I have SNMP outputs like:
IP-MIB::ipNetToMediaPhysAddress.5122.192.19.3.25 = STRING: 34:8:4:56:f4:70
As you can see mac-address output is incorrect, and i fix it with sed:
echo IP-MIB::ipNetToMediaPhysAddress.5122.192.19.3.25 = STRING: 34:8:4:56:f4:70 |
sed -e 's/\b\(\w\)\b/0\1/g'
Output:
IP-MIB::ipNetToMediaPhysAddress.5122.192.19.03.25 = STRING: 34:08:04:56:f4:70
It fixes address but changes IP as well from 192.19.3.25 to 192.19.03.25. How can I avoid it and force to perform sed only after STRING: or only after last space in the string ?
The MAC address is colon-separated. You can use that to limit the substitutions. This will perform the substitutions that you are interested in but only if the word character is next to a colon:
sed -e 's/\b\w:/0&/g; s/:\(\w\)\b/:0\1/g'
For example:
$ echo IP-MIB::ipNetToMediaPhysAddress.5122.192.19.3.25 = STRING: 34:8:4:56:f4:70 | sed -e 's/\b\w:/0&/g; s/:\(\w\)\b/:0\1/g'
IP-MIB::ipNetToMediaPhysAddress.5122.192.19.3.25 = STRING: 34:08:04:56:f4:70
How it works
s/\b\w:/0&/g
This performs the substitution if the word character is preceded by a word break, \b, and followed by a colon. Since we just need to put a zero in front of the entire matched text, not just some section of it, we can omit the parens and just use & to copy the matched text.
s/:\(\w\)\b/:0\1/g
If there are any remaining substitutions that need to be done where the word character is preceded by a colon and followed by a word break, this does them.
Note: We are using GNU extensions that may not be portable.
Another way with sed if the MAC address is at end of line
echo IP-MIB::ipNetToMediaPhysAddress.5122.192.19.3.25 = STRING: 4:8:d:56:f4:7 |
sed -E '
s/$/:/
:A
s/([^[:xdigit:]])([[:xdigit:]]:)/\10\2/
tA
s/:$//'
Related
I have the following string:
|**barak**.version|2001.0132012031539|
in file text.txt.
I would like to replace it with the following:
|**barak**.version|2001.01.2012031541|
So I run:
sed -i "s/\|\*\*$module\*\*.version\|2001.0132012031539/|**$module**.version|$version/" text.txt
but the result is a duplicate instead of replacing:
|**barak**.version|2001.01.2012031541|**barak**.version|2001.0132012031539|
What am I doing wrong?
Here is the value for module and version:
$ echo $module
barak
$ echo $version
2001.01.2012031541
Assumptions:
lines of interest start and end with a pipe (|) and have one more pipe somewhere in the middle of the data
search is based solely on the value of ${module} existing between the 1st/2nd pipes in the data
we don't know what else may be between the 1st/2nd pipes
the version number is the only thing between the 2nd/3rd pipes
we don't know the version number that we'll be replacing
Sample data:
$ module='barak'
$ version='2001.01.2012031541'
$ cat text.txt
**barak**.version|2001.0132012031539| <<<=== leave this one alone
|**apple**.version|2001.0132012031539|
|**barak**.version|2001.0132012031539| <<<=== replace this one
|**chuck**.version|2001.0132012031539|
|**barak**.peanuts|2001.0132012031539| <<<=== replace this one
One sed solution with -Extended regex support enabled and making use of a capture group:
$ sed -E "s/^(\|[^|]*${module}[^|]*).*/\1|${version}|/" text.txt
Where:
\| - first occurrence (escaped pipe) tells sed we're dealing with a literal pipe; follow-on pipes will be treated as literal strings
^(\|[^|]*${module}[^|]*) - first capture group that starts at the beginning of the line, starts with a pipe, then some number of non-pipe characters, then the search pattern (${module}), then more non-pipe characters (continues up to next pipe character)
.* - matches rest of the line (which we're going to discard)
