How to get delete word combination "Name Server:" without quotes but keep 'Name Server:someletters/digits' in sed - bash

I have the following lines:
Name Server:NS92.WORLDNIC.COM(or some other value)
Name Server:
Name Server:
Name Server:
Please see the screenshot for better understanding: http://imgur.com/q6Ir4lo
How do I get rid of the 'Name Server:' line but keep the line with the value?
I tried /Name Server:{0,0}/d but it deletes all lines.
Thanks

I was able to get the following two lines to work:
I believe the [:space:] is POSIX compliant:
cat test |sed '/^Name Server:[[:space:] \t]\?$/d'
An alternative is simply:
cat test |sed '/^Name Server:[ \t]\?$/d'
I've also found in sed, that most of the meta-characters (eg + ? ) need to be escaped for sed to recognize them correctly.

This works for me:
echo "Name Server:NS92.WORLDNIC.COM" | sed 's/^Name Server://'

cut -d ":" -f 2 < ff | sed '/^$/d'
Uses ':' as delimiter and splits the line (-d option), then selects the second field (-f option)

Related

How to get values in a line while looping line by line in a file (shell script)

I have a file which looks like this (file.txt)
{"key":"AJGUIGIDH568","rule":squid:111-some_random_text_here
{"key":"TJHJHJHDH568","rule":squid:111-some_random_text_here
{"key":"YUUUIGIDH566","rule":squid:111-some_random_text_here
{"key":"HJHHIGIDH568","rule":squid:111-some_random_text_here
{"key":"ATYUGUIDH556","rule":squid:111-some_random_text_here
{"key":"QfgUIGIDH568","rule":squid:111-some_random_text_here
I want to loop trough this line by line an extract the key values.
so the result should be like ,
AJGUIGIDH568
AJGUIGIDH568
YUUUIGIDH566
HJHHIGIDH568
ATYUGUIDH556
QfgUIGIDH568
So I wrote a code like this to loop line by line and extract the value between {"key":" and ","rule": because key values is in between these 2 patterns.
while read p; do
echo $p | sed -n "/{"key":"/,/","rule":,/p"
done < file.txt
But this is not working. can someone help me to figure out me this. Thanks in advance.
Your sample input is almost valid json. You could tweak it to make it valid and then extract the values with jq with something like:
sed -e 's/squid/"squid/' -e 's/$/"}/' file.txt | jq -r .key
Or, if your actual input really is valid json, then just use jq:
jq -r .key file.txt
If the "random-txt" may include double quotes, making it difficult to massage the input to make it valid json, perhaps you want something like:
awk '{print $4}' FS='"' file.txt
or
sed -n '/{"key":"\([^"]*\).*/s//\1/p' file.txt
or
while IFS=\" read open_brace key colon val _; do echo "$val"; done < file.txt
For the shown data, you can try this awk:
awk -F '"[:,]"' '{print $2}' file
AJGUIGIDH568
TJHJHJHDH568
YUUUIGIDH566
HJHHIGIDH568
ATYUGUIDH556
QfgUIGIDH568
With the give example you can simple use
cut -d'"' -f4 file.txt
Assumptions:
there may be other lines in the file so we need to focus on just the lines with "key" and "rule"
the only text between "key" and "rule" is the desired string (eg, squid never shows up between the two patterns of interest)
Adding some additional lines:
$ cat file.txt
{"key":"AJGUIGIDH568","rule":squid:111-some_random_text_here
ignore this line}
{"key":"TJHJHJHDH568","rule":squid:111-some_random_text_here
ignore this line}
{"key":"YUUUIGIDH566","rule":squid:111-some_random_text_here
ignore this line}
{"key":"HJHHIGIDH568","rule":squid:111-some_random_text_here
ignore this line}
{"key":"ATYUGUIDH556","rule":squid:111-some_random_text_here
ignore this line}
{"key":"QfgUIGIDH568","rule":squid:111-some_random_text_here
ignore this line}
One sed idea:
$ sed -nE 's/^(.*"key":")([^"]*)(","rule".*)$/\2/p' file.txt
AJGUIGIDH568
TJHJHJHDH568
YUUUIGIDH566
HJHHIGIDH568
ATYUGUIDH556
QfgUIGIDH568
Where:
-E - enable extended regex support (and capture groups without need to escape sequences)
-n - suppress printing of pattern space
^(.*"key":") - [1st capture group] everything from start of line up to and including "key":"
([^"]*) - [2nd capture group] everything that is not a double quote (")
(","rule".*)$ - [3rd capture group] everything from ",rule" to end of line
\2/p - replace the line with the contents of the 2nd capture group and print

