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
Related
I have a file with multiple paths in the same line:
cat modules.dep
kernel/mm/zsmalloc.ko:
kernel/crypto/lzo.ko:
kernel/drivers/char/tpm/tpm_vtpm_proxy.ko: kernel/drivers/char/tpm/tpm.ko
kernel/drivers/block/virtio_blk.ko:
kernel/drivers/block/zram/zram.ko: kernel/mm/zsmalloc.ko
kernel/drivers/nvdimm/virtio_pmem.ko: kernel/drivers/nvdimm/nd_virtio.ko
kernel/drivers/nvdimm/nd_virtio.ko:
kernel/drivers/net/virtio_net.ko: kernel/drivers/net/net_failover.ko kernel/net/core/failover.ko
kernel/drivers/net/net_failover.ko: kernel/net/core/failover.ko
extra/virtio_gpu/virtio-gpu.ko: kernel/drivers/virtio/virtio_dma_buf.ko
extra/wlan_simulation/virt_wifi_sim.ko: kernel/drivers/net/wireless/virt_wifi.ko
I would like to change it to:
/lib/modules/zsmalloc.ko:
/lib/modules/lzo.ko:
/lib/modules/tpm_vtpm_proxy.ko: /lib/modules/tpm.ko
/lib/modules/virtio_blk.ko:
/lib/modules/zram.ko: /lib/modules/zsmalloc.ko
/lib/modules/virtio_pmem.ko: /lib/modules/nd_virtio.ko
/lib/modules/nd_virtio.ko:
/lib/modules/virtio_net.ko: /lib/modules/net_failover.ko /lib/modules/failover.ko
/lib/modules/net_failover.ko: /lib/modules/failover.ko
/lib/modules/virtio-gpu.ko: /lib/modules/virtio_dma_buf.ko
/lib/modules/virt_wifi_sim.ko: /lib/modules/virt_wifi.ko
But my attempt:
sed -i 's/\(.*\)\//\/lib\/modules\//g' modules.load
works only, if there is just one path per line.
How can I achieve this, via sed, with multiple paths per line?
I am using sed from BusyBox in D(ASH) Standalone.
BusyBox v1.32.1-Magisk (2021-01-21 00:17:27 PST) multi-call binary.
Usage: sed [-i[SFX]] [-nrE] [-f FILE]... [-e CMD]... [FILE]...
or: sed [-i[SFX]] [-nrE] CMD [FILE]...
-e CMD Add CMD to sed commands to be executed
-f FILE Add FILE contents to sed commands to be executed
-i[SFX] Edit files in-place (otherwise sends to stdout)
Optionally back files up, appending SFX
-n Suppress automatic printing of pattern space
-r,-E Use extended regex syntax
If no -e or -f, the first non-option argument is the sed command string.
Remaining arguments are input files (stdin if none).
This sed should work:
sed -E 's~[^[:blank:]]+/~/lib/modules/~g' modules.dep
/lib/modules/zsmalloc.ko:
/lib/modules/lzo.ko:
/lib/modules/tpm_vtpm_proxy.ko: /lib/modules/tpm.ko
/lib/modules/virtio_blk.ko:
/lib/modules/zram.ko: /lib/modules/zsmalloc.ko
/lib/modules/virtio_pmem.ko: /lib/modules/nd_virtio.ko
/lib/modules/nd_virtio.ko:
/lib/modules/virtio_net.ko: /lib/modules/net_failover.ko /lib/modules/failover.ko
/lib/modules/net_failover.ko: /lib/modules/failover.ko
/lib/modules/virtio-gpu.ko: /lib/modules/virtio_dma_buf.ko
/lib/modules/virt_wifi_sim.ko: /lib/modules/virt_wifi.ko
[^[:blank:]]+/ finds 1+ non-whitespace characters followed by a / thus matching longest string until it gets a / in each of the multiple string per line.
I want to get the first line of a file that is not commented out with an hash, then append a line of text just after that line just before that line.
I managed to get the number of the line:
sed -n '/^\s*#/!{=;q}' file // prints 2
and also to insert text (specifying the line manually):
sed '2 a extralinecontent' file
I can't get them working together as a one liner or in a batch.
I tried command substitution (with $(command) and also with backticks) but I get an error from bash:
sed '$(sed -n '/^\s*#/!{=;q}' file) a extralinecontent' file
-bash: !{=: event not found
and also tried many other combinations, but no luck.
I'm using gnu-sed (via brew) on macOS.
This might work for you (GNU sed):
sed -e '/^\s*#/b;a extra line content' -e ':a;n;ba' file
Bail out of any lines beginning with a comment at the beginning of the file, append an extra line following the first line that is not a comment and keep fetching/printing all the remaining lines of the file.
Here's a way to do it with GNU sed without reading the file twice
$ cat ip.txt
#comment
foo baz good
123 456 7889
$ sed -e '0,/^\s*[^#[:space:]]/ {// a XYZ' -e '}' ip.txt
#comment
foo baz good
XYZ
123 456 7889
GNU sed allows first address to be 0 if the other address is regex, that way this will work even if first line matches the condition
/^\s*[^#[:space:]]/ as sed doesn't support possessive quantifier, need to ensure that the first character being matched by the character class isn't either a # or a whitespace character
// is a handy shortcut to repeat the last regex
a XYZ your required line to be appended (note that your question mentiones insert, so if you want that, use i instead of a)
Quite certainly I miss something basic. My file contains lines like
fooLOCATION=sdfmsvdnv
fooLOCATION=
barLOCATION=sadssf
barLOCATION=
and I want to delete all lines ending with LOCATION=.
sed -i '/LOCATION=$/d' file
does not do, it deletes nothing, and I have tried endless variations, but I don't get it. What inline sed command can do this?
There are two approaches here, either print all non-matching lines with
sed -in '/LOCATION=$/!p' file
or delete all matching names with
sed -i '/LOCATION=$/d' file
The first uses the n command line option to suppress the default action of printing the line. We then test for lines that end in LOCATION= and invert the pattern (only keeping those that don't match). When we get a desirable line, we print it with the p option.
The second looks for lines matching the end of line pattern, and deletes those that do.
Your file contains blank lines, and both of these keep those. If we don't want to keep those, we can change the first option to
sed -in '/^$/!{/LOCATION=$/!p}' file
which first checks if a line is not empty, and only bothers checking if it should be printed if it isn't empty. We can modify the second option to
sed -i '/^$/d;/LOCATION=$/d' file
which deletes blank lines and then checks about deleting the other pattern.
We can modify the options to work with different line ending by specifying the difference in the pattern. The difference between line endings on Unix/Linux (\n) and Windows (\r\n) is the presence of an extra carriage return on Windows. Modifying the four commands above to accept either, we get
sed -in '/LOCATION=\r\{0,1\}$/!p' file
sed -i '/LOCATION=\r\{0,1\}$/d' file
sed -in '/^\r\{0,1\}$/!{/LOCATION=\r\{0,1\}$/!p}' file
sed -i '/^\r\{0,1\}$/d;/LOCATION=\r\{0,1\}$/d' file
Note that in each of these we allow an optional \r before the end of line. We use the curly bracket notation, as sed does not support the question mark optional quantifier in normal mode (using the r option to GNU sed for enabling extended regular expressions, we can replace \{0,1\} with ?).
On a Windows shell, all of the options above require double quotes instead of single quotes.
Your command does work for me:
$ sed -i '/LOCATION=$/d' file
Results, viewed using cat:
$ cat file
fooLOCATION=sdfmsvdnv
barLOCATION=sadssf
Note
If a file has non-Unix line endings such as files from Windows with DOS-formatted line-endings, it can be a reason for failure. A typical remedy is to use dos2unix:
$ dos2unix file
This converter fixes the newline issues, so that file will now have Unix-style line endings. Sed should now properly recognize those line endings, so retry your sed command and it should work.
This might work for you (GNU sed):
sed -i '/LOCATION=\s*$/d' file
This deletes the line if LOCATION= is at the end of the line or if there is any optional white space following the pattern.
I find many ways to do this, AWK, SED, UNIQ, but none of them are working on my file.
