I need help regarding adding some lines to a file; for expamle, I grep from text this parameters and save it to File1:
cellIdentity="461"
cellIdentity="465"
cellIdentity="462"
cellIdentity="466"
cellIdentity="463"
cellIdentity="467"
And now I need to create another file, File2 that looks like:
cellIdentity="461"
cellIdentity="465"
cellIdentity="468"
cellIdentity="462"
cellIdentity="466"
cellIdentity="469"
cellIdentity="463"
cellIdentity="467"
cellIdentity="460"
Three new lines I spaced out.
So basically, I need to add these 3 lines, the last digit's 8,9,0 are always the same in these three lines, and the first two digit's are like in other lines.
I don't know if it is possible to do it. I was trying with the sed command but no luck.
I'm using /bin/csh in solaris. Any help/tips would be great.
Thanks
Using awk
awk 'BEGIN{split("8 9 0",a," ")}NR%2==0{t=$0;sub(/.$/,a[++i],$2);$0=t RS $0}1' FS=\" OFS=\" file
cellIdentity="461"
cellIdentity="465"
cellIdentity="468"
cellIdentity="462"
cellIdentity="466"
cellIdentity="469"
cellIdentity="463"
cellIdentity="467"
cellIdentity="460"
Explanation
FS=\" OFS=\" define field separator "
BEGIN{split("8 9 0",a," ")} define the 8,9,0 in array a in BEGIN part
NR%2==0 find the odd lines
sub(/.$/,a[++i],$2) replace the last character in column 2 which get from array a one after one
1 same as print
This might work for you (GNU sed):
sed -r '1{x;s/^/890/;x};2~2{p;G;s/(.)("\n)(.)(.*)/\3\2\4\3/;P;s/.*\n//;h;d}' file
Related
I need to find the line with first occurrence of a pattern, then I need to replace the whole line with a completely new one.
I found this command that replaces the first occurrence of a pattern, but not the whole line:
sed -e "0,/something/ s//other-thing/" <in.txt >out.txt
If in.txt is
one two three
four something
five six
something seven
As a result I get in out.txt:
one two three
four other-thing
five six
something seven
However, when I try to modify this code to replace the whole line, as follows:
sed -e "0,/something/ c\COMPLETE NEW LINE" <in.txt >out.txt
This is what I get in out.txt:
COMPLETE NEW LINE
five six
something seven
Do you have any idea why the first line is lost?
The c\ command deletes all lines between and inclusive the first matching address through the second matching address, when used with 2 addresses, and prints out the text specified following the c\ upon matching the second address. If there is no line matching the second address in the input, it just deletes all lines (inclusively) between the first matching address through the last line. Since you want to replace one line only, you shouldn't use the c\ command on an address range. The c\ is immediately followed by a new-line character in normal usage.
The 0,/regexp/ address range is a GNU sed extension, which will try to match regexp in the first input line too, which is different from 1,/regexp/ in that aspect. So, the correct command in GNU sed could be
sed '0,/something/{/something/c\
COMPLETE NEW LINE
}' < in.txt
or simplified as pointed out by Sundeep
sed '0,/something/{//c\
COMPLETE NEW LINE
}' < in.txt
or a one-liner,
sed -e '0,/something/{//cCOMPLETE NEW LINE' -e '}' < in.txt
if a literal new-line character is not desirable.
This one-liner also works as pointed out by potong:
sed '0,/something/!b;//cCOMPLETE NEW LINE' in.txt
This might work for you (GNU sed):
sed '1!b;:a;/something/!{n;ba};cCOMPLETE NEW LINE' file
Set up a loop that will only operate from the first line.
Within in the loop, if the key word is not found in the current line, print the current line, fetch the next and repeat until the end of the file or a match is found.
When a match is found, change the contents of the current line to the required result.
N.B. The c command terminates any further processing of sed commands in the same way the d command does.
If there are lines in the input following the key word match, the negation of address at the start of the sed cycle will capture these lines and result in their printing and no further processing.
