Display matched string to end of line - unix

How to find a particular string in a file and display matched string and rest of the line?
For example- I have a line in a.txt:
This code gives ORA-12345 in my code.
So, I am finding string 'ORA-'
Output Should be:
ORA-12345 in my code
Tried using grep:
grep 'ORA-*' a.txt
but it gives whole line in the output.

# Create test data:
echo "junk ORA-12345 more stuff" > a.tst
echo "junk ORB-12345 another stuff" >> a.tst
# Actually command:
# the -o (--only-matching) flag will print only the matched result, and not the full line
cat a.tst | grep -o 'ORA-.*$' # ORA-12345 more stuff
As fedorqui pointed out you can use:
grep -o 'ORA-.*$' a.tst

An additional answer in awk:
awk '$0 ~ "ORA" {print substr($0, match($0, "ORA"))}' a.tst
From the inside out, here's what's going on:
match($0, "ORA") finds where in the line ORA appears. In this case, it happens to be position 17.
substr($0, match($0, "ORA")) then returns from position 17 to the end of the line.
$0 ~ "ORA" makes sure that the the above is applied only to those lines that contain ORA.

with sed
echo "This code gives ORA-12345 in my code." | sed 's/.*ORA-/ORA-/'

Related

cut command --complement flag equivalent in AWK

I am new to writing shell scripts
I am trying to write an AWK command which does exactly the below
cut --complement -c $IGNORE_RANGE file.txt > tmp
$IGNORE_RANGE can be of any value say, 1-5 or 5-10 etc
i cannot use cut since i am in AIX and AIX does not support --complement, is there any way to achieve this using AWK command
Example:
file.txt
abcdef
123456
Output
cut --complement -c 1-2 file.txt > tmp
cdef
3456
cut --complement -c 4-5 file.txt > tmp
abcf
1236
cut --complement -c 1-5 file.txt > tmp
f
6
Could you please try following, written and tested with shown samples. We have range variable of awk which should be in start_of_position-end_of_position and we could pass it as per need.
awk -v range="4-5" '
BEGIN{
split(range,array,"-")
}
{
print substr($0,1,array[1]-1) substr($0,array[2]+1)
}
' Input_file
OR to make it more clear in understanding wise try following:
awk -v range="4-5" '
BEGIN{
split(range,array,"-")
start=array[1]
end=array[2]
}
{
print substr($0,1,start-1) substr($0,end+1)
}
' Input_file
Explanation: Adding detailed explanation for above.
awk -v range="4-5" ' ##Starting awk program from here creating range variable which has range value of positions which we do not want to print in lines.
BEGIN{ ##Starting BEGIN section of this program from here.
split(range,array,"-") ##Splitting range variable into array with delimiter of - here.
start=array[1] ##Assigning 1st element of array to start variable here.
end=array[2] ##Assigning 2nd element of array to end variable here.
}
{
print substr($0,1,start-1) substr($0,end+1) ##Printing sub-string of current line from 1 to till value of start-1 and then printing from end+1 which basically means will skip that range of characters which OP does not want to print.
}
' Input_file ##Mentioning Input_file name here.
You can do this in awk:
awk -v st=1 -v en=2 '{print substr($0, 1, st-1) substr($0, en+1)}' file
cdef
3456
Or:
awk -v st=4 -v en=5 '{print substr($0, 1, st-1) substr($0, en+1)}' file
abcf
1236

No way to move pound (£) symbol in UNIX?

Need get the delimiters at the starting of each line, below are sample input and output files for reference. In actual Delimiter used are( £{, ^$^)
Note - The file to be rearranged has huge data
Have tried the below but it does not work:
tr £{ \\n
sed 's/£{/\n/g'
awk '{ gsub("£{", "\n") } 1'
Input File:
£{firstlinecontinues£{secondstartsfromhereandit
keepsoncontinueingtillend£{herecomes3rdand£{fi
nallyfourthisalsohere
Output File:
£{firstlinecontinues
£{secondstartsfromhereanditkeepsoncontinueingtillend
£{herecomes3rdand
£{finallyfourthisalsohere
With GNU awk for multi-char RS and \s:
$ awk -v RS='£{' 'NR>1{gsub(/\s/,""); print RS $0}' file
£{firstlinecontinues
£{secondstartsfromhereanditkeepsoncontinueingtillend
£{herecomes3rdand
£{finallyfourthisalsohere
awk 'BEGIN{RS="(£{\|\^\$\^)"; OFS=ORS=""}{$1=$1;print $0 (FNR>1?"\n":"") RT}' file
Since the £ symbol is represented by two Octal codes, 302 and 243, I was able to produce the desired result with this perl command:
perl -pe 's/(\302\243)/\n$1/g' data.txt
NOTE: Here's what I see on my system:
echo "£" | od -c
0000000 302 243 \n
0000003

