How can I extract all the strings inside {} via sed
{}
{A}
{
BC
D
}
expected output:
//empty line
A
BC
D
Try:
echo $str | cut -d "{" -f2 | cut -d "}" -f1
With GNU sed:
sed 's/{\(.*\)}/\1/;/^[{}]$/d' file
Output:
A
BC
D
Related
Can anybody help me please?
grep " 287 " file.txt | grep "HI" | sed -i 's/HIS/HID/g'
sed: no input files
Tried also xargs
grep " 287 " file.txt | grep HI | xargs sed -i 's/HIS/HID/g'
sed: invalid option -- '6'
This works fine
grep " 287 " file.txt | grep HI
If you want to keep your pipeline:
f=file.txt
tmp=$(mktemp)
grep " 287 " "$f" | grep "HI" | sed 's/HIS/HID/g' > "$tmp" && mv "$tmp" "$f"
Or, simplify:
sed -i -n '/ 287 / {/HI/ s/HIS/HID/p}' file.txt
That will filter out any line that does not contain " 287 " and "HI" -- is that what you want? I suspect you really want this:
sed -i '/ 287 / {/HI/ s/HIS/HID/}' file.txt
For lines that match / 287 /, execute the commands in braces. In there, for lines that match /HI/, search for the first "HIS" and replace with "HID". sed implicitly prints all lines if -n is not specified.
Other commands that do the same thing:
awk '/ 287 / && /HI/ {sub(/HIS/, "HID")} {print}' file.txt > new.txt
perl -i -pe '/ 287 / and /HI/ and s/HIS/HID/' file.txt
awk does not have an "in-place" option (except gawk -i inplace for recent gawk versions)
I'm using this code to get all titles from urls with http://something.txt:
#!/usr/bin/perl -w
$output = `cat source.html | grep -o '<a .*href=.*>' | grep -E 'txt' | sed -e 's/<a /\n<a /g' | sed -e 's/<a .*title="//' | cut -f1 -d '"'`;
print("$output");
When i run this on perl i get the error:
sed: -e expression #1, char 6: unterminated `s' command
The error is related with this portion of code:
sed -e 's/<a /\n<a /g'
In backquotes, Perl uses the same rules as in double quotes. Therefore, \n corresponds to a newline; you have to backslash the backslash to pass literal \ to the shell:
`sed -e 's/<a /\\n<a /g'`
I have a file containing:
L1
L2
L3
.
.
.
L512
I want to change its content to :
L1 | L2 | L3 | ... | L512
It seems so easy , but its now 1 hour Im sitting and trying to make it, I tried to do it by sed, but didn't get what I want. It seems that sed just inserts empty lines between the content, any suggestion please?
With sed this requires to read the whole input into a buffer and afterwards replace all newlines by |, like this:
sed ':a;N;$!ba;s/\n/ | /g' input.txt
Part 1 - buffering input
:a defines a label called 'a'
N gets the next line from input and appends it to the pattern buffer
$!ba jumps to a unless the end of input is reached
Part 2 - replacing newlines by |
s/\n/|/ execute the substitute command on the pattern buffern
As you can see, this is very inefficient since it requires to:
read the complete input into memory
operate three times on the input: 1. reading, 2. substituting, 3. printing
Therefore I would suggest to use awk which can do it in one loop:
awk 'NR==1{printf $0;next}{printf " | "$0}END{print ""}' input.txt
Here is one sed
sed ':a;N;s/\n/ | /g;ta' file
L1 | L2 | L3 | ... | L512
And one awk
awk '{printf("%s%s",sep,$0);sep=" | "} END {print ""}' file
L1 | L2 | L3 | ... | L512
perl -pe 's/\n/ |/g unless(eof)' file
if space between | is not mandatory
tr "\n" '|' YourFile
Several options, including those mentioned here:
paste -sd'|' file
sed ':a;N;s/\n/ | /g;ta' file
sed ':a;N;$!ba;s/\n/ | /g' file
perl -0pe 's/\n/ | /g;s/ \| $/\n/' file
perl -0nE 'say join " | ", split /\n/' file
perl -E 'chomp(#x=<>); say join " | ", #x' file
mapfile -t ary < file; (IFS="|"; echo "${ary[*]}")
awk '{printf("%s%s",sep,$0);sep=" | "} END {print ""}' file
I have a example cut down from a log file.
112 172.172.172.1#50912 (ssl.bing.com):
I would like some how to remove the # and numbers after and (): from the url.
Would like the result.
112 172.172.172.1 ssl.bing.com
Here is the sed oneliner I have been working on.
cat newdns.log | sed -e 's/.*query: //' | cut -f 1 -d' ' | sort | uniq -c | sort -k2 > old.log
Thanks
Using sed, you could say:
sed 's/#[0-9]*//;s/(\(.*\)):$/\1/' filename
or, in a single substitution:
sed 's/#[0-9]* *(\(.*\)):$/ \1/' filename
Another sed:
sed -r 's/#[^ ]+|[():]//g'
$ echo '112 172.172.172.1#50912 (ssl.bing.com):' | sed -r 's/#[^ ]+|[():]//g'
112 172.172.172.1 ssl.bing.com
How do I extract 68 from v1+r0.68?
Using awk, returns everything after the last '.'
echo "v1+r0.68" | awk -F. '{print $NF}'
Using sed to get the number after the last dot:
echo 'v1+r0.68' | sed 's/.*[.]\([0-9][0-9]*\)$/\1/'
grep is good at extracting things:
kent$ echo " v1+r0.68"|grep -oE "[0-9]+$"
68
Match the digit string before the end of the line using grep:
$ echo 'v1+r0.68' | grep -Eo '[0-9]+$'
68
Or match any digits after a .
$ echo 'v1+r0.68' | grep -Po '(?<=\.)\d+'
68
Print everything after the . with awk:
echo "v1+r0.68" | awk -F. '{print $NF}'
68
Substitute everything before the . with sed:
echo "v1+r0.68" | sed 's/.*\.//'
68
type man grep
and you will see
...
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
then type echo 'v1+r0.68' | grep -o '68'
if you want it any where special do:
echo 'v1+r0.68' | grep -o '68' > anyWhereSpecial.file_ending