Hidden line in file? - linux

I have a UTF-8/no BOM file (converted from ISO-8859-1) that has 31214 lines. I have already run dos2unix on the file. When I open it in notepad++, I see a blank line underneath. When I remove this blank line, the line count reduces by one. I save it under a different name and when I tail the file, the prompt displays on the same line. From bash, how do I delete the blank line in the 1st file to produce the result displayed below in the 2nd file?
The goal is to do this from bash w/o manually deleting the line in notepad++
1st file:
[user#server]$ cat file1.txt | wc -l
31214
[user#server]$ tail file1.txt
T 31212 Data 20170517
[user#server]$
2nd file (edited with notepad++)
[user#server]$ cat file2.txt | wc -l
31213
[user#server]$ tail file2.txt
T 31212 Data 20170517[user#server]$

That's the trailing newline of the last line. Some editors allow you to go to the nonexisting "empty" line at the end, some don't show it. Again, some programs may allow you to remove the final newline, but note that e.g. POSIX in effect requires it to be there, and some standard utilities act oddly if it isn't present.
E.g. wc -l counts the number of newlines in the input file (printf "foo\nbar" | wc -l shows 1) so removing the final newline does decrease the line count.
Also, Bash prints the prompt wherever it was that the cursor was left on the screen, so if you print something that doesn't have the trailing newline, the prompt will be placed where the final incomplete line ended, as you saw.
There's no need to remove that final newline, just leave it there.

To remove the final newline character it is possible, as explained here, to use
sed -i '$ s/.$//' your.file
which will substitute nothing for the last character in the last line of the file (if you want to delete smth else from the end of the file you can replace the regex .$ with smth-else$). -i means ‘substitute in-place’ (in FreeBSD/MacOS you need to add an empty string as an argument: sed -i "" '$ s/.$//' your.file)

The file2.txt is missing a trailing newline.
Yes, a text file should end on a newline character.
Given that you do know that a trailing newline is missing, this command should be enough to correct the problem:
$ echo >> file2.txt

Related

Filter out only matched values from a text file in each line

I have a file "test.txt" with the lines below and also lot bunch of extra stuff after the "version"
soainfra_metrics{metric_group="sca_composite",partition="test",is_active="true",state="on",is_default="true",composite="test123"} map:stats version:1.0
soainfra_metrics{metric_group="sca_composite",partition="gello",is_active="true",state="on",is_default="true",composite="test234"} map:stats version:1.8
soainfra_metrics{metric_group="sca_composite",partition="bolo",is_active="true",state="on",is_default="true",composite="3415"} map:stats version:3.1
soainfra_metrics{metric_group="sca_composite",partition="solo",is_active="true",state="on",is_default="true",composite="hji"} map:stats version:1.1
I tried:
egrep -r 'partition|is_active|state|is_default|composite' test.txt
It's displaying every line, but I need only specific mentioned fields like this below,ignoring rest of the data/stuff or lines
in a nut shell, i want to display only these fields from a line not the rest
partition="test",is_active="true",state="on",is_default="true",composite="test123"
partition="gello",is_active="true",state="on",is_default="true",composite="test234"
partition="bolo",is_active="true",state="on",is_default="true",composite="3415"
partition="solo",is_active="true",state="on",is_default="true",composite="hji"
If your version of grep supports Perl-style regular expressions, then I'd use this:
grep -oP '.*?,\K[^}]+' file
It removes everything up to the first comma (\K kills any previous output) and prints everything up to the }.
Alternatively, using awk:
awk -F'}' '{ sub(/[^,]+,/, ""); print $1 }' file
This sets the field separator to } so the part you're interested in is the first field. It then uses sub to remove the part up to the first comma.
For completeness, you could also use sed:
sed 's/[^,]*,\([^}]*\).*/\1/' file
This captures the part after the first , up to the } and replaces the content of the line with it.
After the grep to pick out the lines you want, use sed to edit the lines:
sed 's/.*\(partition[^}]*\)} map.*/\1/'
This means: "whenever you see anything .*, followed by partition and
any number of non-}, then } map and anything else, grab the part
from partition up to but not including the brace \(...\) as group 1.
The replacement text is just group 1 \1.
Use a pipe | to connect the output of egrep to the input of sed:
egrep ... | sed ...
As far as i understood your file might have more lines you don't want to see, so i would use:
sed -n 's/.*\(partition.*\)}.*/\1/p' file
we use -n p to show only lines where we made substitution. The substitution part just gets the part of the line you need substituting the whole line with the pattern.
This might work for you (GNU sed):
sed -r 's/(partition|is_active|state|is_default|composite)="[^"]*"/\n&\n/g;s/[^\n]*\n([^\n]*)\n[^\n]*/\1,/g;s/,$//' file
Treat the problem as if it were a "decomposed club sandwich". Identify the fillings, remove the bread and tidy up.