How to swap the first line with last line in a text file using SED/AWK

I am trying to swap the first line with last line in a text file in unix
file has:
line1
line2
line3
line4
line5
line6
I want like this:
line6
line2
line3
line4
line5
line1
I am using sed -n -e '1 s/^.*$/$p' file . Which is not happening.
EDIT2: As per Ed sir's comment adding this solution too here which will be more suitable in case 2nd line is empty then also it will work.
awk 'NR==1{first=$0;next} NR>2{val=val prev ORS} {prev=$0} END{print prev ORS val first} Input_file
EDIT: To exchange only first and last line following may help you(considering that your Input_file is same as shown sample).
awk 'FNR==1{first=$0;next} {val=(val?val ORS prev:prev?$0:"")} {prev=$0} END{print $0 ORS val ORS first}' Input_file
Explanation:
awk '
FNR==1{ ##Checking if line is first line then do following.
first=$0; ##Creating varable first whose value is current line value.
next ##next will skip all further statements from here.
}
{
val=(val?val ORS prev:prev?$0:"") ##Creating variable named val here whoe value is concatenating to its own value with variable prev value.
}
{
prev=$0 ##Creating variable prev whose value will be current value of line but will become previous value for next line.
}
END{
print $0 ORS val ORS first ##Printing current line ORS val ORS and first here.
}' Input_file ##Mentioning Input_file name here.
Could you please try following and let me know if this helps you.
awk -v from=1 -v to=6 'FNR==from{source=$0;next} FNR==to{target=$0;next} {val=val?val ORS $0:$0} END{print target ORS val ORS source}' Input_file
In sed, you'll need to make two passes; it probably makes more sense to use ed instead. Then we just need two move commands - move the last line to just after line 1, then move the first line to the end:
$m1
1m$
Demo
#!/bin/bash
set -euo pipefail
# Create our input file
file=$(mktemp)
trap 'rm "$file"' EXIT
cat >"$file" <<END
line1
line2
line3
line4
line5
line6
END
echo Before:
cat "$file"
echo
# Make the change
ed -s "$file" <<<$'$m1\n1m$\nwq'
# Show the result
echo After:
cat "$file"
If you need to write a different output file, you can of course add a filename argument to the wq command, as usual.
This might work for you (GNU sed):
sed 'H;$!d;x;s/\n\([^\n]*\)\(\n.*\n\)\(.*\)/\3\2\1/' file
Copy the whole file into the hold space (HS) and after the last line, swap to the HS, split the file into first line, middle and last line and substitute the first and last lines.
A two pass alternative solution:
sed -n '1s/.*/$c&/p;$s/.*/1c&/p' file | sed -f - file
Create a sed script to change the first line to the last and last line to the first.
Another solution using cat, head, tail and sed:
cat <(tail -1 file) <(sed '1d;$d' file) <(head -1 file)
Another sed solution:
sed -E '1h;1d;:a;N;$!ba;s/(.*)\n(.*)/\2\n\1/;G' file
Just in case a oneliner mixing, head, tail and sed:
$ FIRST=$(head -1 file.txt) LAST=$(tail -1 file.txt) \
sed "1 s/^.*$/${LAST}/" file.txt | sed "$ s/^.*$/${FIRST}/"
For very large files, you might be interested in a double pass:
awk '(NR!=FNR){print (FNR==1 ? t : (FNR==c ? h : $0) ); next }
(NR==1){h=$0}{t=$0;c=NR}' file file
Using the sponge util:
f=file
{ tail -1 $f ; tail -n +2 $f | head -n -1 ; head -1 $f ; } | sponge $f

