I'm trying swap words around with sed, not replace because that's what I keep finding on Google search.
I don't know if it's the regex that I'm getting wrong. I did a search for everything before a char and everything after a char, so that's how I got the regex.
echo xxx,aaa | sed -r 's/[^,]*/[^,]*$/'
or
echo xxx/aaa | sed -r 's/[^\/]*/[^\/]*$/'
I am getting this output:
[^,]*$,aaa
or this:
[^,/]*$/aaa
What am I doing wrong?
For the first sample, you should use:
echo xxx,aaa | sed 's/\([^,]*\),\([^,]*\)/\2,\1/'
For the second sample, simply use a character other than slash as the delimiter:
echo xxx/aaa | sed 's%\([^/]*\)/\([^/]*\)%\2/\1%'
You can also use \{1,\} to formally require one or more:
echo xxx,aaa | sed 's/\([^,]\{1,\}\),\([^,]\{1,\}\)/\2,\1/'
echo xxx/aaa | sed 's%\([^/]\{1,\}\)/\([^/]\{1,\}\)%\2/\1%'
This uses the most portable sed notation; it should work anywhere. With modern versions that support extended regular expressions (-r with GNU sed, -E with Mac OS X or BSD sed), you can lose some of the backslashes and use + in place of * which is more precisely what you're after (and parallels \{1,\} much more succinctly):
echo xxx,aaa | sed -E 's/([^,]+),([^,]+)/\2,\1/'
echo xxx/aaa | sed -E 's%([^/]+)/([^/]+)%\2/\1%'
With sed it would be:
sed 's#\([[:alpha:]]\+\)/\([[:alpha:]]\+\)#\2,\1#' <<< 'xxx/aaa'
which is simpler to read if you use extended posix regexes with -r:
sed -r 's#([[:alpha:]]+)/([[:alpha:]]+)#\2/\1#' <<< 'xxx/aaa'
I'm using two sub patterns ([[:alpha:]]+) which can contain one or more letters and are separated by a /. In the replacement part I reassemble them in reverse order \2/\1. Please also note that I'm using # instead of / as the delimiter for the s command since / is already the field delimiter in the input data. This saves us to escape the / in the regex.
Btw, you can also use awk for that, which is pretty easy to read:
awk -F'/' '{print $2,$1}' OFS='/' <<< 'xxx/aaa'
Related
I am running a linux bash script. From stout lines like: /gpx/trk/name=MyTrack1, I want to keep only the end of line after =.
I am struggling to understand why the following sed command is not working as I expect:
echo "/gpx/trk/name=MyTrack1" | sed -e "s/^*=//"
(I also tried)
echo "/gpx/trk/name=MyTrack1" | sed -e "s/^*\=//"
The return is always /gpx/trk/name=MyTrack1 and not MyTrack1
An even simpler way if this is the only structure you are concerned about:
echo "/gpx/trk/name=MyTrack1" | cut -d = -f 2
Simply try:
echo "/gpx/trk/name=MyTrack1" | sed 's/.*=//'
Solution 2nd: With another sed.
echo "/gpx/trk/name=MyTrack1" | sed 's/\(.*=\)\(.*\)/\2/'
Explanation: As per OP's request adding explanation for this code here:
s: Means telling sed to do substitution operation.
\(.*=\): Creating first place in memory to keep this regex's value which tells sed to keep everything in 1st place of memory from starting to till = so text /gpx/trk/name= will be in 1 place.
\(.*\): Creating 2nd place in memory for sed telling it to keep everything now(after the match of 1st one, so this will start after =) and have value in it as MyTrack1
/\2/: Now telling sed to substitute complete line with only 2nd memory place holder which is MyTrack1
Solution 3rd: Or with awk considering that your Input_file is same as shown samples.
echo "/gpx/trk/name=MyTrack1" | awk -F'=' '{print $2}'
Solution 4th: With awk's match.
echo "/gpx/trk/name=MyTrack1" | awk 'match($0,/=.*$/){print substr($0,RSTART+1,RLENGTH-1)}'
$ echo "/gpx/trk/name=MyTrack1" | sed -e "s/^.*=//"
MyTrack1
The regular expression ^.*= matches anything up to and including the last = in the string.
