sed replacement for no_proxy cuts number in localhost - bash

I have a no_proxy string I'd like to modify to fit with Java Opts.
❯ export no_proxy_test=localhost,127.0.0.1,.myfancy.domain,.myotherfancy.domain
❯ echo $no_proxy_test
localhost,127.0.0.1,.myfancy.domain,.myotherfancy.domain
I need for each value to be delimited with pipes "|" instead and I need to remove the dots for the wildcard. So should be:
localhost|127.0.0.1|myfancy.domain|myotherfancy.domain
When I use:
❯ echo $no_proxy_test | sed 's/,./|/g'
localhost|27.0.0.1|myfancy.domain|myotherfancy.domain
For some reason it cuts the 1 in 127.0.0.1 and I don't understand why? I thought I may achieve it with a double sed:
❯ echo $no_proxy_test | sed 's/,/|/g' | sed 's/|./|/g'
localhost|27.0.0.1|myfancy.domain|myotherfancy.domain
But same problem there. Does anyone have an idea? I don't want to sed replace 27 with 127. Would be interesting if I ran into a bug or if anyone could explain why 1 is being cut from the string.
Thank you!

You can use
sed -E 's/,\.?/|/g' # POSIX ERE
sed 's/,\.\{0,1\}/|/g' # POSIX BRE
This will replace a comma followed with an optional . char with a pipe symbol.
See the online demo:
#!/bin/bash
s='localhost,127.0.0.1,.myfancy.domain,.myotherfancy.domain'
sed 's/,\.\{0,1\}/|/g' <<< "$s"
# => localhost|127.0.0.1|myfancy.domain|myotherfancy.domain

Related

grep the folder using sed command in shell script

I am trying to grep the folder name from full tar file. Below is the example.
example:
TEST-5.3.0.0-build1.x86_64.tar.gz
I want to grep the folder name (TEST-5.3.0.0-build1) in shell script
So i tried below command for grep
$ package_folder=$(echo TEST-5.3.0.0-build1.x86_64.tar.gz | sed -e "s/.[0-9]*[a-z]*[0-9]*.tar.gz$//" | sed -e 's/\/$//')
But I am getting below output:
$ echo $package_folder
TEST-5.3.0.0-build1.x86
Could you please anyone correct me where I am doing mistake. I need folder name as TEST-5.3.0.0-build1
Thanks in Advance!!!
In your command, you do not match _, x, etc. The [0-9]*[a-z]*[0-9]* only matches a sequence of zero or more digits, zero or more (lowercase) letters, and zero or more digits. It is better to use a [^.]* to match any chars other than . between two . chars. Also, literal dots must be escaped, or an unescaped . will match any single char.
You can use
sed 's/\.[^.]*\.tar\.gz$//'
Or, just use string manipulation if x86_64 is also a constant:
s='TEST-5.3.0.0-build1.x86_64.tar.gz'
s="${s/.x86_64.tar.gz/}"
See the online demo:
#!/bin/bash
s='TEST-5.3.0.0-build1.x86_64.tar.gz'
package_folder=$(sed 's/\.[^.]*\.tar\.gz$//' <<< "$s")
echo "${package_folder}"
# => TEST-5.3.0.0-build1
s="${s/.x86_64.tar.gz/}"
echo "$s"
# => TEST-5.3.0.0-build1
You can use uname -m in replacement part of this string:
s='TEST-5.3.0.0-build1.x86_64.tar.gz'
echo "${s%.$(uname -m)*}"
TEST-5.3.0.0-build1
Or using sed:
sed "s/\.$(uname -m).*//" <<< "$s"
TEST-5.3.0.0-build1

Insert the contents of the variable in SED command [duplicate]