Bash regex: get value in conf file preceded by string with dot

I have to get my db credentials from this configuration file:
# Database settings
Aisse.LocalHost=localhost
Aisse.LocalDataBase=mydb
Aisse.LocalPort=5432
Aisse.LocalUser=myuser
Aisse.LocalPasswd=mypwd
# My other app settings
Aisse.NumDir=../../data/Num
Aisse.NumMobil=3000
# Log settings
#Aisse.Trace_AppliTpv=blabla1.tra
#Aisse.Trace_AppliCmp=blabla2.tra
#Aisse.Trace_AppliClt=blabla3.tra
#Aisse.Trace_LocalDataBase=blabla4.tra
In particular, I want to get the value mydb from line
Aisse.LocalDataBase=mydb
So far, I have developed this
mydbname=$(echo "$my_conf_file.conf" | grep "LocalDataBase=" | sed "s/LocalDataBase=//g" )
that returns
mydb #Aisse.Trace_blabla4.tra
that would be ok if it did not return also the comment string.
Then I have also tryed
mydbname=$(echo "$my_conf_file.conf" | grep "Aisse.LocalDataBase=" | sed "s/LocalDataBase=//g" )
that retruns void string.
How can I get only the value that is preceded by the string "Aisse.LocalDataBase=" ?
Using sed
$ mydbname=$(sed -n 's/Aisse\.LocalDataBase=//p' input_file)
$ echo $mydbname
mydb
I'm afraid you're being incomplete:
You mention you want the line, containing "LocalDataBase", but you don't want the line in comment, let's start with that:
A line which contains "LocalDataBase":
grep "LocalDataBase" conf.conf.txt
A line which contains "LocalDataBase" but who does not start with a hash:
grep "LocalDataBase" conf.conf.txt | grep -v "^ *#"
??? grep -v "^ *#"
That means: don't show (-v) the lines, containing:
^ : the start of the line
* : a possible list of space characters
# : a hash character
Once you have your line, you need to work with it:
You only need the part behind the equality sign, so let's use that sign as a delimiter and show the second column:
cut -d '=' -f 2
All together:
grep "LocalDataBase" conf.conf.txt | grep -v "^ *#" | cut -d '=' -f 2
Are we there yet?
No, because it's possible that somebody has put some comment behind your entry, something like:
LocalDataBase=mydb #some information
In order to prevent that, you need to cut that comment too, which you can do in a similar way as before: this time you use the hash character as a delimiter and you show the first column:
grep "LocalDataBase" conf.conf.txt | grep -v "^ *#" | cut -d '=' -f 2 | cut -d '#' -f 1
Have fun.
You may use this sed:
mydbname=$(sed -n 's/^[^#][^=]*LocalDataBase=//p' file)
echo "$mydbname"
mydb
RegEx Details:
^: Start
[^#]: Matches any character other than #
[^=]*: Matches 0 or more of any character that is not =
LocalDataBase=: Matches text LocalDataBase=
You can use
mydbname=$(sed -n 's/^Aisse\.LocalDataBase=\(.*\)/\1/p' file)
If there can be leading whitespace you can add [[:blank:]]* after ^:
mydbname=$(sed -n 's/^[[:blank:]]*Aisse\.LocalDataBase=\(.*\)/\1/p' file)
See this online demo:
#!/bin/bash
s='# Database settings
Aisse.LocalHost=localhost
Aisse.LocalDataBase=mydb
Aisse.LocalPort=5432
Aisse.LocalUser=myuser
Aisse.LocalPasswd=mypwd
# My other app settings
Aisse.NumDir=../../data/Num
Aisse.NumMobil=3000
# Log settings
#Aisse.Trace_AppliTpv=blabla1.tra
#Aisse.Trace_AppliCmp=blabla2.tra
#Aisse.Trace_AppliClt=blabla3.tra
#Aisse.Trace_LocalDataBase=blabla4.tra'
sed -n 's/^Aisse\.LocalDataBase=\(.*\)/\1/p' <<< "$s"
Output:
mydb
Details:
-n - suppresses default line output in sed
^[[:blank:]]*Aisse\.LocalDataBase=\(.*\) - a regex that matches the start of a string (^), then zero or more whiespaces ([[:blank:]]*), then a Aisse.LocalDataBase= string, then captures the rest of the line into Group 1
\1 - replaces the whole match with the value of Group 1
p - prints the result of the successful substitution.

BASH - replace with variable contain double quotes inside

I have an text file, with line inside...
line: <version="AAA" email="...ANY..." file="BBB">
new desired line in text file to be: <version="AAA" email="NEW_TEXT" file="BBB">
I want to replace the ...ANY... expression with variable (replace entire line)
I have this script text-file script in #!/bin/bash, but I have problem when expanding double quotes in variables.
LINE_NUMBER="$(grep -nr 'email' *.txt | awk '{print $1}' | sed 's/[^0-9]*//g')"
VAR1="$(grep 'email' *.txt | cut -d '"' -f1-3)"
VAR2="$(grep 'email' *.txt | cut -d '"' -f5-)"
VAR3='NEW_TEXT'
NEW_LINE=$VAR1'"'$VAR3'"'$VAR2
new desired line in text file to be... <version="AAA" email="NEW_TEXT" file="BBB">
awk -i inplace 'NR=='"$LINE_NUMBER"'{sub(".*",'"'$NEW_LINE'"')},1' *.txt
but I get this new line:
<version="" email="NEW_TEXT" file="">
what do I do wrong?
How can I prevent expand duouble quotes inside variable?
please better write me an working example, I had tried other topics, forums, posts....but I have no luck.
You cas use sed :
VAR3='NEW_TEXT'
sed -i "s/email=\"[^\"]*\"/email=\"$VAR3\"/" myfile.xml
Suggesting:
var3="text space % special < chars"
Note var3 may not contain & which is special replacement meaning in sed
sed -E 's|email="[^"]*"|email="'"${var3}"'"|' input.1.txt
Explanation
[^"]* : Match longest string not having " till next ".

