Using sed -n "s/.*\(\/.*\/\).*/\1/p on /string1/string2 produces string1, as expected.
However, using the same on /string1/string2/string3 produces string2.
How can I print the first occurrence only, that is string1.
This does exactly what I wanted:
sed -n "s/[^/]*\(\/[a-z]*\).*/\1/p"
You can use this sed:
sed -n 's|/\([^/]*\)/.*|\1|p'
Avoid escaping / by using an alternate delimiter.
This might work for you (GNU sed):
sed -n 's/[^\/]*\(\/[^\/]*\/\).*/\1/p' file
perl's non-greedy regex quantifiers are handy:
perl -pe 's{.*?(/.*?/).*}{$1}' <<END
foobar/string1/string2/string3
END
/string1/
If you want to use bash shell:
str="foobar/string1/string2/string3"
string1=$( IFS=/; set -- $str; echo $2 )
Related
Currently I'm trying to use sed with regex on Solaris but it doesn't work.
I need to show only lines matching to my regex.
sed -n -E '/^[a-zA-Z0-9]*$|^a_[a-zA-Z0-9]*$/p'
input file:
grtad
a_pitr
_aupa
a__as
baman
12353
ai345
ki_ag
-MXx2
!!!23
+_)#*
I want to show only lines matching to above regex:
grtad
a_pitr
baman
12353
ai345
Is there another way to use alternative? Is it possible in perl?
Thanks for any solutions.
With Perl
perl -ne 'print if /^(a_)?[a-zA-Z0-9]*$/' input.txt
The (a_)? matches a_ one-or-zero times, so optionally. It may or may not be there.
The (a_) also captures the match, what is not needed. So you can use (?:a_)? instead. The ?: makes () only group what is inside (so ? applies to the whole thing), but not remember it.
with grep
$ grep -xiE '(a_)?[a-z0-9]*' ip.txt
grtad
a_pitr
baman
12353
ai345
-x match whole line
-i ignore case
-E extended regex, if not available, use grep -xi '\(a_\)\?[a-z0-9]*'
(a_)? zero or one time match a_
[a-z0-9]* zero or more alphabets or numbers
With sed
sed -nE '/^(a_)?[a-zA-Z0-9]*$/p' ip.txt
or, with GNU sed
sed -nE '/^(a_)?[a-z0-9]*$/Ip' ip.txt
Assuming a simple text file:
123.123.123.123
I would like to replace the IP inside of it with 222.222.222.222. I have tried the below but nothing changes, however the same regex seems to work in this Regexr
sed -i '' 's/(\d{1,3}\.){3}\d{1,3}/222.222.222.222/' file.txt
Am I missing something?
Two problems here:
sed doesn't like PCRE digit property \d, use range: [0-9] or POSIX [[:digit:]]
You need to use -r flag for extended regex as well.
This should work:
s='123.123.123.123'
sed -r 's/([0-9]{1,3}\.){3}[0-9]{1,3}/222.222.222.222/' <<< "$s"
222.222.222.222
Better would be to use anchors to avoid matching unexpected input:
sed -r 's/^([0-9]{1,3}\.){3}[0-9]{1,3}$/222.222.222.222/' <<< "$s"
PS: On OSX use -E instead of -r:
sed -E 's/^([0-9]{1,3}\.){3}[0-9]{1,3}$/222.222.222.222/' <<< "$s"
222.222.222.222
You'd better use -r, as indicated by anubhava.
But in case you don't have it, you have to escape every single (, ), { and }. And also, use [0-9] instead of \d:
$ sed 's/\([0-9]\{1,3\}\.\)\{3\}[0-9]\{1,3\}/222.222.222.222/' <<< "123.123.123.123"
222.222.222.222
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 write a sed command to remove a specific string followed by two digits. So far I have:
sed -e 's/bizzbuzz\([0-9][0-9]\)//' file.txt
but I cant seem to get the syntax right. Any suggestions?
sed -re 's/bizzbuzz[0-9]{2}//' file.txt
and
sed -re 's/\bbizzbuzz[0-9]{2}\b//' file.txt
if the searched string have word boundary
sed -e 's/bizzbuzz[0-9]\{2\}//' file.txt
if you don't have GNU sed
Your current approach seems like it should work fine:
$ echo 'FOO bizzbuzz56 BAR' | sed -e 's/bizzbuzz\([0-9][0-9]\)//'
FOO BAR
As said in other answer, the syntax seems to be fine (with unnecesary parenthesis).
But may be you want to replace all the strings found in each line ? In that case, you should add a 'g' at the end of the 's' command:
sed -e 's/bizzbuzz\([0-9][0-9]\)//g' file.txt
I want to match the following line with the regex in sed:
db=connect_str=DBI:SQLAnywhere:ENG=ABC1_hostname12;DBN=ABC12;UID=USERID;PWD=passwd123;LINKS=tcpip{host=10.11.12.13:1234}
The regex I am using is:
sed -n '/ABC1_.+;/p' Config/db_conn.cfg
but this does not work. On the other hand, it works if I use:
sed -n '/ABC1_.*;/p' Config/db_conn.cfg
Can someone please explain why it's not working? Also is there another way to match it?
It's because sed is basic regex by default, which needs + to be escaped or else it represents a literal + instead of a regex +:
sed -n '/ABC1_.\+;/p' Config/db_conn.cfg
To use regex you're familiar with try sed -r -n (extended regex) and then you can do:
sed -r -n '/ABC1_.+;/p' Config/db_conn.cfg