i have file "acl.txt"
192.168.0.1
192.168.4.5
#start_exceptions
192.168.3.34
192.168.6.78
#end_exceptions
192.168.5.55
and another file "exceptions"
192.168.88.88
192.168.76.6
I need to replace everything between #start_exceptions and #end_exceptions with content of exceptions file. I have tried many solutions from this forum but none of them works.
EDITED:
Ok, if you want to retain the #start and #stop, I will revert to awk:
awk '
BEGIN {p=1}
/^#start/ {print;system("cat exceptions");p=0}
/^#end/ {p=1}
p' acl.txt
Thanks to #fedorqui for tweaks in comments below.
Output:
192.168.0.1
192.168.4.5
#start_exceptions
192.168.88.88
192.168.76.6
#end_exceptions
192.168.5.55
p is a flag that says whether or not to print lines. It starts at the beginning as 1, so all lines are printed till I find a line starting with #start. Then I cat the contents of the exceptions file and stop printing lines till I find a line starting with #end, at which point I set the p flag back to 1 so remaining lines get printed.
If you want output to a file, add "> newfile" to the very end of the command like this:
awk '
BEGIN {p=1}
/^#start/ {print;system("cat exceptions");p=0}
/^#end/ {p=1}
p' acl.txt > newfile
YET ANOTHER VERSION IF YOU REALLY WANT TO USE SED
If you really, really want to do it with sed, you can use nested address spaces, firstly to select the lines between #start_exceptions and #end_exceptions, then again to select the first line within that and also lines other than the #end_exceptions line:
sed '
/^#start/,/^#end/{
/^#start/{
n
r exceptions
}
/^#end/!d
}
' acl.txt
Output:
192.168.0.1
192.168.4.5
#start_exceptions
192.168.88.88
192.168.76.6
#end_exceptions
192.168.5.55
ORIGINAL ANSWER
I think this will work:
sed -e '/^#end/r exceptions' -e '/^#start/,/^#end/d' acl.txt
When it finds /^#end/ it reads in the exceptions file. And it also deletes everything between /#start/ and /#end/.
I have left the matching slightly "loose" for clarity of expressing the technique.
You can use the following, based on Replace string with contents of a file using sed:
$ sed $'/end/ {r exceptions\n} ; /start/,/end/ {d}' acl.txt
192.168.0.1
192.168.4.5
192.168.88.88
192.168.76.6
192.168.5.55
Explanation
sed $'one_thing; another_thing' ac1.txt performs the two actions.
/end/ {r exceptions\n} if the line contains end, then read the file exceptions and append it.
/start/,/end/ {d} from a line containing start to a line containing end, delete all the lines.
I had problem with Mark Setchell's solution in MINGW. The caret was not picking up the beginning of line. Indeed, is the detection of the separator dependent on it being at the beginning of the line?
I came up with this awk alternative...
$ awk -v data="$(<exceptions)" '
BEGIN {p=1}
/#start_exceptions/ {print; print data;p=0}
/#end_exceptions/ {p=1}
p
' acl.txt
Related
I need to filter the output of a command.
I tried this.
bpeek | grep nPDE
My problem is that I need all matches of nPDE and the line after the found file. So the output would be like:
iteration nPDE
1 1
iteration nPDE
2 4
The best case would be if it would show me the found line only once and then only the line after it.
I found solutions with awk, But as far as I know awk can only read files.
There is an option for that.
grep --help
...
-A, --after-context=NUM print NUM lines of trailing context
Therefore:
bpeek | grep -A 1 'nPDE'
With awk (for completeness since you have grep and sed solutions):
awk '/nPDE/{c=2} c&&c--'
grep -A works if your grep supports it (it's not in POSIX grep). If it doesn't, you can use sed:
bpeek | sed '/nPDE/!d;N'
which does the following:
/nPDE/!d # If the line doesn't match "nPDE", delete it (starts new cycle)
N # Else, append next line and print them both
Notice that this would fail to print the right output for this file
nPDE
nPDE
context line
If you have GNU sed, you can use an address range as follows:
sed '/nPDE/,+1!d'
Addresses of the format addr1,+N define the range between addr1 (in our case /nPDE/) and the following N lines. This solution is easier to adapt to a different number of context lines, but still fails with the example above.