AWK to print field $2 first, then field $1

Here is the input(sample):
name1#gmail.com|com.emailclient.account
name2#msn.com|com.socialsite.auth.account
I'm trying to achieve this:
Emailclient name1#gmail.com
Socialsite name2#msn.com
If I use AWK like this:
cat foo | awk 'BEGIN{FS="|"} {print $2 " " $1}'
it messes up the output by overlaying field 1 on the top of field 2.
Any tips/suggestions? Thank you.
A couple of general tips (besides the DOS line ending issue):
cat is for concatenating files, it's not the only tool that can read files! If a command doesn't read files then use redirection like command < file.
You can set the field separator with the -F option so instead of:
cat foo | awk 'BEGIN{FS="|"} {print $2 " " $1}'
Try:
awk -F'|' '{print $2" "$1}' foo
This will output:
com.emailclient.account name1#gmail.com
com.socialsite.auth.accoun name2#msn.com
To get the desired output you could do a variety of things. I'd probably split() the second field:
awk -F'|' '{split($2,a,".");print a[2]" "$1}' file
emailclient name1#gmail.com
socialsite name2#msn.com
Finally to get the first character converted to uppercase is a bit of a pain in awk as you don't have a nice built in ucfirst() function:
awk -F'|' '{split($2,a,".");print toupper(substr(a[2],1,1)) substr(a[2],2),$1}' file
Emailclient name1#gmail.com
Socialsite name2#msn.com
If you want something more concise (although you give up a sub-process) you could do:
awk -F'|' '{split($2,a,".");print a[2]" "$1}' file | sed 's/^./\U&/'
Emailclient name1#gmail.com
Socialsite name2#msn.com
Use a dot or a pipe as the field separator:
awk -v FS='[.|]' '{
printf "%s%s %s.%s\n", toupper(substr($4,1,1)), substr($4,2), $1, $2
}' << END
name1#gmail.com|com.emailclient.account
name2#msn.com|com.socialsite.auth.account
END
gives:
Emailclient name1#gmail.com
Socialsite name2#msn.com
Maybe your file contains CRLF terminator. Every lines followed by \r\n.
awk recognizes the $2 actually $2\r. The \r means goto the start of the line.
{print $2\r$1} will print $2 first, then return to the head, then print $1. So the field 2 is overlaid by the field 1.
The awk is ok. I'm guessing the file is from a windows system and has a CR (^m ascii 0x0d) on the end of the line.
This will cause the cursor to go to the start of the line after $2.
Use dos2unix or vi with :se ff=unix to get rid of the CRs.

