Good Evening,
I'm trying to replace a trailing minus sign with a leading minus sign. The link below was very helpful, but I'm having a hard time dealing with both a comma and period. For example:
Input Data:
|76534| 253,453.86-| 6/4/2012| 56487-56987|
|32567| 36,000.00| 7/8/2012| 45684-4541|
|58531| 400.56-| 10/5/2012| 15232-1254|
|12584| 5.56-| 12/12/2013| 125565-451|
Desired Results (all other columns remain the same):
-253,453.86
36,000.00
-400.56
-5.56
Using sed -r 's/([[:digit:]]+(\,[[:digit:]]+(\.[[:digit:]]+)?))-/-\1/' I've been able to treat numbers with a comma, but I would like to treat all types in one command if possible.
Helpful Link:
Replace a trailing minus with leading minus
This seems to work with your multi-column data:
sed -r '{s#([0-9][0-9,.]*)-\|#-\1\|#g}'
Use [[:digit:],.] to create a character class with numbers, commas and periods. The rest you already know how to do.
sed -r 's/([[:digit:].,])-/-\1/'
You can do:
$ sed '/-$/{s/-$//;s/[0-9]/-&/;b};s/^/ /' file
-253,453.86
36,000.00
-400.56
-5.56
Updated Answer:
$ awk 'BEGIN{FS=OFS="|"}{for(i=1;i<=NF;i++)if($i~/-$/){sub(/-$/,"",$i);sub(/[0-9]/,"-&",$i)}}1' file
|76534| -253,453.86| 6/4/2012| 56487-56987|
|32567| 36,000.00| 7/8/2012| 45684-4541|
|58531| -400.56| 10/5/2012| 15232-1254|
|12584| -5.56| 12/12/2013| 125565-451|
$ awk 'BEGIN{FS=OFS="|"} sub(/-$/,"",$3){sub(/[^ ]/,"-&",$3)}1' file
|76534| -253,453.86| 6/4/2012| 56487-56987|
|32567| 36,000.00| 7/8/2012| 45684-4541|
|58531| -400.56| 10/5/2012| 15232-1254|
|12584| -5.56| 12/12/2013| 125565-451|
Another one through GNU sed,
$ sed -r 's/^(\|[0-9]+\|)( *?)([^|]*?)(-)(.*)$/\1\2\4\3\5/g' file
|76534| -253,453.86| 6/4/2012| 56487-56987|
|32567| 36,000.00| 7/8/2012| 45684-4541|
|58531| -400.56| 10/5/2012| 15232-1254|
|12584| -5.56| 12/12/2013| 125565-451|
A simple solution:
sed 's/\([^ ]*\)\-|/-\1|/' input
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 on Linux OS. I have a file to modify in my bash script.
My original file is like that:
...
ERIC-1898
HELENE-5456
THOMAS-54565
IRON-06516
...
And I'd like to modify this file with duplicate words (and -SYSTEM- word in second field), and add double quotes.
So, the result has to be like that:
...
"ERIC-1898" "ERIC-SYSTEM-1898"
"HELENE-5456" "HELENE-SYSTEM-5456"
"THOMAS-54565" "THOMAS-SYSTEM-54565"
"IRON-06516" "IRON-SYSTEM-06516"
...
How can I do that, for example with sed?
With sed and two capture groups:
$ sed 's/\(.*-\)\(.*\)/"&" "\1SYSTEM-\2"/' infile
"ERIC-1898" "ERIC-SYSTEM-1898"
"HELENE-5456" "HELENE-SYSTEM-5456"
"THOMAS-54565" "THOMAS-SYSTEM-54565"
"IRON-06516" "IRON-SYSTEM-06516"
Assuming that there is exactly one hyphen per input line.
awk solution:
awk -F'-' '{printf("\"%s\" \"%s-SYSTEM-%s\"\n", $1FS$2,$1,$2)}' file
The output would be like:
"ERIC-1898" "ERIC-SYSTEM-1898"
"HELENE-5456" "HELENE-SYSTEM-5456"
"THOMAS-54565" "THOMAS-SYSTEM-54565"
"IRON-06516" "IRON-SYSTEM-06516"
Not using external program:
#!/bin/bash
IFS=$'-'
while read -r first second;do
echo "\"$first-$second\" \"$first-SYSTEM-$second\""
done <infile
awk '{sub(/ /,"\" \"");print "\042" $0 "\042"}' file
"ERIC-1898" "ERIC-SYSTEM-1898"
"HELENE-5456" "HELENE-SYSTEM-5456"
"THOMAS-54565" "THOMAS-SYSTEM-54565"
"IRON-06516" "IRON-SYSTEM-06516"
I need help in processing data from STDIN (data is taken from another file with 'tail -f' plus grepped to filter out garbage). There are several lines between patterns:
<DN> 589</DN>
<DD>03.12.2014</DD>
<ST> </ST>
<STC>0</STC>
<STT>0</STT>
<PU>5</PU>
<OT>01</OT>
<DSN></DSN>
<NRA>40807,40820,426,30231,40818,30230</NRA>
<GR>300 000-00

</GR>
then next block with DN/GR starts
I need to convert lines between and to a single line, comma-separated:
<DN> 589</DN>,<DD>03.12.2014</DD>,<ST> </ST>,<STC>0</STC>,<STT>0</STT>,<PU>5</PU>,<OT>01</OT>,<DSN></DSN>,<NRA>40807,40820,426,30231,40818,30230</NRA>,<GR>300 000-00

</GR>
I need a one-liner with awk or sed or perl to do it and put result to STDOUT.
I've tried to do it, but failed due to lack of experience. Also tried to google and didn't find a working solution.
whatever..| awk '{sub(/^\s*/,"");printf "%s%s",$0,(/\/GR>\s*$/?"\n":",")}'
this line does:
remove the leading spaces from each line
join all line with sep , till the block end /GR>
if you have x data blocks, it gives you x long lines.
sed -nr '/<DN>/,/<GR>/{ H; /<GR>/{ g; s%\n%,%g; s%^,%%; p; s%.*%%; h }; }' <<'EOSEQ'
<DN> 589</DN>
<DD>03.12.2014</DD>
<STC>0</STC>
<GR>300 000-00

</GR>
<DN>900</DN>
<DD>20.11.2014</DD>
<OT>01</OT>
<NRA>40807,40820,426,30231,40818,30230</NRA>
<GR>300 000-00

</GR>
EOSEQ
SED one-liner, as you wish :)
Using awk you could do the following:
awk '{printf ("%s,", $NF)}' test.txt ##Will have comma at the end which may/may not be ok for you.
You can use the following one in sed.
sed -r ':loop ;N;s/(.*)\n(.*)/\1,\2/ ; t loop ' file name.
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.
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.