Your regular expression ^*= would match the literal string *= at the start of a string, e.g.
$ echo "*=/gpx/trk/name=MyTrack1" | sed -e "s/^*=//"
/gpx/trk/name=MyTrack1
The * character in a regular expression usually modifies the immediately previous expression so that zero or more of it may be matched. When * occurs at the start of an expression on the other hand, it matches the character *.
Not to take you off the sed track, but this is easy with Bash alone:
$ echo "$s"
/gpx/trk/name=MyTrack1
$ echo "${s##*=}"
MyTrack1
The ##*= pattern removes the maximal pattern from the beginning of the string to the last =:
$ s="1=2=3=the rest"
$ echo "${s##*=}"
the rest
The equivalent in sed would be:
$ echo "$s" | sed -E 's/^.*=(.*)/\1/'
the rest
Where #*= would remove the minimal pattern:
$ echo "${s#*=}"
2=3=the rest
And in sed:
$ echo "$s" | sed -E 's/^[^=]*=(.*)/\1/'
2=3=the rest
Note the difference in * in Bash string functions vs a sed regex:
The * in Bash (in this context) is glob like - itself means 'any character'
The * in a regex refers to the previous pattern and for 'any character' you need .*
Bash has extensive string manipulation functions. You can read about Bash string patterns in BashFAQ.
I would like to print directly with sed a HEX value translation by isolating the HEX values in capture groups. This works:
echo bbb3Accc | sed -n 's/3A/\x3A/p'
bbb:ccc
...but this doesn't work:
echo bbb3Accc | sed 's/\(3A\)/\x\1/'
bbbx3Accc
...or an actual capture group REGEX matching based on URL encoded strings:
echo bbb%3Accc | sed 's/%\([A-Za-z0-9]\)/\x\1/'
bbbx3Accc
Apparently sed no longer interprets and translates the HEX value if it is constructed from a REGEX capture group, together with the \x escape.
But I am wondering if there's a workaround that I am not aware of, to make this work only with sed. Note that I am aware that I can do a bash command substitution and wrap the sed syntax in a echo -e but I would like to avoid that.
Your question isn't clear but maybe this is what you're trying to do using GNU awk for multi-char RS, RT, and strtonum():
$ echo 'bbb%3Accc%21ddd' |
gawk -v RS='%[[:xdigit:]]{2}' 'sub(/%/,"0x",RT){RT=sprintf("%c",strtonum(RT))} {ORS=RT} 1'
bbb:ccc!ddd
As mentioned in the comments, \xAB is interpreted by sed's parser, rather than as an expression, so \x won't work in the way you were trying.
sed is pretty primitive and your example is beyond what it is intended for, so you'd be better off using something more general purpose. For example, in Perl:
$ echo bbb3Accc | perl -ple 's/([0-9A-F]{2})/chr(hex($1))/ge'
bbb:ccc
Using Sed I tried but it did not worked out.
Basically, I have a string say:-
Input:-
'http://www.google.com/photos'
Output required:-
http://www.google.com
I tried using sed but escaping ' is not possible.
what i did was:-
sed 's/\'//' | sed 's/photos//'
sed for photos worked but for ' it didn't.
Please suggest what can be the solution.
Escaping ' in sed is possible via a workaround:
sed 's/'"'"'//g'
# |^^^+--- bash string with the single quote inside
# | '--- return to sed string
# '------- leave sed string and go to bash
But for this job you should use tr:
tr -d "'"
Perl Replacements have a syntax identical to sed, works better than sed, is installed almost in every system by default and works for all machines the same way (portability):
$ echo "'http://www.google.com/photos'" |perl -pe "s#\'##g;s#(.*//.*/)(.*$)#\1#g"
http://www.google.com/
Mind that this solution will keep only the domain name with http in front, discarding all words following http://www.google.com/
If you want to do it with sed , you can use sed "s/'//g" as advised by Wiktor Stribiżew in comments.
