I can print the current filename with GNU sed using "F". I can make changes to the pattern space with "s" (and other commands). Is there any way to put the filename into the pattern space so I can manipulate this as well.
For example:
09:24:25:~ $ sed -n 'F ; s/hello/goodbye/ig ; p' hello.txt
hello.txt
goodbye world
09:24:42:~ $
Is there a thing I could put instead of "F" which would get this to result in:
goodbye.txt
goodbye world
Piping back to sed should work
$ sed 'F' hello.txt | sed -n 's/hello/goodbye/p'
goodbye.txt
goodbye world
This can be done in a single gnu awk like this:
awk -v IGNORECASE=1 '
NR==1{$0 = FILENAME "\n" $0} {gsub(/hello/, "goodbye")} 1' hello.txt
goodbye.txt
goodbye world
This might work for you (GNU sed):
sed '1{h;s/.*/sed -n 1F file/e;x};G' file
This appends the file name to each line i.e. the file name is held in the hold space.
For line one only, copy current pattern space to the hold space. Replace the line by another sed scripts output which evaluates to the current file name. Swap the hold space for the pattern space.
For each line append the hold space.
Related
I have this content in a file where I want to replace spaces at certain positions with pipe symbol (|). I used sed for this, but it is replacing all the spaces in the string. But I don't want to replace the space for the 3rd and 4th string.
How to achieve this?
Input:
test test test test
My attempt:
sed -e 's/ /|/g file.txt
Expected Output:
test|test|test test
Actual Output:
test|test|test|test
sed 's/ /\
/3;y/\n / |/'
As newline cannot appear in a sed pattern space, you can change the third space to a newline, then change all newlines and spaces to spaces and pipes.
GNU sed can use \n in the replacement text:
sed 's/ /\n/3;y/\n / |/'
If the original input doesn't contain any pipe characters, you can do
sed -e 's/ /|/g' -e 's/|/ /3' file
to retain the third white space. Otherwise see other answers.
You could replace the 'first space' twice, e.g.
sed -e 's/ /|/' -e 's/ /|/' file.txt
Or, if you want to specify the positions (e.g. the 2nd and 1st spaces):
sed -e 's/ /|/2' -e 's/ /|/1' file.txt
Using GNU sed to replace the first and second one or more whitespace chunks:
sed -i -E 's/\s+/|/;s/\s+/|/' file
See the online demo.
Details
-i - inline replacements on
-E - POSIX ERE syntax enabled
s/\s+/|/ - replaces the first one or more whitespace chars
; - and then
s/\s+/|/ the second one or more whitespace chars on each line (if present).
Keep it simple and use awk, e.g. using any awk in any shell on every Unix box no matter what other characters your input contains:
$ awk '{for (i=1;i<NF;i++) sub(/ /,"|")} 1' file
test|test|test test
The above replaces all but the last " " on each line. If you want to replace a specific number, e.g. 2, then just change NF to 2.
Have a file that has been created incorrectly. There are several space delimited fields in the file but one text field has some unwanted newlines. This is causing a big problem.
How can I remove these characters but not the wanted line ends?
file is:
'Number field' 'Text field' 'Number field'
1 Some text 999999
2 more
text 111111111
3 Even more text 8888888888
EOF
So there is a NL after the word "more".
I've tried sed:
sed 's/.$//g' test.txt > test.out
and
sed 's/\n//g' test.txt > test.out
But none of these work. The newlines do not get removed.
tr -d '\n' does too much - I need to remove ONLY the newlines that are preceded by a space.
How can I delete newlines that follow a space?
SunOS 5.10 Generic_144488-09 sun4u sparc SUNW,Sun-Fire-V440
A sed solution is
sed '/ $/{N;s/\n//}'
Explanation:
/ $/: whenever the line ends in space, then
N: append a newline and the next line of input, and
s/\n//: delete the newline.
It might be simplest with Perl:
perl -p0 -e 's/ \n/ /g'
The -0 flag makes Perl read the entire file as one line. Then we can substitute using s in the usual way. You can, of course, also add the -i option to edit the file in-place.
How can I delete newlines that follow a space?
If you want every occurrence of $' \n' in the original file to be replaced by a space ($' '), and if you know of a character (e.g. a control character) that does not appear in the file, then the task can be accomplished quite simply using sed and tr (as you requested). Let's suppose, for example, that control-A is a character that is not in the file. For the sake of simplicity, let's also assume we can use bash. Then the following script should do the job:
#!/bin/bash
A=$'\01'
tr '\n' "$A" | sed "s/ $A/ /g" | tr "$A" '\n'
I have a file in which some lines start by a >
For these lines, and only these ones, I want to keep the first eleven characters.
How can I do that using sed ?
Or maybe something else is better ?
Thanks !
Muriel
Let's start with this test file:
$ cat file
line one with something or other
>1234567890abc
other line in file
To keep only the first 11 characters of lines starting with > while keeping all other lines:
$ sed -r '/^>/ s/(.{11}).*/\1/' file
line one with something or other
>1234567890
other line in file
To keep only the first eleven characters of lines starting with > and deleting all other lines:
$ sed -rn '/^>/ s/(.{11}).*/\1/p' file
>1234567890
The above was tested with GNU sed. For BSD sed, replace the -r option with -E.
Explanation:
/^>/ is a condition. It means that the command which follows only applies to lines that start with >
s/(.{11}).*/\1/ is a substitution command. It replaces the whole line with just the first eleven characters.
-r turns on extended regular expression format, eliminating the need for some escape characters.
