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)=$(NF-1) " jjjjj" I am just appending jjjjj to last-1 field.
Related
I have the following string:
|**barak**.version|2001.0132012031539|
in file text.txt.
I would like to replace it with the following:
|**barak**.version|2001.01.2012031541|
So I run:
sed -i "s/\|\*\*$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|/|$version|/" text.txt
I have a string like this:
/var/cpanel/users/joebloggs:DNS9=domain.example
I need to extract the username (joebloggs) from this string and store it in a variable.
The format of the string will always be the same with exception of joebloggs and domain.example so I am thinking the string can be split twice using cut?
The first split would split by : and we would store the first part in a variable to pass to the second split function.
The second split would split by / and store the last word (joebloggs) into a variable
I know how to do this in PHP using arrays and splits but I am a bit lost in bash.
To extract joebloggs from this string in bash using parameter expansion without any extra processes...
MYVAR="/var/cpanel/users/joebloggs:DNS9=domain.example"
NAME=${MYVAR%:*} # retain the part before the colon
NAME=${NAME##*/} # retain the part after the last slash
echo $NAME
Doesn't depend on joebloggs being at a particular depth in the path.
Summary
An overview of a few parameter expansion modes, for reference...
${MYVAR#pattern} # delete shortest match of pattern from the beginning
${MYVAR##pattern} # delete longest match of pattern from the beginning
${MYVAR%pattern} # delete shortest match of pattern from the end
${MYVAR%%pattern} # delete longest match of pattern from the end
So # means match from the beginning (think of a comment line) and % means from the end. One instance means shortest and two instances means longest.
You can get substrings based on position using numbers:
${MYVAR:3} # Remove the first three chars (leaving 4..end)
${MYVAR::3} # Return the first three characters
${MYVAR:3:5} # The next five characters after removing the first 3 (chars 4-9)
You can also replace particular strings or patterns using:
${MYVAR/search/replace}
The pattern is in the same format as file-name matching, so * (any characters) is common, often followed by a particular symbol like / or .
Examples:
Given a variable like
MYVAR="users/joebloggs/domain.example"
Remove the path leaving file name (all characters up to a slash):
echo ${MYVAR##*/}
domain.example
Remove the file name, leaving the path (delete shortest match after last /):
echo ${MYVAR%/*}
users/joebloggs
Get just the file extension (remove all before last period):
echo ${MYVAR##*.}
example
NOTE: To do two operations, you can't combine them, but have to assign to an intermediate variable. So to get the file name without path or extension:
NAME=${MYVAR##*/} # remove part before last slash
echo ${NAME%.*} # from the new var remove the part after the last period
domain
Define a function like this:
getUserName() {
echo $1 | cut -d : -f 1 | xargs basename
}
And pass the string as a parameter:
userName=$(getUserName "/var/cpanel/users/joebloggs:DNS9=domain.example")
echo $userName
What about sed? That will work in a single command:
sed 's#.*/\([^:]*\).*#\1#' <<<$string
The # are being used for regex dividers instead of / since the string has / in it.
.*/ grabs the string up to the last backslash.
\( .. \) marks a capture group. This is \([^:]*\).
The [^:] says any character _except a colon, and the * means zero or more.
.* means the rest of the line.
\1 means substitute what was found in the first (and only) capture group. This is the name.
Here's the breakdown matching the string with the regular expression:
/var/cpanel/users/ joebloggs :DNS9=domain.example joebloggs
sed 's#.*/ \([^:]*\) .* #\1 #'
Using a single Awk:
... | awk -F '[/:]' '{print $5}'
That is, using as field separator either / or :, the username is always in field 5.
To store it in a variable:
username=$(... | awk -F '[/:]' '{print $5}')
A more flexible implementation with sed that doesn't require username to be field 5:
... | sed -e s/:.*// -e s?.*/??
That is, delete everything from : and beyond, and then delete everything up until the last /. sed is probably faster too than awk, so this alternative is definitely better.
Using a single sed
echo "/var/cpanel/users/joebloggs:DNS9=domain.example" | sed 's/.*\/\(.*\):.*/\1/'
I like to chain together awk using different delimitators set with the -F argument. First, split the string on /users/ and then on :
txt="/var/cpanel/users/joebloggs:DNS9=domain.com"
echo $txt | awk -F"/users/" '{print$2}' | awk -F: '{print $1}'
$2 gives the text after the delim, $1 the text before it.
I know I'm a little late to the party and there's already good answers, but here's my method of doing something like this.
DIR="/var/cpanel/users/joebloggs:DNS9=domain.example"
echo ${DIR} | rev | cut -d'/' -f 1 | rev | cut -d':' -f1
I have files that need to be removed from comments and white space until keyword . Line number varies . Is it possible to limit multiple continued sed substitutions based on Keyword ?
