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.
It is simple - I have a data stream with IPv4 addresses encoded into hexadecimal representation like for example 0c22384e which stands for 12.34.56.78.
I figured out sed command with substitution of captured octets into decimal numbers separated by dot.
echo "0c22384e" | sed -E 's/([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})/printf "%d.%d.%d.%d" 0x\1 0x\2 0x\3 0x\4/eg'
This works with a single number BUT as soon I add some text that is not supposed to be matched, it is also passed for the execution - via printf in this case.
How can I preserve the unmatched part of the line without being passed for the execution?
With only one address in a line you could use
echo "Something 0c22384e more" |
sed -r 's/(.*)([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})(.*)/"\1" 0x\2 0x\3 0x\4 0x\5 "\6"/' |
xargs -n6 printf '%s%d.%d.%d.%d%s\n'
EDIT:
Replaced solution for one line and more addresses
with solution for more lines (assuming no '\r' in the stream):
echo "Something 0c22384e more 0c22385e
Second line: 0c22386e and 0c223870
Third line: 0c22388e and 0c223890
4th line: 0c2238ae and 0c2238b0" |
sed 's/$/\r/' |
sed -r 's/[0-9a-f]{8}/\n&\n/g' |
sed -r 's/([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})/printf '%d.%d.%d.%d' 0x\1 0x\2 0x\3 0x\4/e' |
tr -d '\n' |
tr '\r' '\n'
I have a file with below records
user1,fuser1,luser1,user1#test.com,data,user1
user2,fuser2,luser2,user2#test.com,data,user2
user3,fuser3,luser3,user3#test.com,data,user3
I wanted to perform some text replacements from
user1,fuser1,luser1,user1#test.com,data,user1
to
New_user1,New_fuser1,New_luser1,New_user1#test.com,data,New_user1
so I wrote below sed script.
sed -i -e 's/user/New_user/g; s/fuser/New_fuser/g; s/luser/New_luser/g' file
This works perfect. Now I have a requirement that I want to replace in specific line range.
start=2
end=3
sed -i -e ''${start},${end}'s/user/New_user/g; s/fuser/New_fuser/g; s/luser/New_luser/g' file
but this command is replacing pattern in all lines. example output is,
user1,New_fuser1,New_luser1,user1#test.com,data,New_user1
user2,New_fuser2,New_luser2,user2#test.com,data,New_user2
user3,New_fuser3,New_luser3,user3#test.com,data,New_user3
Looks like range is getting applied only to first expression and remaining expressions are getting applied on whole file. How to apply this range to all expressions?
You can use awk variables to use for this functionality, controlling the row and column numbers used for replacing
awk -vFS="," -vOFS="," -v columnStart=2 -v columnEnd=3 -v rowStart=1 -v rowEnd=2 \
'NR>=rowStart&&NR<=rowEnd{for(i=columnStart; i<=columnEnd; i++) \
$i="New_"$i; print }' file
where the awk variables columnStart, columnEnd, rowStart and rowStart determine which columns and rows to replace with , as the de-limiter adopted.
For your input file:-
$ cat input-file
user1,fuser1,luser1,user1#test.com,data,user1
user2,fuser2,luser2,user2#test.com,data,user2
user3,fuser3,luser3,user3#test.com,data,user3
Assuming I want to do replacement in lines 2 and 3 from columns 3-4, I can set-up my awk as
awk -vFS="," -vOFS="," -v columnStart=3 -v columnEnd=4 -v rowStart=2 -v rowEnd=3 \
'NR>=rowStart&&NR<=rowEnd{for(i=columnStart; i<=columnEnd; i++) \
$i="New_"$i; print }' file
user2,fuser2,New_luser2,New_user2#test.com,data,user2
user3,fuser3,New_luser3,New_user3#test.com,data,user3
To apply on the say the last column, set the columnStart and columnEnd to the same value e.g. say on column 6 and on last line only.
awk -vFS="," -vOFS="," -v columnStart=6 -v columnEnd=6 -v rowStart=3 -v rowEnd=3 \
'NR>=rowStart&&NR<=rowEnd{for(i=columnStart; i<=columnEnd; i++) \
$i="New_"$i; print }' file
user3,fuser3,luser3,user3#test.com,data,New_user3
When using GNU Sed (present on Ubuntu, probably Debian, and probably others).
There is a feature which makes this easy:
https://www.gnu.org/software/sed/manual/sed.html#Common-Commands
A group of commands may be enclosed between { and } characters. This
is particularly useful when you want a group of commands to be
triggered by a single address (or address-range) match.
Example: perform substitution then print the second input line:
$ seq 3 | sed -n '2{s/2/X/ ; p}'
X
Given the original question, this should do the trick:
sed -i -e '2,3 {s/user/New_user/g; s/fuser/New_fuser/g; s/luser/New_luser/g}' file
The following works for me:
START=2
NUM=1
sed -i -e "$START,+${NUM} s/user/New_user/g; $START,+${NUM} s/fuser/New_fuser/g; $START,+${NUM} s/luser/New_luser/g" file
As you can see, there are several changes:
The line range has to be present at each expression
The range should be represented (in this case) as the start line number and number of lines (the number of affected lines is NUM+1)
You put extra apostrophe symbols.
Using a single s command:
start=1
end=2
sed -e "$start,$end s/\([fl]*\)user/New_\1user/g" file
[fl]*user will match user with optional f or l first letter
output:
New_user1,New_fuser1,New_luser1,New_user1#test.com,data,New_user1
New_user2,New_fuser2,New_luser2,New_user2#test.com,data,New_user2
user3,fuser3,luser3,user3#test.com,data,user3
Suppose that I have a file like this:
tst.txt
fName1 lName1-a 222
fname1 lName1-b 22
fName1 lName1 2
And I want to get the 3rd column only for "fName1 lName1", using this command:
var=`grep -i -w "fName1 lName1" tst.txt`
However this returns me every line that starts with "fName1 lName1", how can I look for the exact match?
Here you go:
#!/bin/bash
var=$(grep -Po '(?<=fName1 lName1 ).+' tst.txt)
echo $var
The trick is to use the o option of the grep command. The P option tells the interpreter to use Perl-compatible regular expression syntax when parsing the pattern.
var=$(grep "fName1 lName1 " tst.txt |cut -d ' ' -f 3)
you can try this method:
grep -i -E "^fName1 lName1\s" tst.txt | cut -f3,3- -d ' '
But you must be sure that line starts with fName1 and you have space after lName1.
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.