An alternative:
sed 'x;/./{x;b};x;/something/h;//cCOMPLETE NEW LINE' file
Or (specific to GNU and bash):
sed $'0,/something/{//cCOMPLETE NEW LINE\n}' file
Just use awk:
$ awk '!done && sub(/something/,"other-thing"){done=1} {print}' file
one two three
four other-thing
five six
something seven
$ awk '!done && sub(/.*something.*/,"other-thing"){done=1} {print}' file
one two three
other-thing
five six
something seven
$ awk '!done && /something/{$0="other-thing"; done=1} {print}' file
one two three
other-thing
five six
something seven
and look what you can trivially do if you want to replace the Nth occurrence of something:
$ awk -v n=1 '/something/ && (++cnt == n){$0="other-thing"} {print}' file
one two three
other-thing
five six
something seven
$ awk -v n=2 '/something/ && (++cnt == n){$0="other-thing"} {print}' file
one two three
four something
five six
other-thing
given text like
_adsp TXT "dkim=all"
VVKMU6SE3C2MF88BG4DJQAECMR9SIIF0 NSEC3 1 1 10 C4F407437E8EA4C5 (
175MCHR31K25LP89OVJI5LCE0JA2N2AP
A MX TXT AAAA RRSIG SPF )
RRSIG NSEC3 7 3 1800 (
20200429171433 20200330161758 11672 example.com
H3l26qmtkuiFZCeSYCCAo5krFE3gjM0I8UeQ9jhj3STy
X6fM0YizCHEuv4VZynOJGJc1XJnHRHI+p7yLlZ+OVseK
UfIkPVP+VOmlerwozEpM+Tnt8evwnMTDbcn0zxf/6YJx
kZeO2AszWkRZ0bctqW7INYo8YuyyuTSxSr8se27fiaPA
4GXQymepGgv/JGqargzHbyhhkDhENmNo7Qwkjl+a0kI4
6qqKcEWCsDvnlYUQiDFzc5oRs2j7TT9uybTfwUDQxV+t
MQFMhzu7LNbRIUuOb16sAEGSdl9mWQ4sZRJ9wuXJWbso
G+3tY0pBbq4ffScz/JKcrJ0qAuBF1F5JcQ== )
$TTL 1800
I want to get rid by the part with the "(not beginning with whitespace) NSEC3 " until the first line not beginning with a whitespace character.
resulting
_adsp TXT "dkim=all"
$TTL 1800
in the example.
I tried sed '/^[^\s].*\sNSEC3\s/,/^[^\s]/d;' filename but that doesn't work as expected, example results in
_adsp TXT "dkim=all"
H3l26qmtkuiFZCeSYCCAo5krFE3gjM0I8UeQ9jhj3STy
X6fM0YizCHEuv4VZynOJGJc1XJnHRHI+p7yLlZ+OVseK
UfIkPVP+VOmlerwozEpM+Tnt8evwnMTDbcn0zxf/6YJx
kZeO2AszWkRZ0bctqW7INYo8YuyyuTSxSr8se27fiaPA
4GXQymepGgv/JGqargzHbyhhkDhENmNo7Qwkjl+a0kI4
6qqKcEWCsDvnlYUQiDFzc5oRs2j7TT9uybTfwUDQxV+t
MQFMhzu7LNbRIUuOb16sAEGSdl9mWQ4sZRJ9wuXJWbso
G+3tY0pBbq4ffScz/JKcrJ0qAuBF1F5JcQ== )
$TTL 1800
so resuming printout way too early?
what do I miss?
thank you
P.S.:
you maybe see what I want to do is removing DNSSEC parts out of an named zone. didn't find any other way to remove RRSIG and NSEC3 entries, yet. If someone has an idea, I would appreciate that too.
[\s] matches a literal \ or s characters. It doesn't match whitespace.
The /^[^\s]/d; (if [\s] would work as you expect) will also include removing the last line with non-leading whitespaces. I think you have to loop manually.