\1|${version}| - replace line with our first capture group, then a pipe, then the new replacement value (${version}), then the final pipe
The above generates:
**barak**.version|2001.0132012031539|
|**apple**.version|2001.0132012031539|
|**barak**.version|2001.01.2012031541| <<<=== replaced
|**chuck**.version|2001.0132012031539|
|**barak**.peanuts|2001.01.2012031541| <<<=== replaced
An awk alternative using GNU awk:
awk -v mod="$module" -v vers="$version" -F \| '{ OFS=FS;split($2,map,".");inmod=substr(map[1],3,length(map[1])-4);if (inmod==mod) { $3=vers } }1' file
Pass two variables mod and vers to awk using $module and $version. Set the field delimiter to |. Split the second field into array map using the split function and using . as the delimiter. Then strip the leading and ending "**" from the first index of the array to expose the module name as inmod using the substr function. Compare this to the mod variable and if there is a match, change the 3rd delimited field to the variable vers. Print the lines with short hand 1
Pipe is only special when you're using extended regular expressions: sed -E
There's no reason why you need extended here, stick with basic regex:
sed "
# for lines matching module.version
/|\*\*$module\*\*.version|/ {
# replace the version
s/|2001.0132012031539|/|$version|/
}
" text.txt
or as an unreadable one-liner
sed "/|\*\*$module\*\*.version|/ s/|2001.0132012031539|/|$version|/" text.txt
I'm trying to remove all white spaces before and after 3 consecutive periods and replace it with the actual ellipse symbol.
I've tried the following code:
sed 's/[[:space:]]*\.\.\.[[:space:]]*/…/g'
It replaces the 3 periods with the ellipse symbol, but the spaces before and after remain.
Sample Input.
hello ... world
Desired output
hello…world
Expression you are using is ERE(extended regular expressions) you have to add -E option to sed as follows to allow it, since you are using character classes in your code [[:space:]].
sed -E 's/[[:space:]]*\.\.\.[[:space:]]*/.../g' Input_file
Without -E try:
sed 's/ *\.\.\. */.../g' Input_file
Here is another sed
echo "hello ... world" | sed -E 's/ +(\.\.\.) +/\1/g'
hello...world
4 dots, do nothing?
echo "hello .... world" | sed -E 's/ +(\.\.\.) +/\1/g'
hello .... world
In bash, just use parameter substitution...
foo="hello ... world"
foo="${foo//+( )...+( )/...}"
Now, echo "$foo", outputs:
hello...world
The syntax for BaSH regex variable substitution are as follows:
${var-name/search/replace}
A single /replaces only the first occurrence from the left, while a double //replaces every occurrence.
One of ?*+#! followed by (pattern-list) replaces a specified number of occurrences of the patterns in pattern-list as follows:
? Zero or one occurrence
* Zero or more occurrences
+ One or more occurrences
# A single occurence
! Anything that *doesn't* match one of the occurrences
Pattern list can be any combination of literal strings, or character classes, separated by the pipe character |
I am trying to change the following string
FROM java_jre_8#sha256:92f22331226b9b3c43a15eeeb304dd7
to
FROM docker-registry.service.consul:5000/java_jre_8#sha256:92f22331226b9b3c43a15eeeb304dd7
but am having difficult with sed as a result of / character
This is for a build server.
There are two ways of doing this. The first is to escape each / in the string you're replacing:
sed 's/from/to with \/ ... /'
The other, more simple way is to use a delimiter other than /. While most sed examples use / as a delimiter, you can use any character:
sed 's|from|to with / ...|'
Here, the | is the first character following s, and therefore sed knows to use this as a delimiter.