Removing trailing / starting newlines with sed, awk, tr, and friends

I would like to remove all of the empty lines from a file, but only when they are at the end/start of a file (that is, if there are no non-empty lines before them, at the start; and if there are no non-empty lines after them, at the end.)
Is this possible outside of a fully-featured scripting language like Perl or Ruby? I’d prefer to do this with sed or awk if possible. Basically, any light-weight and widely available UNIX-y tool would be fine, especially one I can learn more about quickly (Perl, thus, not included.)
From Useful one-line scripts for sed:
# Delete all leading blank lines at top of file (only).
sed '/./,$!d' file
# Delete all trailing blank lines at end of file (only).
sed -e :a -e '/^\n*$/{$d;N;};/\n$/ba' file
Therefore, to remove both leading and trailing blank lines from a file, you can combine the above commands into:
sed -e :a -e '/./,$!d;/^\n*$/{$d;N;};/\n$/ba' file
So I'm going to borrow part of #dogbane's answer for this, since that sed line for removing the leading blank lines is so short...
tac is part of coreutils, and reverses a file. So do it twice:
tac file | sed -e '/./,$!d' | tac | sed -e '/./,$!d'
It's certainly not the most efficient, but unless you need efficiency, I find it more readable than everything else so far.
here's a one-pass solution in awk: it does not start printing until it sees a non-empty line and when it sees an empty line, it remembers it until the next non-empty line
awk '
/[[:graph:]]/ {
# a non-empty line
# set the flag to begin printing lines
p=1
# print the accumulated "interior" empty lines
for (i=1; i<=n; i++) print ""
n=0
# then print this line
print
}
p && /^[[:space:]]*$/ {
# a potentially "interior" empty line. remember it.
n++
}
' filename
Note, due to the mechanism I'm using to consider empty/non-empty lines (with [[:graph:]] and /^[[:space:]]*$/), interior lines with only whitespace will be truncated to become truly empty.
As mentioned in another answer, tac is part of coreutils, and reverses a file. Combining the idea of doing it twice with the fact that command substitution will strip trailing new lines, we get
echo "$(echo "$(tac "$filename")" | tac)"
which doesn't depend on sed. You can use echo -n to strip the remaining trailing newline off.
Here's an adapted sed version, which also considers "empty" those lines with just spaces and tabs on it.
sed -e :a -e '/[^[:blank:]]/,$!d; /^[[:space:]]*$/{ $d; N; ba' -e '}'
It's basically the accepted answer version (considering BryanH comment), but the dot . in the first command was changed to [^[:blank:]] (anything not blank) and the \n inside the second command address was changed to [[:space:]] to allow newlines, spaces an tabs.
An alternative version, without using the POSIX classes, but your sed must support inserting \t and \n inside […]. GNU sed does, BSD sed doesn't.
sed -e :a -e '/[^\t ]/,$!d; /^[\n\t ]*$/{ $d; N; ba' -e '}'
Testing:
prompt$ printf '\n \t \n\nfoo\n\nfoo\n\n \t \n\n'
foo
foo
prompt$ printf '\n \t \n\nfoo\n\nfoo\n\n \t \n\n' | sed -n l
$
\t $
$
foo$
$
foo$
$
\t $
$
prompt$ printf '\n \t \n\nfoo\n\nfoo\n\n \t \n\n' | sed -e :a -e '/[^[:blank:]]/,$!d; /^[[:space:]]*$/{ $d; N; ba' -e '}'
foo
foo
prompt$
using awk:
awk '{a[NR]=$0;if($0 && !s)s=NR;}
END{e=NR;
for(i=NR;i>1;i--)
if(a[i]){ e=i; break; }
for(i=s;i<=e;i++)
print a[i];}' yourFile
this can be solved easily with sed -z option
sed -rz 's/^\n+//; s/\n+$/\n/g' file
Hello
Welcome to
Unix and Linux
For an efficient non-recursive version of the trailing newlines strip (including "white" characters) I've developed this sed script.
sed -n '/^[[:space:]]*$/ !{x;/\n/{s/^\n//;p;s/.*//;};x;p;}; /^[[:space:]]*$/H'
It uses the hold buffer to store all blank lines and prints them only after it finds a non-blank line. Should someone want only the newlines, it's enough to get rid of the two [[:space:]]* parts:
sed -n '/^$/ !{x;/\n/{s/^\n//;p;s/.*//;};x;p;}; /^$/H'
I've tried a simple performance comparison with the well-known recursive script
sed -e :a -e '/^\n*$/{$d;N;};/\n$/ba'
on a 3MB file with 1MB of random blank lines around a random base64 text.
shuf -re 1 2 3 | tr -d "\n" | tr 123 " \t\n" | dd bs=1 count=1M > bigfile
base64 </dev/urandom | dd bs=1 count=1M >> bigfile
shuf -re 1 2 3 | tr -d "\n" | tr 123 " \t\n" | dd bs=1 count=1M >> bigfile
The streaming script took roughly 0.5 second to complete, the recursive didn't end after 15 minutes. Win :)
For completeness sake of the answer, the leading lines stripping sed script is already streaming fine. Use the most suitable for you.
sed '/[^[:blank:]]/,$!d'
sed '/./,$!d'
Using bash
$ filecontent=$(<file)
$ echo "${filecontent/$'\n'}"