A solution that manages cases like
blah
nPDE
context
blah
blah
nPDE
nPDE
context
nPDE
would like like
sed -n '/nPDE/{$p;:a;N;/\n[^\n]*nPDE[^\n]*$/!{p;b};ba}'
doing the following:
/nPDE/ { # If the line matches "nPDE"
$p # If we're on the last line, just print it
:a # Label to jump to
N # Append next line to pattern space
/\n[^\n]*nPDE[^\n]*$/! { # If appended line does not contain "nPDE"
p # Print pattern space
b # Branch to end (start new loop)
}
ba # Branch to label (appended line contained "nPDE")
}
All other lines are not printed because of the -n option.
As pointed out in Ed's comment, this is neither readable nor easily extended to a larger amount of context lines, but works correctly for one context line.
I'm trying to create a little script that basically uses dig +short to find the IP of a website, and then pipe that to sed/awk/grep to replace a line. This is what the current file looks like:
#Server
123.455.1.456
246.523.56.235
So, basically, I want to search for the '#Server' line in a text file, and then replace the two lines underneath it with an IP address acquired from dig.
I understand some of the syntax of sed, but I'm really having trouble figuring out how to replace two lines underneath a match. Any help is much appreciated.
Based on the OP, it's not 100% clear exactly what needs to replaced where, but here's a a one-liner for the general case, using GNU sed and bash. Replace the two lines after "3" with standard input:
echo Hoot Gibson | sed -e '/3/{r /dev/stdin' -e ';p;N;N;d;}' <(seq 7)
Outputs:
1
2
3
Hoot Gibson
6
7
Note: sed's r command is opaquely documented (in Linux anyway). For more about r, see:
"5.9. The 'r' command isn't inserting the file into the text" in this sed FAQ.
here's how in awk:
newip=12.34.56.78
awk -v newip=$newip '{
if($1 == "#Server"){
l = NR;
print $0
}
else if(l>0 && NR == l+1){
print newip
}
else if(l==0 || NR != l+2){
print $0
}
}' file > file.tmp
mv -f file.tmp file
explanation:
pass $newip to awk
if the first field of the current line is #Server, let l = current line number.
else if the current line is one past #Server, print the new ip.
else if the current row is not two past #Server, print the line.
overwrite original file with modified version.
I want to delete a blank line only if this one is after the line of my pattern using sed or awk
for example if I have
G
O TO P999-ERREUR
END-IF.
the pattern in this case is G
I want to have this output
G
O TO P999-ERREUR
END-IF.
This will do the trick:
$ awk -v n=-2 'NR==n+1 && !NF{next} /G/ {n=NR}1' file
G
O TO P999-ERREUR
END-IF.
Explanation:
-v n=-2 # Set n=-2 before the script is run to avoid not printing the first line
NR == n+1 # If the current line number is equal to the matching line + 1
&& !NF # And the line is empty
{next} # Skip the line (don't print it)
/G/ # The regular expression to match
{n = NR} # Save the current line number in the variable n
1 # Truthy value used a shorthand to print every (non skipped) line
Using sed
sed '/GG/{N;s/\n$//}' file
If it sees GG, gets the next line, removes the newline between them if the next line is empty.
Note this will only remove one blank line after, and the line must be blank i.e not spaces or tabs.
This might work for you (GNU sed):
sed -r 'N;s/(G.*)\n\s*$/\1/;P;D' file
Keep a moving window of two lines throughout the length of the file and remove a newline (and any whitespace) if it follows the intended pattern.
Using ex (edit in-place):
ex +'/G/j' -cwq foo.txt
or print to the standard output (from file or stdin):
ex -s +'/GG/j|%p|q!' file_or_/dev/stdin
where:
/GG/j - joins the next line when the pattern is found
%p - prints the buffer
q! - quits
For conditional checking (if there is a blank line), try:
ex -s +'%s/^\(G\)\n/\1/' +'%p|q!' file_or_/dev/stdin
I have a log file containing the following info:
<msisdn>37495989804</msisdn>
<address>10.14.14.26</address>
<msisdn>37495371855</msisdn>
<address>10.14.0.172</address>
<msisdn>37495989832</msisdn>
<address>10.14.14.29</address>
<msisdn>37495479810</msisdn>
<address>10.14.1.11</address>
<msisdn>37495429157</msisdn>
<address>10.14.0.213</address>
<msisdn>37495275824</msisdn>
<msisdn>37495739176</msisdn>
<address>10.14.2.86</address>
<msisdn>37495479840</msisdn>
<address>10.14.1.12</address>
<msisdn>37495706059</msisdn>
<msisdn>37495619889</msisdn>
<address>10.14.1.198</address>
<msisdn>37495574341</msisdn>
<address>10.14.1.148</address>
<msisdn>37495391624</msisdn>
<address>10.14.0.188</address>
<msisdn>37495989796</msisdn>
<address>10.14.14.24</address>
<msisdn>37495835940</msisdn>
<address>10.14.2.164</address>
<msisdn>37495743249</msisdn>
<address>10.14.2.94</address>
<msisdn>37495674117</msisdn>
<address>10.14.1.236</address>
<msisdn>37495754536</msisdn>
<address>10.14.2.120</address>
<msisdn>37495576434</msisdn>
<msisdn>37495823889</msisdn>
<address>10.14.2.159</address>
There are some lines where the 'msisdn' line is not followed by an 'address' line, like this:
<msisdn>37495576434</msisdn>
<msisdn>37495823889</msisdn>
I would like to write a script which will output only the lines ('msisdn' lines), that aren't followed by 'address'. Expected output:
<msisdn>37495275824</msisdn>
<msisdn>37495706059</msisdn>
<msisdn>37495576434</msisdn>
If it will be smth with awk/sed, it will be perfect.