If I run these commands from a script:
#my.sh
PWD=bla
sed 's/xxx/'$PWD'/'
...
$ ./my.sh
xxx
bla
it is fine.
But, if I run:
#my.sh
sed 's/xxx/'$PWD'/'
...
$ ./my.sh
$ sed: -e expression #1, char 8: Unknown option to `s'
I read in tutorials that to substitute environment variables from shell you need to stop, and 'out quote' the $varname part so that it is not substituted directly, which is what I did, and which works only if the variable is defined immediately before.
How can I get sed to recognize a $var as an environment variable as it is defined in the shell?
Your two examples look identical, which makes problems hard to diagnose. Potential problems:
You may need double quotes, as in sed 's/xxx/'"$PWD"'/'
$PWD may contain a slash, in which case you need to find a character not contained in $PWD to use as a delimiter.
To nail both issues at once, perhaps
sed 's#xxx#'"$PWD"'#'
In addition to Norman Ramsey's answer, I'd like to add that you can double-quote the entire string (which may make the statement more readable and less error prone).
So if you want to search for 'foo' and replace it with the content of $BAR, you can enclose the sed command in double-quotes.
sed 's/foo/$BAR/g'
sed "s/foo/$BAR/g"
In the first, $BAR will not expand correctly while in the second $BAR will expand correctly.
Another easy alternative:
Since $PWD will usually contain a slash /, use | instead of / for the sed statement:
sed -e "s|xxx|$PWD|"
You can use other characters besides "/" in substitution:
sed "s#$1#$2#g" -i FILE
一. bad way: change delimiter
sed 's/xxx/'"$PWD"'/'
sed 's:xxx:'"$PWD"':'
sed 's#xxx#'"$PWD"'#'
maybe those not the final answer,
you can not known what character will occur in $PWD, / : OR #.
if delimiter char in $PWD, they will break the expression
the good way is replace(escape) the special character in $PWD.
二. good way: escape delimiter
for example:
try to replace URL as $url (has : / in content)
x.com:80/aa/bb/aa.js
in string $tmp
URL
A. use / as delimiter
escape / as \/ in var (before use in sed expression)
## step 1: try escape
echo ${url//\//\\/}
x.com:80\/aa\/bb\/aa.js #escape fine
echo ${url//\//\/}
x.com:80/aa/bb/aa.js #escape not success
echo "${url//\//\/}"
x.com:80\/aa\/bb\/aa.js #escape fine, notice `"`
## step 2: do sed
echo $tmp | sed "s/URL/${url//\//\\/}/"
URL
echo $tmp | sed "s/URL/${url//\//\/}/"
URL
OR
B. use : as delimiter (more readable than /)
escape : as \: in var (before use in sed expression)
## step 1: try escape
echo ${url//:/\:}
x.com:80/aa/bb/aa.js #escape not success
echo "${url//:/\:}"
x.com\:80/aa/bb/aa.js #escape fine, notice `"`
## step 2: do sed
echo $tmp | sed "s:URL:${url//:/\:}:g"
x.com:80/aa/bb/aa.js
With your question edit, I see your problem. Let's say the current directory is /home/yourname ... in this case, your command below:
sed 's/xxx/'$PWD'/'
will be expanded to
sed `s/xxx//home/yourname//
which is not valid. You need to put a \ character in front of each / in your $PWD if you want to do this.
Actually, the simplest thing (in GNU sed, at least) is to use a different separator for the sed substitution (s) command. So, instead of s/pattern/'$mypath'/ being expanded to s/pattern//my/path/, which will of course confuse the s command, use s!pattern!'$mypath'!, which will be expanded to s!pattern!/my/path!. I’ve used the bang (!) character (or use anything you like) which avoids the usual, but-by-no-means-your-only-choice forward slash as the separator.
Dealing with VARIABLES within sed
[root#gislab00207 ldom]# echo domainname: None > /tmp/1.txt
[root#gislab00207 ldom]# cat /tmp/1.txt
domainname: None
[root#gislab00207 ldom]# echo ${DOMAIN_NAME}
dcsw-79-98vm.us.oracle.com
[root#gislab00207 ldom]# cat /tmp/1.txt | sed -e 's/domainname: None/domainname: ${DOMAIN_NAME}/g'
--- Below is the result -- very funny.
domainname: ${DOMAIN_NAME}
--- You need to single quote your variable like this ...
[root#gislab00207 ldom]# cat /tmp/1.txt | sed -e 's/domainname: None/domainname: '${DOMAIN_NAME}'/g'
--- The right result is below
domainname: dcsw-79-98vm.us.oracle.com
VAR=8675309
echo "abcde:jhdfj$jhbsfiy/.hghi$jh:12345:dgve::" |\
sed 's/:[0-9]*:/:'$VAR':/1'
where VAR contains what you want to replace the field with
I had similar problem, I had a list and I have to build a SQL script based on template (that contained #INPUT# as element to replace):
for i in LIST
do
awk "sub(/\#INPUT\#/,\"${i}\");" template.sql >> output
done
If your replacement string may contain other sed control characters, then a two-step substitution (first escaping the replacement string) may be what you want:
PWD='/a\1&b$_' # these are problematic for sed
PWD_ESC=$(printf '%s\n' "$PWD" | sed -e 's/[\/&]/\\&/g')
echo 'xxx' | sed "s/xxx/$PWD_ESC/" # now this works as expected
for me to replace some text against the value of an environment variable in a file with sed works only with quota as the following:
sed -i 's/original_value/'"$MY_ENVIRNONMENT_VARIABLE"'/g' myfile.txt
BUT when the value of MY_ENVIRONMENT_VARIABLE contains a URL (ie https://andreas.gr) then the above was not working.
THEN use different delimiter:
sed -i "s|original_value|$MY_ENVIRNONMENT_VARIABLE|g" myfile.txt