I want to delete duplicate lines. Here is an example of part of my file:
KTBX
KFSO
KCLK
KTBX
KFSO
KCLK
PAJZ
PAJZ
NOTE: I had to manually add line feeds when I cut and pasted from the file...for some reason it was putting all the variables on one line. Makes me think that my 44,000 line text file actually has only "1" line? Is there a way to modify it so I can delete dups?
You can see all non-printed characters with this command:
od -c oldfile
If all your records are on one line, you can use sed to replace a whitespace (space, tab, newline) with a linebreak:
sed -e 's/\s\+/\n/g' oldfile > oldfile.1
Once you have multiple lines, this awk one-liner:
awk '!x[$0]++' oldfile.1 > newfile
my outfile:
KTBX
KFSO
KCLK
PAJZ
Perl One-Liner:
perl -nle 'unless($hash{$_}++){print $_}' file
how to remove comment lines (as # bal bla ) and empty lines (lines without charecters) from file with one sed command?
THX
lidia
If you're worried about starting two sed processes in a pipeline for performance reasons, you probably shouldn't be, it's still very efficient. But based on your comment that you want to do in-place editing, you can still do that with distinct commands (sed commands rather than invocations of sed itself).
You can either use multiple -e arguments or separate commands with a semicolon, something like (just one of these, not both):
sed -i 's/#.*$//' -e '/^$/d' fileName
sed -i 's/#.*$//;/^$/d' fileName
The following transcript shows this in action:
pax> printf 'Line # with a comment\n\n# Line with only a comment\n' >file
pax> cat file
Line # with a comment
# Line with only a comment
pax> cp file filex ; sed -i 's/#.*$//;/^$/d' filex ; cat filex
Line
pax> cp file filex ; sed -i -e 's/#.*$//' -e '/^$/d' filex ; cat filex
Line
Note how the file is modified in-place even with two -e options. You can see that both commands are executed on each line. The line with a comment first has the comment removed then all is removed because it's empty.
In addition, the original empty line is also removed.
#paxdiablo has a good answer but it can be improved.
(1) The '/^$/d' clause only matches 100% blank lines.
If you want to also match lines that are entirely whitespace (spaces, tabs etc.) use this instead:
'/^\s*$/d'
(2) The 's/#.*$//' clause only matches lines that start with the # character in column 0.
If you want to also match lines that have only whitespace before the first # use this instead:
'/^\s*#.*$/d'
The above criteria may not be universal (e.g. within a HEREDOC block, or in a Python multi-line string the different approaches could be significant), but in many cases the conventional definition of "blank" lines include whitespace-only, and "comment" lines include whitespace-then-#.
(3) Lastly, on OSX at least, the #paxdiablo solution in which the first clause turns comment lines into blank lines, and the second clause strips blank lines (including what were originally comments) doesn't work. It seems to be more portable to make both clauses /d delete actions as I've done.
The revised command incorporating the above is:
sed -e '/^\s*#.*$/d' -e '/^\s*$/d' inputFile
This tiny jewel removes all # comments, no matter where they begin in a line (see caution below):
sed -e 's/\s*#.*$//'
Example:
text="
this is a # test
#this is a test
#this is a #test
this is # another #test
"
$echo "$text" | sed -e 's/\s*#.*$//'
this is a
this is
Next this removes any resulting blank lines:
$echo "$text" | sed -e 's/\s*#.*$//' | sed -e '/^\s*$/d'
Caution: Depending on the syntax and/or interpretation of the lines your processing, this might not be an appropriate solution, as it just stupidly removes end of lines, even if the '#' is part of your data or code. However, for use cases where you'll never use a hash except for as an end of line comment then it works fine. So just as with all coding, context must be taken into consideration.
Alternative variant, using grep:
cat file.txt | grep -Ev '(#.*$)|(^$)'
you can use awk
awk 'NF{gsub(/^[ \t]*#/,"");print}' file
First example(paxdiablo) is very good except its not change file, just output result. If you want to change it inline:
sudo sed -i 's/#.*$//;/^$/d' inputFile
On (one of) my linux boxes, sed understands extended regular expressions with the -r option, so:
sed -r '/(^\s*#)|(^\s*$)/d' squid.conf.installed
is very useful for showing all non-blank, non comment lines.
The regex matches either start of line followed by zero or more spaces or tabs followed by either a hash or end of line, and deletes those matching lines from the input.