Thanks.
One way with awk:
awk '/address/{p=0}p{print a;p=0}/msisdn/{a=$0;p=1}' log
you can use pcregrep to match next line is not adddress and use awk show it
pcregrep -M '(.*</msisdn>)\n.*<msi' | awk 'NR % 2 == 1'
This might work for you (GNU sed):
sed -r '$!N;/(<msisdn>).*\n.*\1/P;D' file
This reads 2 lines into the pattern space and trys to match the pattern <msisdn> in both the 2 lines. If the pattern matchs it prints out the first line. The first line is then deleted and the process begins again, however since the pattern space contains the second line (now the first), the automatic reading of a line is forgone and process begins as of $!N.
Perl has its own way to do this:
perl -lne 'if($prev && $_!~/\./){print $prev}unless(/\./){$prev=$_}else{undef $prev}' your_file
Tested Below:
> perl -lne 'if($prev && $_!~/\./){print $prev}unless(/\./){$prev=$_}else{undef $prev}' temp
<msisdn>37495275824</msisdn>
<msisdn>37495706059</msisdn>
<msisdn>37495576434</msisdn>
>
Does anyone know how to replace line a with line b and line b with line a in a text file using the sed editor?
I can see how to replace a line in the pattern space with a line that is in the hold space (i.e., /^Paco/x or /^Paco/g), but what if I want to take the line starting with Paco and replace it with the line starting with Vinh, and also take the line starting with Vinh and replace it with the line starting with Paco?
Let's assume for starters that there is one line with Paco and one line with Vinh, and that the line Paco occurs before the line Vinh. Then we can move to the general case.
#!/bin/sed -f
/^Paco/ {
:notdone
N
s/^\(Paco[^\n]*\)\(\n\([^\n]*\n\)*\)\(Vinh[^\n]*\)$/\4\2\1/
t
bnotdone
}
After matching /^Paco/ we read into the pattern buffer until s// succeeds (or EOF: the pattern buffer will be printed unchanged). Then we start over searching for /^Paco/.
cat input | tr '\n' 'ç' | sed 's/\(ç__firstline__\)\(ç__secondline__\)/\2\1/g' | tr 'ç' '\n' > output
Replace __firstline__ and __secondline__ with your desired regexps. Be sure to substitute any instances of . in your regexp with [^ç]. If your text actually has ç in it, substitute with something else that your text doesn't have.
try this awk script.
s1="$1"
s2="$2"
awk -vs1="$s1" -vs2="$s2" '
{ a[++d]=$0 }
$0~s1{ h=$0;ind=d}
$0~s2{
a[ind]=$0
for(i=1;i<d;i++ ){ print a[i]}
print h
delete a;d=0;
}
END{ for(i=1;i<=d;i++ ){ print a[i] } }' file
output
$ cat file
1
2
3
4
5
$ bash test.sh 2 3
1
3
2
4
5
$ bash test.sh 1 4
4
2
3
1
5
Use sed (or not at all) for only simple substitution. Anything more complicated, use a programming language
A simple example from the GNU sed texinfo doc:
Note that on implementations other than GNU `sed' this script might
easily overflow internal buffers.
#!/usr/bin/sed -nf
# reverse all lines of input, i.e. first line became last, ...
# from the second line, the buffer (which contains all previous lines)
# is *appended* to current line, so, the order will be reversed
1! G
# on the last line we're done -- print everything
$ p
# store everything on the buffer again
h