sed: remove all characters except for last n characters

I am trying to remove every character in a text string except for the remaining 11 characters. The string is Sample Text_that-would$normally~be,here--pe_-l4_mBY and what I want to end up with is just -pe_-l4_mBY.
Here's what I've tried:
$ cat food
Sample Text_that-would$normally~be,here--pe_-l4_mBY
$ cat food | sed 's/^.*(.{3})$/\1/'
sed: 1: "s/^.*(.{3})$/\1/": \1 not defined in the RE
Please note that the text string isn't really stored in a file, I just used cat food as an example.
OS is macOS High Sierra 10.13.6 and bash version is 3.2.57(1)-release
You can use this sed with a capture group:
sed -E 's/.*(.{11})$/\1/' file
-pe_-l4_mBY
Basic regular expressions (used by default by sed) require both the parentheses in the capture group and the braces in the brace expression to be escaped. ( and { are otherwise treated as literal characters to be matched.
$ cat food | sed 's/^.*\(.\{3\}\)$/\1/'
mBY
By contrast, explicitly requesting sed to use extended regular expressions with the -E option reverses the meaning, with \( and \{ being the literal characters.
$ cat food | sed -E 's/^.*(.{3})$/\1/'
mBY
Try this also:
grep -o -E '.{11}$' food
grep, like sed, accepts an arbitrary number of file name arguments, so there is no need for a separate cat. (See also useless use of cat.)
You can use tail or Parameter Expansion :
string='Sample Text_that-would$normally~be,here--pe_-l4_mBY'
echo "$string" | tail -c 11
echo "${string#${string%??????????}}"
pe_-l4_mBY
pe_-l4_mBY
also with rev/cut/rev
$ echo abcdefghijklmnopqrstuvwxyz | rev | cut -c1-11 | rev
pqrstuvwxyz
man rev => rev - reverse lines characterwise