-n turns off automatic printing. With -n in effect, lines are only printed if we explicitly ask them to be printed. In the second case above, that is done by adding a p after the substitute command.
Other forms:
$ sed -r 's/(>.{10}).*/\1/' file
line one with something or other
>1234567890
other line in file
And:
$ sed -rn 's/(>.{10}).*/\1/p' file
>1234567890
I have 'file1' with (say) 100 lines. I want to use sed or awk to print lines 23, 71 and 84 (for example) to 'file2'. Those 3 line numbers are in a separate file, 'list', with each number on a separate line.
When I use either of these commands, only line 84 gets printed:
for i in $(cat list); do sed -n "${i}p" file1 > file2; done
for i in $(cat list); do awk 'NR==x {print}' x=$i file1 > file2; done
Can a for loop be used in this way to supply line addresses to sed or awk?
This might work for you (GNU sed):
sed 's/.*/&p/' list | sed -nf - file1 >file2
Use list to build a sed script.
You need to do > after the loop in order to capture everything. Since you are using it inside the loop, the file gets overwritten. Inside the loop you need to do >>.
Good practice is to or use > outside the loop so the file is not open for writing during every loop iteration.
However, you can do everything in awk without for loop.
awk 'NR==FNR{a[$1]++;next}FNR in a' list file1 > file2
You have to >>(append to the file) . But you are overwriting the file. That is why, You are always getting 84 line only in the file2.
Try use,
for i in $(cat list); do sed -n "${i}p" file1 >> file2; done
With sed:
sed -n $(sed -e 's/^/-e /' -e 's/$/p/' list) input
given the example input, the inner command create a string like this: `
-e 23p
-e 71p
-e 84p
so the outer sed then prints out given lines
You can avoid running sed/awk in a for/while loop altgether:
# store all lines numbers in a variable using pipe
lines=$(echo $(<list) | sed 's/ /|/g')
# print lines of specified line numbers and store output
awk -v lineS="^($lines)$" 'NR ~ lineS' file1 > out
We have a process which can use a file containing sed commands to alter piped input.
I need to replace a placeholder in the input with a variable value, e.g. in a single -e type of command I can run;
$ echo "Today is XX" | sed -e "s/XX/$(date +%F)/"
Today is 2012-10-11
However I can only specify the sed aspects in a file (and then point the process at the file), E.g. a file called replacements.sed might contain;
s/XX/Thursday/
So obviously;
$ echo "Today is XX" | sed -f replacements.sed
Today is Thursday
If I want to use an environment variable or shell value, though, I can't find a way to make it expand, e.g. if replacements.txt contains;
s/XX/$(date +%F)/
Then;
$ echo "Today is XX" | sed -f replacements.sed
Today is $(date +%F)
Including double quotes in the text of the file just prints the double quotes.
Does anyone know a way to be able to use variables in a sed file?
This might work for you (GNU sed):
cat <<\! > replacements.sed
/XX/{s//'"$(date +%F)"'/;s/.*/echo '&'/e}
!
echo "Today is XX" | sed -f replacements.sed
If you don't have GNU sed, try:
cat <<\! > replacements.sed
/XX/{
s//'"$(date +%F)"'/
s/.*/echo '&'/
}
!
echo "Today is XX" | sed -f replacements.sed | sh
AFAIK, it's not possible. Your best bet will be :
INPUT FILE
aaa
bbb
ccc
SH SCRIPT
#!/bin/sh
STRING="${1//\//\\/}" # using parameter expansion to prevent / collisions
shift
sed "
s/aaa/$STRING/
" "$#"
COMMAND LINE
./sed.sh "fo/obar" <file path>
OUTPUT
fo/obar
bbb
ccc
As others have said, you can't use variables in a sed script, but you might be able to "fake" it using extra leading input that gets added to your hold buffer. For example:
[ghoti#pc ~/tmp]$ cat scr.sed
1{;h;d;};/^--$/g
[ghoti#pc ~/tmp]$ sed -f scr.sed <(date '+%Y-%m-%d'; printf 'foo\n--\nbar\n')
foo
2012-10-10
bar
[ghoti#pc ~/tmp]$
In this example, I'm using process redirection to get input into sed. The "important" data is generated by printf. You could cat a file instead, or run some other program. The "variable" is produced by the date command, and becomes the first line of input to the script.
The sed script takes the first line, puts it in sed's hold buffer, then deletes the line. Then for any subsequent line, if it matches a double dash (our "macro replacement"), it substitutes the contents of the hold buffer. And prints, because that's sed's default action.
Hold buffers (g, G, h, H and x commands) represent "advanced" sed programming. But once you understand how they work, they open up new dimensions of sed fu.
Note: This solution only helps you replace entire lines. Replacing substrings within lines may be possible using the hold buffer, but I can't imagine a way to do it.
(Another note: I'm doing this in FreeBSD, which uses a different sed from what you'll find in Linux. This may work in GNU sed, or it may not; I haven't tested.)
I am in agreement with sputnick. I don't believe that sed would be able to complete that task.
However, you could generate that file on the fly.
You could change the date to a fixed string, like
__DAYOFWEEK__.
Create a temp file, use sed to replace __DAYOFWEEK__ with $(date +%Y).
Then parse your file with sed -f $TEMPFILE.
sed is great, but it might be time to use something like perl that can generate the date on the fly.
To add a newline in the replacement expression using a sed file, what finally worked for me is escaping a literal newline. Example: to append a newline after the string NewLineHere, then this worked for me:
#! /usr/bin/sed -f
s/NewLineHere/NewLineHere\
/g
Not sure it matters but I am on Solaris unix, so not GNU sed for sure.