This removes all comments and white spaces from file :
sed -i -e 's/#.*$//' -e 's/;.*$//' -e '/^$/d' file
For example something like this :
# string1
# string2
some string
; string3
; string4
####
<Keyword_Keep_this_line_and_comments_white_space_after_this>
# More comments that need to be here
; etc.
sed -i '1,/keyword/{/^[#;]/d;/^$/d;}' file
I would suggest using awk and setting a flag when you reach your keyword:
awk '/Keyword/ { stop = 1 } stop || !/^[[:blank:]]*([;#]|$)/' file
Set stop to true when the line contains Keyword. Do the default action (print the line) when stop is true or when the line doesn't match the regex. The regex matches lines whose first non-blank character is a semicolon or hash, or blank lines. It's slightly different to your condition but I think it does what you want.
The command prints to standard output so you should redirect to a new file and then overwrite the original to achieve an "in-place edit":
awk '...' input > tmp && mv tmp input
Use grep -n keyword to get the line number that contains the keyword.
Use sed -i -e '1,N s/#..., when N is the line number that contains the keyword, to only remove comments on the lines 1 to N.
For example, I have a file containing a line as below:
"abc":"def"
I need to insert 123 between "abc":" and def" so that the line will become: "abc":"123def".
As "abc" appears only once so I think I can just search it and do the insertion.
How to do this with bash script such as sed or awk?
AMD$ sed 's/"abc":"/&123/' File
"abc":"123def"
Match "abc":", then append this match with 123 (& will contain the matched string "abc":")
If you want to take care of space before and after :, you can use:
sed 's/"abc" *: *"/&123/'
For replacing all such patterns, use g with sed.
sed 's/"abc" *: *"/&123/g' File
sed:
$ sed -E 's/(:")(.*)/\1123\2/' <<<'"abc":"def"'
"abc":"123def"
(:") gets :" and put in captured group 1
(.*) gets the remaining portion and put in captured group 2
in the replacement, \1123\2 puts 123 between the groups
awk:
$ awk -F: 'sub(".", "&123", $2)' <<<'"abc":"def"'
"abc" "123def"
In the sub() function, the second ($2) field is being operated on, pattern is used as . (which would match "), and in the replacement the matched portion (&) is followed by 123.
echo '"abc":"def"'| awk '{sub(/def/,"123def")}1'
"abc":"123def"
I have a tab deliminated file which I want to add "$" end of each variable, Can I do that with awk,sed or anything else?
Example
input:
a seq1 anot1
b seq2 anot2
c seq3 anot3
d seq4 anot4
I neet to have this:
output:
a$ seq1$ anot1$
b$ seq2$ anot2$
c$ seq3$ anot3$
d$ seq4$ anot4$
Any answer will be appreciated,
Thanks
In bash alone:
while read line; do echo "${line//$'\t'/\$$'\t'}\$"; done < file
This hackish solution relies on two "special" things -- parameter expansion to do the replacement, and format expansion to allow the tabs to be parsed.
In awk, you can process fields much more safely:
awk -F'\t' 'BEGIN{OFS=FS} {for(n=1;n<=NF;n++){$n=$n "$"}} 1' file
This works by stepping through each line of input and replacing each field with itself plus the dollar sign. The BEGIN block insures that your output will use the same field separators as your input. The 1 at the end is awk short-hand for "print the current line".
late to the party...
another awk solution. Prefix field and record separators with "$"
$ awk -F'\t' 'BEGIN{OFS="$"FS; ORS="$"RS} {$1=$1}1' file
With sed:
sed 's/[^ ]*/&$/g' filename
which replaces any non-space words with the word (&) followed by a $.
Oops! You said tabs. You can replace the above space with "\t" to use tab delimited.
sed 's/[^\t]*/&$/g' filename
Actually, even better, for tabs OR spaces:
sed 's/[^[:blank:]]*/&$/g' filename
awk is your friend :
awk '{for(i=1;i<=NF;i++)sub(/$/,"$",$i);print}' file
or
awk '{for(i=1;i<=NF;i++)sub(/$/,"$",$i);}1' file
Sample Output
a$ seq1$ anot1$
b$ seq2$ anot2$
c$ seq3$ anot3$
d$ seq4$ anot4$
What is happening here?
Using a for-loop we iterate thru all the fields in a record.
We use the awk sub function to replace the end ie (/$/) with a $ ie ("$") for each record ($i).
Use print explicitly to print the record. Numeric 1 also represents the default action that is to print the record.
awk '{gsub(/ /,"$ ")}{print $0 "$\r"}' file
a$ seq1$ anot1$
b$ seq2$ anot2$
c$ seq3$ anot3$
d$ seq4$ anot4$
What happens?
First replace spaces with dollar sign and new space.
Last insert dollar sign before the carriage return.