On the example you've given, the following seems to work:
sed -n '/^[^ \t].*\sNSEC3\s/{ :a; n; /^[^ \t]/bb; ba}; :b; p'
This might work for you (GNU sed):
sed -n '/^\S.*NSEC/{:a;n;/^\S/!ba};p' file
Turn off implicit printing by using the -n option.
Throw away lines between one starting with a non-space and containing the string NSEC and any lines not starting with a non-space.
Print all other lines.
Alternative:
sed '/^\S.*NSEC/,/^\S/{/^\S.*NSEC\|^\s/d}' file
Yet another alternative:
sed '/^\S.*NSEC/{:a;N;/\n\S/!ba;s/.*\n//}' file
And another:
sed '/^\S.*NSEC/{:a;N;/\n\S/!s/\n//;ta;D}' file
N.B. The first two solutions will delete lines regardless of a line delimiting the end of the deletions. Whereas the last two solutions will only delete lines if there is a line delimiting the end of the deletions.
I have file that looks like this :
1,2,3,4
5,6,7,8
I want to substitute 2rd column containing 6 to 89. The desired output is
1,2,3,4
5,89,7,8
But if I type
index=2
cat file | sed 's/[^,]*/89/'$index
I get
1,89,3,4
5,89,7,8
and if I type
index=2
cat file | sed 's/[^,]6/89/'$index
nothing changes.
Why is it like this? How can I fix this? Thank you.
Since you want to change the second column containing a 6 and you have a comma as field separator it is actually very easy with sed:
sed 's/^\([^,]*\),6,/\1,89,/`
Here we make use of back-referencing to remember the first column.
If you want to replace the 6 in the 5th column, you can do something like:
sed 's/^\(\([^,]*,\)\{4\}\)6,/\189,/'
It is, however, much more comfortable using awk:
awk 'BEGIN{FS=OFS=","}($2==6){$2=89}1'
I solved this by using awk
awk 'BEGIN{FS=OFS=","} {if ($2==6) $2=89}1' file >file1
I have a simple text file in below format.
1 12658003Y
2 34345345N
3 34653785Y
4 36452342N
5 86747488Y
6 34634543Y
so on
10 37456338Y
11 33535555Y
12 37456378Y
so on
100 23432434Y
As you can see there are two white spaces after first number.
I'm trying to write SED command to remove the digits before whitespaces. Is there any SED command to remove spaces and number before spaces?
Output file should look like below.
12658003Y
34345345N
34653785Y
36452342N
so on..
Please assist. I'm very new to shell scripting.
sed 's/[0-9]\+\s\+//' infile > outfile
Explanation:
s: we want to use substitution
/: mark start and end of the expression we want to match
[0-9]: match any digit
+: match the previous one or more time
\s: space
+: match the previous one or more time
/: mark start of what we want to change our matches to (which is nothing)
/: some special operators goes after this (we use no such)
infile: the file we want to change
>: pipe stdout to
outfile: where we want to store output
Your sed command would be,
sed 's/.* //g' file
This would remove the first numbers along with the space followed.
Remove leading digits, then following spaces:
sed 's/^[0-9]* *//' file
sed 's/^[0-9]*[ ]*//g' input.txt
I am looking for two consecutive lines matching a certain pattern, say containing word 'pat' using sed and have noticed that I am able to detect it sometimes with this command:
sed -n 'N; /.*pat.*\n.*pat.*/p'
but this command fails if the line numbers for the duplicates are not of the same parity and I assume it's because we're searching lines 1+2, 3+4, 5+6 etc.. if this is the case, what would be the correct way to do this?
Why does it need to be sed? May I suggest awk?
awk '{/pat/?f++:f=0} f==2' file
If pat is found, increment f with 1
If pat is not found, reset f to 0
If f==2 print the line.
This might work for you (GNU sed):
sed '$!N;/pattern.*\n.*pattern/p;D' file
This keeps 2 lines in the pattern space and prints both of them out if the regexp matches.