I want to print next line of matching word with sed.
I tried this command but it gives error :
sed -n '/<!\[CDATA\[\]\]>/ { N p}/' test.xml
what about grep -e -A 1 regex? It will print line below regex.
With sed, looking for pattern "dd", below works fine as you would:
sed -n '/dd/ {n;p}' file
For file content:
dd
aa
ss
aa
It prints:
aa
use awk
awk '/pattern/{getline;print}' file
Related
I have a tab file with two columns like below
BB_12 100_AA
BB_13 101_AB
BB_14 102_AD
BB_15 103_AC
I wish to remove the number_ in second column (replace number_ with nothing). For this I tried sed replace in the following ways unsuccessfully.
sed 's/\d+\_//g' infile
sed 's/(\d+\_)//g' infile
But none of the tweaks worked. It looks like it is not searching in 2nd column. How to modify this ? The expected output is
BB_12 AA
BB_13 AB
BB_14 AD
BB_15 AC
Thanks in advance.
You may just process the last column with sed:
sed -E 's/[^ ]*_([^ ]*) *$/\1/' file
The output:
BB_12 AA
BB_13 AB
BB_14 AD
BB_15 AC
Awk alternative:
awk '{ sub(/^[^ ]+_/, "", $2) }1' OFS='\t' file
Following simple sed may help you in same.
sed 's/\([^ ]*\) \([^_]*\)_\(.*\)/\1 \3/g' Input_file
Output will be as follows.
BB_12 AA
BB_13 AB
BB_14 AD
BB_15 AC
I am trying replace a block of code between two patterns with blank lines
Tried using below command
sed '/PATTERN-1/,/PATTERN-2/d' input.pl
But it only removes the lines between the patterns
PATTERN-1 : "=head"
PATTERN-2 : "=cut"
input.pl contains below text
=head
hello
hello world
world
morning
gud
=cut
Required output :
=head
=cut
Can anyone help me on this?
$ awk '/=cut/{f=0} {print (f ? "" : $0)} /=head/{f=1}' file
=head
=cut
To modify the given sed command, try
$ sed '/=head/,/=cut/{//! s/.*//}' ip.txt
=head
=cut
//! to match other than start/end ranges, might depend on sed implementation whether it dynamically matches both the ranges or statically only one of them. Works on GNU sed
s/.*// to clear these lines
awk '/=cut/{found=0}found{print "";next}/=head/{found=1}1' infile
# OR
# ^ to take care of line starts with regexp
awk '/^=cut/{found=0}found{print "";next}/^=head/{found=1}1' infile
Explanation:
awk '/=cut/{ # if line contains regexp
found=0 # set variable found = 0
}
found{ # if variable found is nonzero value
print ""; # print ""
next # go to next line
}
/=head/{ # if line contains regexp
found=1 # set variable found = 1
}1 # 1 at the end does default operation
# print current line/row/record
' infile
Test Results:
$ cat infile
=head
hello
hello world
world
morning
gud
=cut
$ awk '/=cut/{found=0}found{print "";next}/=head/{found=1}1' infile
=head
=cut
This might work for you (GNU sed):
sed '/=head/,/=cut/{//!z}' file
Zap the lines between =head and =cut.
How to print only string figure with the following line :
\begin{figure}[h!]
I tried :
firstLine='\begin{figure}[h!]'
echo $firstLine | sed -n 's/\\begin{\(.*\)}/\1/p'
but returns :
figure[h!] instead of figure
It seems that issue comes from [] or ! character.
firstLine='\begin{figure}[h!]'
echo "$firstLine" | sed 's/.*{\(.*\)}.*/\1/'
Output:
figure
With your code (add .*):
echo $firstLine | sed -n 's/\\begin{\(.*\)}.*/\1/p'
This might work for you (GNU sed):
sed 's/.*{\(.*\)}.*/\1/' file
This assumes there is only one {...} expression and one line.
A more rigorous solution would be:
sed -n 's/.*\\begin{\([^}]*\)}.*/\1/p' file
However nothing would be output if no match was found.
When I want to print an output like this
./myScript (prints some lines)
or
cat myFile
I want the output to show with linebreakers , for example each line will include not more than 100 chars.
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffff
vbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbf
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
there is something I can add to the command line to get this result ?
Thanks.
You can use sed if you want the line terminator as ,.
$ cat myfile
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffffvbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbfaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
$ sed -r 's/.{50}/&,\n/g' myfile
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa,
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffffvbbb,
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbfaaaaaa,
aaaaaaaaaaaaaaaaaaaaaaaaaaaaa
fold is another utility but won't add a , at the end
$ fold -w50 myfile
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffffvbbb
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbfaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaa
I have a file which looks like below:
memory=500G
brand=HP
color=black
battery=5 hours
For every line, I want to remove everything after = and also the =.
Eventually, I want to get something like:
memory:brand:color:battery:
(All on one line with colons after every word)
Is there a one-line sed command that I can use?
sed -e ':a;N;$!ba;s/=.\+\n\?/:/mg' /my/file
Adapted from this fine answer.
To be frank, however, I'd find something like this more readable:
cut -d = -f 1 /my/file | tr \\n :
Here's one way using GNU awk:
awk -F= '{ printf "%s:", $1 } END { printf "\n" }' file.txt
Result:
memory:brand:color:battery:
If you don't want a colon after the last word, you can use GNU sed like this:
sed -n 's/=.*//; H; $ { g; s/\n//; s/\n/:/g; p }' file.txt
Result:
memory:brand:color:battery
This might work for you (GNU sed):
sed -i ':a;$!N;s/=[^\n]*\n\?/:/;ta' file
perl -F= -ane '{print $F[0].":"}' your_file
tested below:
> cat temp
abc=def,100,200,dasdas
dasd=dsfsf,2312,123,
adasa=sdffs,1312,1231212,adsdasdasd
qeweqw=das,13123,13,asdadasds
dsadsaa=asdd,12312,123
> perl -F= -ane '{print $F[0].":"}' temp
abc:dasd:adasa:qeweqw:dsadsaa:
My command is
First step:
sed 's/([a-z]+)(\=.*)/\1:/g' Filename |cat >a
cat a
memory:
brand:
color:
battery:
Second step:
sed -e 'N;s/\n//' a | sed -e 'N;s/\n//'
My output is
memory:brand:color:battery: