I'm trying to find replace the following
<a href=\"test\">
with
<a href="test">
using sed.
I understand that both the \ and the " need to be escaped, so I do the following:
sed -i "s|a href=\\\"test\\\"|a href=\"test\"|g"
but this doesn't seem to work. Any idea what I'm doing wrong?
Just use single quotes instead of doubles.
$ echo '<a href=\"test\">' | sed "s|a href=\\\"test\\\"|a href=\"test\"|g"
<a href=\"test\">
$ echo '<a href=\"test\">' | sed 's|a href=\\\"test\\\"|a href=\"test\"|g'
<a href="test">
Ori, if you can't use single quotes, just add even more escapes :-)
$ echo '<a href=\"test\">' | sed "s|a href=\\\\\"test\\\\\"|a href=\"test\"|g"
<a href="test">
Related
I have lines that start like this: 2141058222 11/22/2017 and I want to append a ; at the end of the ten digit number like this: 2141058222; 11/22/2017.
I've tried sed with sed -i 's/^[0-9]\{10\}\\$/;&/g' which does nothing.
What am I missing?
Try this:
echo "2141058222 11/22/2017" | sed -r 's/^([0-9]{10})/&;/'
echo "2141058222 11/22/2017" | sed 's/ /; /'
Output:
2141058222; 11/22/2017
If the input is always in the format specified, GNU cut works, and might even be more efficient than sed:
cut -c -10,11- --output-delimiter ';' <<< "2141058222 11/22/2017"
Output:
2141058222; 11/22/2017
For an input file that'd be:
cut -c -10,11- --output-delimiter ';' file
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.
I'm trying to extract data/urls (in this case - someurl) from a file that contains them within some tag ie.
xyz>someurl>xyz
I don't mind using either awk or sed.
I think the best, easiest, way is with cut:
$ echo "xyz>someurl>xyz" | cut -d'>' -f2
someurl
With awk can be done like:
$ echo "xyz>someurl>xyz" | awk 'BEGIN { FS = ">" } ; { print $2 }'
someurl
And with sed is a little bit more tricky:
$ echo "xyz>someurl>xyz" | sed 's/\(.*\)>\(.*\)>\(.*\)/\2/g'
someurl
we get blocks of something1<something2<something3 and print the 2nd one.
grep was born to extract things:
kent$ echo "xyz>someurl>xyz"|grep -Po '>\K[^>]*(?=>)'
someurl
you could kill a fly with a bomb of course:
kent$ echo "xyz>someurl>xyz"|awk -F\> '$0=$2'
someurl
If your grep supports P option then you can use lookahead and lookbehind regular expression to identify the url.
$ echo "xyz>someurl>xyz" | grep -oP '(?<=xyz>).*(?=>xyz)'
someurl
This is just a sample to get you started not the final answer.
How to cut off known substring from the string in sh?
For example, I have string "http://www.myserver.org/very/very/long/path/mystring"
expression "http://www.myserver.org/very/very/long/path/" is known. How can I get "mystring"?
Thanks.
E.g. using perl:
echo "http://www.myserver.org/very/very/long/path/mystring" | perl -pe 's|^http://www.myserver.org/very/very/long/path/(.*)$|\1|'
E.g. using sed:
echo "http://www.myserver.org/very/very/long/path/mystring" | sed 's|^http://www.myserver.org/very/very/long/path/\(.*\)$|\1|'
E.g. when the search string is held in a variable, here named variable. Use double quotes to expand the variable.
echo "http://www.myserver.org/very/very/long/path/mystring" | sed "s|^${variable}\(.*\)$|\1|"
Tested under /bin/dash
$ S="http://www.myserver.org/very/very/long/path/mystring" && echo ${S##*/}
mystring
where
S is the variable-name
## remove largest prefix pattern
*/ upto the last slash
For further reading, search "##" in man dash
Some more illustrations:
$ S="/mystring/" ; echo ${S##*/}
$ S="/mystring" ; echo ${S##*/}
mystring
$ S="mystring" ; echo ${S##*/}
mystring
The sed below will output the input exactly. What I'd like to do is replace all occurrences of _ with - in the first matching group (\1), but not in the second. Is this possible?
echo 'abc_foo_bar=one_two_three' | sed 's/\([^=]*\)\(=.*\)/\1\2/'
abc_foo_bar=one_two_three
So, the output I'm hoping for is:
abc-foo-bar=one_two_three
I'd prefer not to resort to awk since I'm doing a string of other sed commands too, but I'll resort to that if I have to.
Edit: Minor fix to RE
You can do this in sed using the hold space:
$ echo 'abc_foo_bar=one_two_three' | sed 'h; s/[^=]*//; x; s/=.*//; s/_/-/g; G; s/\n//g'
abc-foo-bar=one_two_three
You could use awk instead of sed as follows:
echo 'abc_foo_bar=one_two_three' | awk -F= -vOFS== '{gsub("_", "-", $1); print $1, $2}'
The output would be, as expected:
abc-foo-bar=one_two_three
You could use ghc instead of sed as follows:
echo "abc_foo_bar=one_two_three" | ghc -e "getLine >>= putStrLn . uncurry (++) . (map (\x -> if x == '_' then '-' else x) *** id) . break (== '=')"
The output would be, as expected:
abc-foo-bar=one_two_three
This might work for you:
echo 'abc_foo_bar=one_two_three' |
sed 's/^/\n/;:a;s/\n\([^_=]*\)_/\1-\n/;ta;s/\n//'
abc-foo-bar=one_two_three
Or this:
echo 'abc_foo_bar=one_two_three' |
sed 'h;s/=.*//;y/_/-/;G;s/\n.*=/=/'
abc-foo-bar=one_two_three