How to replace "\n" string with a new line in Unix Bash script

Cannot seem to find an answer to this one online...
I have a string variable (externally sourced) with new lines "\n" encoded as strings.
I want to replace those strings with actual new line carriage returns. The code below can achieve this...
echo $EXT_DESCR | sed 's/\\n/\n/g'
But when I try to store the result of this into it's own variable, it converts them back to strings
NEW_DESCR=`echo $EXT_DESCR | sed 's/\\n/\n/g'`
How can this be achieved, or what I'm I doing wrong?
Here's my code I've been testing to try get the right results
EXT_DESCR="This is a text\nWith a new line"
echo $EXT_DESCR | sed 's/\\n/\n/g'
NEW_DESCR=`echo $EXT_DESCR | sed 's/\\n/\n/g'`
echo ""
echo "$NEW_DESCR"
No need for sed, using parameter expansion:
$ foo='1\n2\n3'; echo "${foo//'\n'/$'\n'}"
1
2
3
With bash 4.4 or newer, you can use the E operator in ${parameter#operator}:
$ foo='1\n2\n3'; echo "${foo#E}"
1
2
3
Other answers contain alternative solutions. (I especially like the parameter expansion one.)
Here's what's wrong with your attempt:
In
echo $EXT_DESCR | sed 's/\\n/\n/g'
the sed command is in single quotes, so sed gets s/\\n/\n/g as is.
In
NEW_DESCR=`echo $EXT_DESCR | sed 's/\\n/\n/g'`
the whole command is in backticks, so a round of backslash processing is applied. That leads to sed getting the code s/\n/\n/g, which does nothing.
A possible fix for this code:
NEW_DESCR=`echo $EXT_DESCR | sed 's/\\\\n/\\n/g'`
By doubling up the backslashes, we end up with the right command in sed.
Or (easier):
NEW_DESCR=$(echo $EXT_DESCR | sed 's/\\n/\n/g')
Instead of backticks use $( ), which has less esoteric escaping rules.
Note: Don't use ALL_UPPERCASE for your shell variables. UPPERCASE is (informally) reserved for system variables such as HOME and special built-in variables such as IFS or RANDOM.
Depending on what exactly you need it for:
echo -e $EXT_DESCR
might be all you need.
From echo man page:
-e
enable interpretation of backslash escapes
This printf would do the job by interpreting all escaped constructs:
printf -v NEW_DESCR "%b" "$EXT_DESCR"
-v option will store output in a variable so no need to use command substitution here.
Problem with your approach is use of old back-ticks. You could do:
NEW_DESCR=$(echo "$EXT_DESCR" | sed 's/\\n/\n/g')
Assuming you're using gnu sed as BSD sed won't work with this approach.

Adding double quotes to beginning, end and around comma's in bash variable

I have a shell script that accepts a parameter that is comma delimited,
-s 1234,1244,1567
That is passed to a curl PUT json field. Json needs the values in a "1234","1244","1567" format.
Currently, I am passing the parameter with the quotes already in it:
-s "\"1234\",\"1244\",\"1567\"", which works, but the users are complaining that its too much typing and hard to do. So I'd like to just take a comma delimited list like I had at the top and programmatically stick the quotes in.
Basically, I want a parameter to be passed in as 1234,2345 and end up as a variable that is "1234","2345"
I've come to read that easiest approach here is to use sed, but I'm really not familiar with it and all of my efforts are failing.
You can do this in BASH:
$> arg='1234,1244,1567'
$> echo "\"${arg//,/\",\"}\""
"1234","1244","1567"
awk to the rescue!
$ awk -F, -v OFS='","' -v q='"' '{$1=$1; print q $0 q}' <<< "1234,1244,1567"
"1234","1244","1567"
or shorter with sed
$ sed -r 's/[^,]+/"&"/g' <<< "1234,1244,1567"
"1234","1244","1567"
translating this back to awk
$ awk '{print gensub(/([^,]+)/,"\"\\1\"","g")}' <<< "1234,1244,1567"
"1234","1244","1567"
you can use this:
echo QV=$(echo 1234,2345,56788 | sed -e 's/^/"/' -e 's/$/"/' -e 's/,/","/g')
result:
echo $QV
"1234","2345","56788"
just add double quotes at start, end, and replace commas with quote/comma/quote globally.
easy to do with sed
$ echo '1234,1244,1567' | sed 's/[0-9]*/"\0"/g'
"1234","1244","1567"
[0-9]* zero more consecutive digits, since * is greedy it will try to match as many as possible
"\0" double quote the matched pattern, entire match is by default saved in \0
g global flag, to replace all such patterns
In case, \0 isn't recognized in some sed versions, use & instead:
$ echo '1234,1244,1567' | sed 's/[0-9]*/"&"/g'
"1234","1244","1567"
Similar solution with perl
$ echo '1234,1244,1567' | perl -pe 's/\d+/"$&"/g'
"1234","1244","1567"
Note: Using * instead of + with perl will give
$ echo '1234,1244,1567' | perl -pe 's/\d*/"$&"/g'
"1234""","1244""","1567"""
""$
I think this difference between sed and perl is similar to this question: GNU sed, ^ and $ with | when first/last character matches
Using sed:
$ echo 1234,1244,1567 | sed 's/\([0-9]\+\)/\"\1\"/g'
"1234","1244","1567"
ie. replace all strings of numbers with the same strings of numbers quoted using backreferencing (\1).

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