In bash, using cat, wc, grep, sed, tail and head:
# number of first line that contains non-empty character
i=`grep -n "^[^\B*]" <your_file> | sed -e 's/:.*//' | head -1`
# number of hte last one
j=`grep -n "^[^\B*]" <your_file> | sed -e 's/:.*//' | tail -1`
# overall number of lines:
k=`cat <your_file> | wc -l`
# how much empty lines at the end of file we have?
m=$(($k-$j))
# let strip last m lines!
cat <your_file> | head -n-$m
# now we have to strip first i lines and we are done 8-)
cat <your_file> | tail -n+$i
Man, it's definitely worth to learn "real" programming language to avoid that ugliness!
#dogbane has a nice simple answer for removing leading empty lines. Here's a simple awk command which removes just the trailing lines. Use this with #dogbane's sed command to remove both leading and trailing blanks.
awk '{ LINES=LINES $0 "\n"; } /./ { printf "%s", LINES; LINES=""; }'
This is pretty simple in operation.
Add every line to a buffer as we read it.
For every line which contains a character, print the contents of the buffer and then clear it.
So the only things that get buffered and never displayed are any trailing blanks.
I used printf instead of print to avoid the automatic addition of a newline, since I'm using newlines to separate the lines in the buffer already.
This AWK script will do the trick:
BEGIN {
ne=0;
}
/^[[:space:]]*$/ {
ne++;
}
/[^[:space:]]+/ {
for(i=0; i < ne; i++)
print "";
ne=0;
print
}
The idea is simple: empty lines do not get echoed immediately. Instead, we wait till we get a non-empty line, and only then we first echo out as much empty lines as seen before it, and only then echo out the new non-empty line.
perl -0pe 's/^\n+|\n+(\n)$/\1/gs'
Here's an awk version that removes trailing blank lines (both empty lines and lines consisting of nothing but white space).
It is memory efficient; it does not read the entire file into memory.
awk '/^[[:space:]]*$/ {b=b $0 "\n"; next;} {printf "%s",b; b=""; print;}'
The b variable buffers up the blank lines; they get printed when a non-blank line is encountered. When EOF is encountered, they don't get printed. That's how it works.
If using gnu awk, [[:space:]] can be replaced with \s. (See full list of gawk-specific Regexp Operators.)
If you want to remove only those trailing lines that are empty, see #AndyMortimer's answer.
A bash solution.
Note: Only useful if the file is small enough to be read into memory at once.
[[ $(<file) =~ ^$'\n'*(.*)$ ]] && echo "${BASH_REMATCH[1]}"
$(<file) reads the entire file and trims trailing newlines, because command substitution ($(....)) implicitly does that.
=~ is bash's regular-expression matching operator, and =~ ^$'\n'*(.*)$ optionally matches any leading newlines (greedily), and captures whatever comes after. Note the potentially confusing $'\n', which inserts a literal newline using ANSI C quoting, because escape sequence \n is not supported.
Note that this particular regex always matches, so the command after && is always executed.
Special array variable BASH_REMATCH rematch contains the results of the most recent regex match, and array element [1] contains what the (first and only) parenthesized subexpression (capture group) captured, which is the input string with any leading newlines stripped. The net effect is that ${BASH_REMATCH[1]} contains the input file content with both leading and trailing newlines stripped.
Note that printing with echo adds a single trailing newline. If you want to avoid that, use echo -n instead (or use the more portable printf '%s').
I'd like to introduce another variant for gawk v4.1+
result=($(gawk '
BEGIN {
lines_count = 0;
empty_lines_in_head = 0;
empty_lines_in_tail = 0;
}
/[^[:space:]]/ {
found_not_empty_line = 1;
empty_lines_in_tail = 0;
}
/^[[:space:]]*?$/ {
if ( found_not_empty_line ) {
empty_lines_in_tail ++;
} else {
empty_lines_in_head ++;
}
}
{
lines_count ++;
}
END {
print (empty_lines_in_head " " empty_lines_in_tail " " lines_count);
}
' "$file"))
empty_lines_in_head=${result[0]}
empty_lines_in_tail=${result[1]}
lines_count=${result[2]}
if [ $empty_lines_in_head -gt 0 ] || [ $empty_lines_in_tail -gt 0 ]; then
echo "Removing whitespace from \"$file\""
eval "gawk -i inplace '
{
if ( NR > $empty_lines_in_head && NR <= $(($lines_count - $empty_lines_in_tail)) ) {
print
}
}
' \"$file\""
fi
Because I was writing a bash script anyway containing some functions, I found it convenient to write those:
function strip_leading_empty_lines()
{
while read line; do
if [ -n "$line" ]; then
echo "$line"
break
fi
done
cat
}
function strip_trailing_empty_lines()
{
acc=""
while read line; do
acc+="$line"$'\n'
if [ -n "$line" ]; then
echo -n "$acc"
acc=""
fi
done
}

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