I am trying to remove the last space of a file.txt which contains many rows.
I just need to remove "only the last space" after the third column/each line.
My file looks like this:
3 180 120
3 123 145
6 234 0
4 122 12
I have been trying with the following script but it does not work, so far. Somebody can help me, please?
#!/bin/bash
var="val1 val2 val3 "
var="${var%"${var##*[![:space:]]}"}"
echo "===$var===" <Antart_csv1_copy.txt> trimmed.txt
You can use sed:
sed -i -e 's/ $//g' filename.txt
-i will make the command inplace (change the original file)
-e 's/ $//g' will take regular expression <space><endline> and change it to nothing. Modifier g makes it for all lines in the file
You can try it first without -i and redirect output:
sed -e 's/ $//g' filename.txt > trimmed.txt
Another solution removing all trailing spaces from each line :
while read line; do echo "${line%%*( )}" >> trimmed.txt; done < Antart_csv1_copy.txt
Related
I have many files named a, b, c and so on. These files contain line like this:-
11.077-105.882
-22.134-302.321
-1.011-201.254
I want to add a space when - sign come in mid of line. I want my output file look like this:-
11.077 -105.882
-22.134 -302.321
-1.011 -201.254
I have tried this command:-
cat a |sed 's/-/ -/g' >out.txt
But it do not give desired result
Require (and capture) a character before each - to replace:
$ sed 's/\(.\)-/\1 -/g' < tmp.txt
11.077 -105.882
-22.134 -302.321
-1.011 -201.254
This will only match a - that is not line-initial, and will include the preceding character in the replacement text.
You could combine 2 sed commands:
$ sed 's/-/ -/g' a | sed 's/^ //'
11.077 -105.882
-22.134 -302.321
-1.011 -201.254
Or, in a single line solution add whitespaces only before - that come after a digit:
$ sed 's,\([0-9]\)-,\1 -,' a
11.077 -105.882
-22.134 -302.321
-1.011 -201.254
I am trying to get a substring between &DEST= and the next & or a line break.
For example :
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
In this I need to extract "SFO"
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
In this I need to extract "SANFRANSISCO"
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
In this I need to extract "SANJOSE"
I am reading a file line by line, and I need to update the text after &DEST= and put it back in the file. The modification of the text is to mask the dest value with X character.
So, SFO should be replaced with XXX.
SANJOSE should be replaced with XXXXXXX.
Output :
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
Please let me know how to achieve this in script (Preferably shell or bash script).
Thanks.
$ cat file
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=PORTORICA
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
$ sed -E 's/^.*&DEST=([^&]*)[&]*.*$/\1/' file
SFO
PORTORICA
SANFRANSISCO
SANJOSE
should do it
Replacing airports with an equal number of Xs
Let's consider this test file:
$ cat file
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
To replace the strings after &DEST= with an equal length of X and using GNU sed:
$ sed -E ':a; s/(&DEST=X*)[^X&]/\1X/; ta' file
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
To replace the file in-place:
sed -i -E ':a; s/(&DEST=X*)[^X&]/\1X/; ta' file
The above was tested with GNU sed. For BSD (OSX) sed, try:
sed -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta file
Or, to change in-place with BSD(OSX) sed, try:
sed -i '' -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta file
If there is some reason why it is important to use the shell to read the file line-by-line:
while IFS= read -r line
do
echo "$line" | sed -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta
done <file
How it works
Let's consider this code:
search_str="&DEST="
newfile=chart.txt
sed -E ':a; s/('"$search_str"'X*)[^X&]/\1X/; ta' "$newfile"
-E
This tells sed to use Extended Regular Expressions (ERE). This has the advantage of requiring fewer backslashes to escape things.
:a
This creates a label a.
s/('"$search_str"'X*)[^X&]/\1X/
This looks for $search_str followed by any number of X followed by any character that is not X or &. Because of the parens, everything except that last character is saved into group 1. This string is replaced by group 1, denoted \1 and an X.
ta
In sed, t is a test command. If the substitution was made (meaning that some character needed to be replaced by X), then the test evaluates to true and, in that case, ta tells sed to jump to label a.
This test-and-jump causes the substitution to be repeated as many times as necessary.
Replacing multiple tags with one sed command
$ name='DEST|ORIG'; sed -E ':a; s/(&('"$name"')=X*)[^X&]/\1X/; ta' file
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=XXXX
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=XXXX
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
Answer for original question
Using shell
$ s='MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546'
$ s=${s#*&DEST=}
$ echo ${s%%&*}
SFO
How it works:
${s#*&DEST=} is prefix removal. This removes all text up to and including the first occurrence of &DEST=.
${s%%&*} is suffix removal_. It removes all text from the first & to the end of the string.
Using awk
$ echo 'MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546' | awk -F'[=\n]' '$1=="DEST"{print $2}' RS='&'
SFO
How it works:
-F'[=\n]'
This tells awk to treat either an equal sign or a newline as the field separator
$1=="DEST"{print $2}
If the first field is DEST, then print the second field.
