worked is :
sed -i "1c $(sed 1q file1)" file2
Thanks to #Sundeed for the link explaining why some commands were only displaying the results in the shell but not writing in the file : https://mywiki.wooledge.org/BashPitfalls#cat_file_.7C_sed_s.2Ffoo.2Fbar.2F_.3E_file
This might work for you (GNU sed):
sed -e '1R file1' -e '1d' file2
Read the first line of file2. Read the first line of file1 and insert it into the output, then delete the first line of file2. Now read and output the rest of file2.
Simply use head and tail for this task:
head -n 1 Input_file1 && tail -n 2 Input_file2
Output will be as follows:
aaaaaaaa
yyyyyyyy
wwwwwwww
You certainly can do this with sed, but why would you?
sed "1c\\
$(sed 1q file1)
" file2
Or with ed
f1="file1";f2="file2";printf "%s\n" '2,$d' "r $f2" '2d' "wq $f2" | ed -s "$f1"
This question already has answers here:
Bash script prints "Command Not Found" on empty lines
(17 answers)
Closed 6 years ago.
I have a file store version information and I wrote a shell to read two fields and combine them. But when I concatenate those two fields, it show me a werid result.
version file:
buildVer = 3
version = 1.0.0
script looks like:
#!bin/bash
verFile='version'
sdk_ver=`cat $verFile | sed -nE 's/version = (.*)/\1/p'`
build_ver=`cat $verFile | sed -nE 's/buildVer = (.*)/\1/p'`
echo $sdk_ver
echo $build_ver
tagname="$sdk_ver.$build_ver"
echo $tagname
The output shows
1.0.0
3
.30.0
I tried to echo the sdk_ver directly without read it from file, this piece of script works well. So I think it may relate to the sed, but I couldn't figure out how to fix it.
Does anyone know why it acts like that?
You're getting this problem because of presence of DOS line ending i.e. \r in each line of version file.
Use dos2unix or this sed command to remove \r first:
sed -i 's/\r//' version
btw you can also simplify your script using pure BASH constructs like this:
#!/bin/bash
while IFS='= ' read -r k v; do
declare $k="$v"
done < <(sed $'s/\r//' version)
tagname="$version.$buildVer"
echo "$tagname"
This will give output:
1.0.0.3
Alternate solution, with awk:
awk '/version/{v=$3} /buildVer/{b=$3} END{print v "." b}' version.txt
Example:
$ cat file.txt
buildVer = 3
version = 1.0.0
$ awk '/version/{v=$3} /buildVer/{b=$3} END{print v "." b}' file.txt
1.0.0.3
This question already has answers here:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
Closed 6 years ago.
I'm creating a variable from an array which build up multiple -e clauses for a sed command.
The resulting variable is something like:
sedArgs="-e 's/search1/replace1/g' -e 's/search2/replace2/g' -e 's/search3/replace3/g'"
But when I try to call sed with this as the argument I get the error sed: -e expression #1, char 1: unknown command: ''
I've tried to call sed the following ways:
cat $myFile | sed $sedArgs
cat $myFile | sed ${sedArgs}
cat $myFile | sed `echo $sedArgs`
cat $myFile | sed "$sedArgs"
cat $myFile | sed `echo "$sedArgs"`
and all give the same error.
UPDATE - Duplicate question
As has been identified, this is a 'quotes expansion' issue - I thought it was something sed specific, but the duplicate question that has been identified put me on the right track.
I managed to resolve the issue by creating the sedArgs string as:
sedArgs="-e s/search1/replace1/g -e s/search2/replace2/g -e s/search3/replace3/g"
and calling it with:
cat $myFile | sed $sedArgs
which works perfectly.
Then I took the advice of tripleee and kicked the useless cat out!
sed $sedArgs $myFile
also works perfectly.
Use BASH arrays instead of simple string:
# sed arguments in an array
sedArgs=(-e 's/search1/replace1/g' -e 's/search2/replace2/g' -e 's/search3/replace3/g')
# then use it as
sed "${sedArgs[#]}" file
Here is no sane way to do that, but you can pass the script as a single string.
sedArgs='s/search1/replace1/g
s/search2/replace2/g
s/search3/replace3/g'
: then
sed "$sedArgs" "$myFile"
The single-quoted string spans multiple lines; this is scary when you first see it, but perfectly normal shell script. Notice also how the cat is useless as ever, and how the file name needs to be quoted, too.
I have been trying to solve a simple sed line deletion problem.
Looked here and there. It didn't solve my problem.
My problem could simply be achieved by using sed -i'{/^1\|^2\|^3/d;}' infile.txt which deletes lines beginning with 1,2 and 3 from the infile.txt.
But what I want instead is to take the starting matching patterns from a file than manually feeding into the stream editor.
E.g: deletePattern
1
3
2
infile.txt
1 Line here
2 Line here
3 Line here
4 Line here
Desired output
4 Line here
Thank you in advance,
This grep should work:
grep -Fvf deletePattern infile.txt
4 Line here
But this will skip a line if patterns in deletePattern are found anywhere in the 2nd file.
More accurate results can be achieved by using this awk command:
awk 'FILENAME == ARGV[1] && FNR==NR{a[$1];next} !($1 in a)' deletePattern infile.txt
4 Line here
Putting together a quick command substitution combined with a character class will allow a relatively short oneliner:
$ sed -e "/^[$( while read -r ch; do a+=$ch; done <pattern.txt; echo "$a" )]/d" infile.txt
4 Line here
Of course, change the -e to -i for actual in-place substitution.
With GNU sed (for -f -):
sed 's!^[0-9][0-9]*$!/^&[^0-9]/d!' deletePattern | sed -f - infile.txt
The first sed transforms deletePattern into a sed script, then the second sed applies this script.
Try this:
sed '/^[123]/ d' infile.txt