You can use # as the delimiter as it doesn't appear in your string (you can still use / but then you'll have to quote the actual /s that are part of the string).
sed "s#FROM java_jre_8##FROM docker-registry.service.consul:5000/java_jre_8##'
Example:
$ echo "FROM java_jre_8#sha256:92f22331226b9b3c43a15eeeb304dd7" | sed "s#FROM java_jre_8##FROM docker-registry.service.consul:5000/java_jre_8##"
FROM docker-registry.service.consul:5000/java_jre_8#sha256:92f22331226b9b3c43a15eeeb304dd7
I have strings which contain double quotes like this one:
"[{"clientid":"*", "identityzone":"*"}]"
I would like to use set or grep to delete the double quotes at the beginning and at the end of it, the output should look like :
[{"clientid":"*", "identityzone":"*"}]
I have used : sed -e 's/\"//g' but this deletes all the " in a string
You need to use line anchors
$ echo '"[{"clientid":"*", "identityzone":"*"}]"' | sed 's/^"//; s/"$//'
[{"clientid":"*", "identityzone":"*"}]
^" match " only at start of line
"$ match " only at end of line
You can also combine them using | as sed 's/^"\|"$//g'
See Overview of basic regular expression syntax
easy:
sed 's/^\"\(.*\)\"$/\1/g' <<<'"[{"clientid":"*", "identityzone":"*"}]"'
I have a string which i need to insert at a specific position in a file :
The file contains multiple semicolons(;) i need to insert the string just before the last ";"
Is this possible with SED ?
Please do post the explanation with the command as I am new to shell scripting
before :
adad;sfs;sdfsf;fsdfs
string = jjjjj
after
adad;sfs;sdfsf jjjjj;fsdfs
Thanks in advance
This might work for you:
echo 'adad;sfs;sdfsf;fsdfs'| sed 's/\(.*\);/\1 jjjjj;/'
adad;sfs;sdfsf jjjjj;fsdfs
The \(.*\) is greedy and swallows the whole line, the ; makes the regexp backtrack to the last ;. The \(.*\) make s a back reference \1. Put all together in the RHS of the s command means insert jjjjj before the last ;.
sed 's/\([^;]*\)\(;[^;]*;$\)/\1jjjjj\2/' filename
(substitute jjjjj with what you need to insert).
Example:
$ echo 'adad;sfs;sdfsf;fsdfs;' | sed 's/\([^;]*\)\(;[^;]*;$\)/\1jjjjj\2/'
adad;sfs;sdfsfjjjjj;fsdfs;
Explanation:
sed finds the following pattern: \([^;]*\)\(;[^;]*;$\). Escaped round brackets (\(, \)) form numbered groups so we can refer to them later as \1 and \2.
[^;]* is "everything but ;, repeated any number of times.
$ means end of the line.
Then it changes it to \1jjjjj\2.
\1 and \2 are groups matched in first and second round brackets.
For now, the shorter solution using sed : =)
sed -r 's#;([^;]+);$#; jjjjj;\1#' <<< 'adad;sfs;sdfsf;fsdfs;'
-r option stands for extented Regexp
# is the delimiter, the known / separator can be substituted to any other character
we match what's finishing by anything that's not a ; with the ; final one, $ mean end of the line
the last part from my explanation is captured with ()
finally, we substitute the matching part by adding "; jjjj" ans concatenate it with the captured part
Edit: POSIX version (more portable) :
echo 'adad;sfs;sdfsf;fsdfs;' | sed 's#;\([^;]\+\);$#; jjjjj;\1#'
echo 'adad;sfs;sdfsf;fsdfs;' | sed -r 's/(.*);(.*);/\1 jjjj;\2;/'
You don't need the negation of ; because sed is by default greedy, and will pick as much characters as it can.
sed -e 's/\(;[^;]*\)$/ jjjj\1/'
Inserts jjjj before the part where a semicolon is followed by any number of non-semicolons ([^;]*) at the end of the line $. \1 is called a backreference and contains the characters matched between \( and \).
UPDATE: Since the sample input has no longer a ";" at the end.
Something like this may work for you:
echo "adad;sfs;sdfsf;fsdfs"| awk 'BEGIN{FS=OFS=";"} {$(NF-1)=$(NF-1) " jjjjj"; print}'
OUTPUT:
adad;sfs;sdfsf jjjjj;fsdfs
Explanation: awk starts with setting FS (field separator) and OFS (output field separator) as semi colon ;. NF in awk stands for number of fields. $(NF-1) thus means last-1 field. In this awk command {$(NF-1)=$(NF-1) " jjjjj" I am just appending jjjjj to last-1 field.