RS='&'
This sets the record separator to &.
With GNU bash:
while IFS= read -r line; do
[[ $line =~ (.*&DEST=)(.*)((&.*|$)) ]] && echo "${BASH_REMATCH[1]}fooooo${BASH_REMATCH[3]}"
done < file
Output:
MYREQUESTISTO8764GETTHIS&DEST=fooooo&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=fooooo&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=fooooo
Replace the characters between &DEST and & (or EOL) with x's:
awk -F'&DEST=' '{
printf("%s&DEST=", $1);
xlen=index($2,"&");
if ( xlen == 0) xlen=length($2)+1;
for (i=0;i<xlen;i++) printf("%s", "X");
endstr=substr($2,xlen);
printf("%s\n", endstr);
}' file
How can I remove the set of characters from a specified length in a file using shell script.
Example:
Filename : abc.txt
helloshell
Now how can I remove characters starting from 8 to 10 (the ell at the end)?
I have tried sed -r command on Linux servers but it's not working on AIX servers.
Linux command:
sed -r 's/.(.{3}).*/\1/' filename.txt
With Bash (extract substring from 0 to 7th character):
str="helloshell"
echo ${str:0:7}
With sed (removes 3 characters starting from 7th position) :
str="helloshell"
startpos=7;
nbchar=3;
echo "$str" | sed "s/^\(.\{$startpos\}\).\{$nbchar\}\(.*\)/\1\2/"
$ sed -r 's/^(.{7}).{0,3}(.*)$/\1\2/g'
helloshell
hellosh
1234567890
1234567
$
{0,3} ensures 3 or less chars from 8th position (0,3 implies remove only if present and hence also removes 1/2/3 chars from 8th position) i.e match and remove minimum 0 chars(for no chars) and maximum 3 chars
If you want exactly only 3 chars to removed from 8th position use {3} but it wont remove chars from 8th position if there are less than 3 chars, eg:
$ sed -r 's/^(.{7}).{3}(.*)$/\1\2/g'
123456789
123456789
$
Edit1:
you can use this instead without the -r switch and some escaping: sed 's/^\(.\{7\}\).\{0,3\}\(.*\)$/\1\2/g'
for performing the above operation only lines starting with BH, you can add a restriction for the substitute command like this:
sed '/^BH/s/^\(.\{7\}\).\{0,3\}\(.*\)$/\1\2/g'
/^BH/s.. would ensure substitution is performed only on lines starting with BH
$ sed '/^BH/s/^\(.\{7\}\).\{0,3\}\(.*\)$/\1\2/g'
BHhelloshell
BHhelloll
BH123456789
BH123459
helloshell
helloshell
$
To exclude BH while counting you can use:
$ sed '/^BH/s/^BH\(.\{7\}\).\{0,3\}\(.*\)$/BH\1\2/g'
BHhelloshell
BHhellosh
BH123456789
BH1234567
helloshell
helloshell
$
Try this if you are always expecting the last 3 to be deleted.
echo helloshell | sed 's/...$//'
For 8-10 try this:
echo helloshell | sed 's/\(.\{7\}\).../\1/'
echo helloshellHowAreYou | sed 's/\(.\{7\}\).../\1/'
For AIX you may need to remove the \ from \{
sed 's/\(.{7}\).../\1/'
And if there is a pattern that you want in the search string you have to adjust the value within in the \( and \)
this question originated from string pattaren-matching using awk , basically we are splitting a line of text in multiple groups based on a regex pattern, and then printing two groups only. Now the question is can we right align a group while printing through sed?
below is an example
$cat input.txt
it is line one
it is longggggggg one
itttttttttt is another one
now
$sed -e 's/\(.*\) \(.*\) \(.*\) \(.*\)/\1 \3/g' input.txt
it splits and prints group 1 and 3, but the output is
it line
it longggggggg
itttttttttt another
my question is can we do it through sed so that the output comes as
it line
it longggggggg
itttttttttt another
I did it with awk but I feel it can be done through sed, but I am not able to get how I am going to get the length of the second group and then pad correct number of spaces in between the groups, I am open to any suggestions to try out.
This might work for you (GNU sed):
sed -r 's/^(.*) .* (.*) .*$/\1 \2/;:a;s/^.{1,40}$/ &/;ta;s/^( *)(\S*)/\2\1/' file
or:
sed -r 's/^(.*) .* (.*) .*$/printf "%-20s%20s" \1 \2/e' file
You can use looping in sed to achieve what you want:
#!/bin/bash
echo 'aa bb cc dd
11 22 33333333 44
ONE TWO THREEEEEEEEE FOUR' | \
sed -e 's/\(.*\) \(.*\) \(.*\) \(.*\)/\1 \3/g' \
-e '/\([^ ]*\) \([^ ]*\)/ { :x ; s/^\(.\{1,19\}\) \(.\{1,19\}\)$/\1 \2/g ; tx }'
The two 19's control the width of your columns. The :x is a label which is looped to by tx whenever the preceding substitution succeeded. (You could add a p; before tx to "debug" it.