Bash command to extract characters in a string

I want to write a small script to generate the location of a file in an NGINX cache directory.
The format of the path is:
/path/to/nginx/cache/d8/40/32/13febd65d65112badd0aa90a15d84032
Note the last 6 characters: d8 40 32, are represented in the path.
As an input I give the md5 hash (13febd65d65112badd0aa90a15d84032) and I want to generate the output: d8/40/32/13febd65d65112badd0aa90a15d84032
I'm sure sed or awk will be handy, but I don't know yet how...
This awk can make it:
awk 'BEGIN{FS=""; OFS="/"}{print $(NF-5)$(NF-4), $(NF-3)$(NF-2), $(NF-1)$NF, $0}'
Explanation
BEGIN{FS=""; OFS="/"}. FS="" sets the input field separator to be "", so that every char will be a different field. OFS="/" sets the output field separator as /, for print matters.
print ... $(NF-1)$NF, $0 prints the penultimate field and the last one all together; then, the whole string. The comma is "filled" with the OFS, which is /.
Test
$ awk 'BEGIN{FS=""; OFS="/"}{print $(NF-5)$(NF-4), $(NF-3)$(NF-2), $(NF-1)$NF, $0}' <<< "13febd65d65112badd0aa90a15d84032"
d8/40/32/13febd65d65112badd0aa90a15d84032
Or with a file:
$ cat a
13febd65d65112badd0aa90a15d84032
13febd65d65112badd0aa90a15f1f2f3
$ awk 'BEGIN{FS=""; OFS="/"}{print $(NF-5)$(NF-4), $(NF-3)$(NF-2), $(NF-1)$NF, $0}' a
d8/40/32/13febd65d65112badd0aa90a15d84032
f1/f2/f3/13febd65d65112badd0aa90a15f1f2f3
With sed:
echo '13febd65d65112badd0aa90a15d84032' | \
sed -n 's/\(.*\([0-9a-f]\{2\}\)\([0-9a-f]\{2\}\)\([0-9a-f]\{2\}\)\)$/\2\/\3\/\4\/\1/p;'
Having GNU sed you can even simplify the pattern using the -r option. Now you won't need to escape {} and () any more. Using ~ as the regex delimiter allows to use the path separator / without need to escape it:
sed -nr 's~(.*([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2}))$~\2/\3/\4/\1~p;'
Output:
d8/40/32/13febd65d65112badd0aa90a15d84032
Explained simple the pattern does the following: It matches:
(all (n-5 - n-4) (n-3 - n-2) (n-1 - n-0))
and replaces it by
/$1/$2/$3/$0
You can use a regular expression to separate each of the last 3 bytes from the rest of the hash.
hash=13febd65d65112badd0aa90a15d84032
[[ $hash =~ (..)(..)(..)$ ]]
new_path="/path/to/nginx/cache/${BASH_REMATCH[1]}/${BASH_REMATCH[2]}/${BASH_REMATCH[3]}/$hash"
Base="/path/to/nginx/cache/"
echo '13febd65d65112badd0aa90a15d84032' | \
sed "s|\(.*\(..\)\(..\)\(..\)\)|${Base}\2/\3/\4/\1|"
# or
# sed sed 's|.*\(..\)\(..\)\(..\)$|${Base}\1/\2/\3/&|'
Assuming info is a correct MD5 (and only) string
First of all - thanks to all of the responders - this was extremely quick!
I also did my own scripting meantime, and came up with this solution:
Run this script with a parameter of the URL you're looking for (www.example.com/article/76232?q=hello for example)
#!/bin/bash
path=$1
md5=$(echo -n "$path" | md5sum | cut -f1 -d' ')
p3=$(echo "${md5:0-2:2}")
p2=$(echo "${md5:0-4:2}")
p1=$(echo "${md5:0-6:2}")
echo "/path/to/nginx/cache/$p1/$p2/$p3/$md5"
This assumes the NGINX cache has a key structure of 2:2:2.

sed edit of text with variables and special characters

I'm on OS X and writing a bash script to edit text in a file which includes some known text with special characters. There will be a variable too which needs to be retained and some text entered or replaced.
Here is the input file contents:
user_pref("intl.charsetmenu.browser.cache", "UTF-8");
user_pref("network.automatic-ntlm-auth.trusted-uris", "search.co.za");
user_pref("network.cookie.prefsMigrated", true);
I currently have this code:
existingTrusts=`more ~/prefs.js | grep "network.automatic-ntlm-auth.trusted-uris" | awk '{print $2}' | sed 's/);//g' | sed 's/"//g'`
trustSites="company.com,organisation.co.uk,$existingTrusts"
replacementValue='"user_pref("network.automatic-ntlm-auth.trusted-uris", "$trustSites");"'
sed -i 's/^user_pref("network.automatic-ntlm-auth.trusted-uris/$replacementValue/' ~/prefs.js > ~/newPrefs.js
Any help appreciated.
You are using too many pipes to set your existingTrusts variable. Set your variables like this:
existingTrusts=$(awk '/network.automatic-ntlm-auth.trusted-uris/ {gsub(/"|\);/, "", $2); print $2}' ~/prefs.js)
trustSites="company.com,organisation.co.uk,$existingTrusts"
replacementValue='user_pref("network.automatic-ntlm-auth.trusted-uris", "'$trustSites'");'
# and now finally your sed command
sed 's/^user_pref("network.automatic-ntlm-auth.trusted-uris".*$/'"$replacementValue"'/' ~/prefs.js > ~/newPrefs.js
Why so complicated?
trustedSites='company.com,organisation.co.uk,'
sed -i '' -e '/network.automatic-ntlm-auth.trusted-uris/s/, "\([^"]*\)/, "'"${trustedSites}"'\1/' prefs.js
This is imperfect because
It uses unescaped . in a pattern where a literal . is presumed
It presumes , " will appear exactly as that exactly where expected
These things could be fixed.

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