PS: I sometimes refer to special chars with their ascii hex code of the special char as advised by man ascii, which is \x27 for '
So for sed you can do it:
$ echo "'http://www.google.com/photos'" |sed -r "s#'##g; s#(.*//.*/)(.*$)#\1#g;"
http://www.google.com/
# sed "s#\x27##g' will also remove the single quote using hex ascii code.
$ echo "'http://www.google.com/photos'" |sed -r "s#'##g; s#(.*//.*)(/.*$)#\1#g;"
http://www.google.com #Without the last slash
If your string is stored in a variable, you can achieve above operations with pure bash, without the need of external tools like sed or perl like this:
$ a="'http://www.google.com/photos'" && a="${a:1:-1}" && echo "$a"
http://www.google.com/photos
# This removes 1st and last char of the variable , whatever this char is.
$ a="'http://www.google.com/photos'" && a="${a:1:-1}" && echo "${a%/*}"
http://www.google.com
#This deletes every char from the end of the string up to the first found slash /.
#If you need the last slash you can just add it to the echo manually like echo "${a%/*}/" -->http://www.google.com/
It's unclear if the ' are actually around your string, although this should take care it:
str="'http://www.google.com/photos'"
echo "$str" | sed s/\'//g | sed 's/\/photos//g'
Combined:
echo "$str" | sed -e "s/'//g" -e 's/\/photos//g'
Using tr:
echo "$str" | sed -e "s/\/photos//g" | tr -d \'
Result:
http://www.google.com
If the single quotes are not around your string it should work regardless.
Everyone!!
I want to get specific substring from stdout of command.
stdout:
{"response":
{"id":"110200dev1","success":"true","token":"09ad7cc7da1db13334281b84f2a8fa54"},"success":"true"}
I need to get a hex string after token without quotation marks, the length of hex string is 32 letters.I suppose it can be done by sed or egrep. I don't want to use awk here. Because the stdout is being changed very often.
This is an alternate gnu-awk solution when grep -P isn't available:
awk -F: '{gsub(/"/, "")} NF==2&&$1=="token"{print $2}' RS='[{},]' <<< "$string"
09ad7cc7da1db13334281b84f2a8fa54
grep's nature is extracting things:
grep -Po '"token":"\K[^"]+'
-P option interprets the pattern as a Perl regular expression.
-o option shows only the matching part that matches the pattern.
\K throws away everything that it has matched up to that point.
Or an option using sed...
sed 's/.*"token":"\([^"]*\)".*/\1/'
With sed:
your-command | sed 's/.*"token":"\([^"]*\)".*/\1/'
YourStreamOrFile | sed -n 's/.*"token":"\([a-f0-9]\{32\}\)".*/\1/p'
doesn not return a full string if not corresponding
I'm trying to replace /./ or /././ or /./././ to / only in bash script. I've managed to create regex for sed but it doesn't work.
variable="something/./././"
variable=$(echo $variable | sed "s/\/(\.\/)+/\//g")
echo $variable # this should output "something/"
When I tried to replace only /./ substring it worked with regex in sed \/\.\/. Does sed regex requires more flags to use multiplication of substring with + or *?
Use -r option to make sed to use extended regular expression:
$ variable="something/./././"
$ echo $variable | sed -r "s/\/(\.\/)+/\//g"
something/
Any sed:
sed 's|/\(\./\)\{1,\}|/|g'
But a + or \{1,\} would not even be required in this case, a * would do nicely, so
sed 's|/\(\./\)*|/|g'
should suffice
Two things to make it simple:
$ variable="something/./././"
$ sed -r 's#(\./){1,}##' <<< "$variable"
something/
Use {1,} to indicate one or more patterns. You won't need g with this.
Use different delimiterers # in above case to make it readable
+ is ERE so you need to enable -E or -r option to use it
You can also do this with bash's built-in parameter substitution. This doesn't require sed, which doesn't accept -r on a Mac under OS X:
variable="something/./././"
a=${variable/\/*/}/ # Remove slash and everything after it, then re-apply slash afterwards
echo $a
something/
See here for explanation and other examples.