It most easy to use awk in this case...
You could too use a bash loop to calculate the number of space and run this command on the line covered :
while read; do
# ... calculate $SPACE ...
echo $REPLY|sed "s/\([^\ ]*\)\ *[^\ ]*\ *\([^\ ]*\)/\1$SPACES\2/g"
done < file
But I prefer use awk for do all that (or other advanced shell languages such as Perl, Python, PHP shell mode, ...)
TemplateSpace=" "
TemplateSize=${#TemplateSpace}
sed "
# split your group (based on word here but depend on your real need)
s/^ *\(\w\) \(\w\) \(\w\) \(\w\).*$/\1 \3/
# align
s/$/${TemplateSpace}/
s/^\(.\{${TemplateSize}\}\).*$/\1/
s/\(\w\) \(\w\)\( *\)/\1 \3\2/
"
or more simple for avoiding TemplateSize (and there are no dot in content)
TemplateSpace="............................................................."
and replace
s/^\(.\{${TemplateSize}\}.*$/\1/
by
s/^\(${TemplateSpace}\).*$/\1/
s/\./ /g
Del columns 2 and 4. Right justify resulting col 2 at line length of 23 chars.
sed -e '
s/[^ ]\+/ /4;
s/[^ ]\+//2;
s/^\(.\{23\}\).*$/\1/;
s/\(^[^ ]\+[ ]\+\)\([^ ]\+\)\([ ]\+\)/\1\3\2/;
'
or gnu sed with extended regex:
sed -r '
s/\W+\w+\W+(\w+)\W+\w+$/\1 /;
s/^(.{23}).*/\1/;
s/(+\W)(\w+)(\W+)$/\1\3\2/
'
This question is old, but I like to see it as a puzzle.
While I love the loop solution for its brevity, here is one without a loop or shell help.
sed -E "s/ \w+ (\w+) \w+$/ \1/;h;s/./ /g;s/$/# /;s/( *)#\1//;x;H;x;s/\n//;s/^( *)(\w+)/\2\1/"
or without extended regex
sed "s/ .* \(.*\) .*$/ \1/;h;s/./ /g;s/$/# /;s/\( *\)#\1//;x;H;x;s/\n//;s/^\( *\)\([^ ]*\)/\2\1/"
I would like to insert lines into a file in bash starting from a specific line.
Each line is a string which is an element of an array
line[0]="foo"
line[1]="bar"
...
and the specific line is 'fields'
file="$(cat $myfile)"
for p in $file; do
if [ "$p" = 'fields' ]
then insertlines() #<- here
fi
done
This can be done with sed: sed 's/fields/fields\nNew Inserted Line/'
$ cat file.txt
line 1
line 2
fields
line 3
another line
fields
dkhs
$ sed 's/fields/fields\nNew Inserted Line/' file.txt
line 1
line 2
fields
New Inserted Line
line 3
another line
fields
New Inserted Line
dkhs
Use -i to save in-place instead of printing to stdout
sed -i 's/fields/fields\nNew Inserted Line/'
As a bash script:
#!/bin/bash
match='fields'
insert='New Inserted Line'
file='file.txt'
sed -i "s/$match/$match\n$insert/" $file
Or anoter one example with the sed:
Prepare a test.txt file:
echo -e "line 1\nline 2\nline 3\nline 4" > /tmp/test.txt
cat /tmp/test.txt
line 1
line 2
line 3
line 4
Add a new line into the test.txt file:
sed -i '2 a line 2.5' /tmp/test.txt
# sed for in-place editing (-i) of the file: 'LINE_NUMBER a-ppend TEXT_TO_ADD'
cat /tmp/test.txt
line 1
line 2
line 2.5
line 3
line 4
This is definitely a case where you want to use something like sed (or awk or perl) rather than readling one line at a time in a shell loop. This is not the sort of thing the shell does well or efficiently.
You might find it handy to write a reusable function. Here's a simple one, though it won't work on fully-arbitrary text (slashes or regular expression metacharacters will confuse things):
function insertAfter # file line newText
{
local file="$1" line="$2" newText="$3"
sed -i -e "/^$line$/a"$'\\\n'"$newText"$'\n' "$file"
}
Example:
$ cat foo.txt
Now is the time for all good men to come to the aid of their party.
The quick brown fox jumps over a lazy dog.
$ insertAfter foo.txt \
"Now is the time for all good men to come to the aid of their party." \
"The previous line is missing 'bjkquvxz.'"
$ cat foo.txt
Now is the time for all good men to come to the aid of their party.
The previous line is missing 'bjkquvxz.'
The quick brown fox jumps over a lazy dog.
$
sed is your friend:
:~$ cat text.txt
foo
bar
baz
~$
~$ sed '/^bar/\na this is the new line/' text.txt > new_text.txt
~$ cat new_text.txt
foo
bar
this is the new line
baz
~$