Replacing multiple line using sed command

I have a text file file.log contains following text
file.log
ab
cd
ef
I want to replace "ab\ncd" with "ab\n" and the final file.log should look like this:
ab
ef
This is the sed command I am using but it couldn't recognize the newline character to match the pattern:
sed -i 's/\(.*\)\r \(.*\)/\1\r/g' file.log
with 3 character space after '\r' but no change is made with this.
\(.*\) - This matches any character(.) followed by 0 or more (*) of the preceding character
\r - For newline
\1 - Substitution for the first matching pattern. In this case, it's 'ab'
Can you help me out what's wrong with the above command.
The issue is that, the sed is a stream editor, which reads line by line from the input file
So when it reads line
ab
from the input file, it doesnt know whether the line is followed by a line
cd
When it reads the line cd it sed will habe removed the line ab from the pattern space, this making the pattern invalid for the current pattern space.
Solution
A solution can be to read the entire file, and append them into the hold space, and then replace the hold space. As
$ sed -n '1h; 1!H;${g;s/ab\ncd/ab\n/g;p}' input
ab
ef
What it does
1h Copies the first line into the hold space.
1!H All lines excpet the first line (1!) appends the line to the hold space.
$ matches the last line, performs the commands in {..}
g copies the contents of hold space back to pattern space
s/ab\ncd/ab\n/g makes the substitution.
p Prints the entire patterns space.
Sed processes the input file line by line. So can't do like the above . You need to include N, so that it would append the next line into pattern space.
$ sed 'N;s~ab\ncd~ab\n~g' file
ab
ef
A couple of other options:
perl -i -0pe 's/^ab\n\Kcd$//mg' file.log
which will change any such pattern in the file
If there's just one, good ol' ed
ed file.log <<END_SCRIPT
/^ab$/+1 c
.
wq
END_SCRIPT

How can I remove the last character of a file in unix?

Say I have some arbitrary multi-line text file:
sometext
moretext
lastline
How can I remove only the last character (the e, not the newline or null) of the file without making the text file invalid?
A simpler approach (outputs to stdout, doesn't update the input file):
sed '$ s/.$//' somefile
$ is a Sed address that matches the last input line only, thus causing the following function call (s/.$//) to be executed on the last line only.
s/.$// replaces the last character on the (in this case last) line with an empty string; i.e., effectively removes the last char. (before the newline) on the line.
. matches any character on the line, and following it with $ anchors the match to the end of the line; note how the use of $ in this regular expression is conceptually related, but technically distinct from the previous use of $ as a Sed address.
Example with stdin input (assumes Bash, Ksh, or Zsh):
$ sed '$ s/.$//' <<< $'line one\nline two'
line one
line tw
To update the input file too (do not use if the input file is a symlink):
sed -i '$ s/.$//' somefile
Note:
On macOS, you'd have to use -i '' instead of just -i; for an overview of the pitfalls associated with -i, see the bottom half of this answer.
If you need to process very large input files and/or performance / disk usage are a concern and you're using GNU utilities (Linux), see ImHere's helpful answer.
truncate
truncate -s-1 file
Removes one (-1) character from the end of the same file. Exactly as a >> will append to the same file.
The problem with this approach is that it doesn't retain a trailing newline if it existed.
The solution is:
if [ -n "$(tail -c1 file)" ] # if the file has not a trailing new line.
then
truncate -s-1 file # remove one char as the question request.
else
truncate -s-2 file # remove the last two characters
echo "" >> file # add the trailing new line back
fi
This works because tail takes the last byte (not char).
It takes almost no time even with big files.
Why not sed
The problem with a sed solution like sed '$ s/.$//' file is that it reads the whole file first (taking a long time with large files), then you need a temporary file (of the same size as the original):
sed '$ s/.$//' file > tempfile
rm file; mv tempfile file
And then move the tempfile to replace the file.
Here's another using ex, which I find not as cryptic as the sed solution:
printf '%s\n' '$' 's/.$//' wq | ex somefile
The $ goes to the last line, the s deletes the last character, and wq is the well known (to vi users) write+quit.
After a whole bunch of playing around with different strategies (and avoiding sed -i or perl), the best way i found to do this was with:
sed '$! { P; D; }; s/.$//' somefile
If the goal is to remove the last character in the last line, this awk should do:
awk '{a[NR]=$0} END {for (i=1;i<NR;i++) print a[i];sub(/.$/,"",a[NR]);print a[NR]}' file
sometext
moretext
lastlin
It store all data into an array, then print it out and change last line.
Just a remark: sed will temporarily remove the file.
So if you are tailing the file, you'll get a "No such file or directory" warning until you reissue the tail command.
EDITED ANSWER
I created a script and put your text inside on my Desktop. this test file is saved as "old_file.txt"
sometext
moretext
lastline
Afterwards I wrote a small script to take the old file and eliminate the last character in the last line
#!/bin/bash
no_of_new_line_characters=`wc '/root/Desktop/old_file.txt'|cut -d ' ' -f2`
let "no_of_lines=no_of_new_line_characters+1"
sed -n 1,"$no_of_new_line_characters"p '/root/Desktop/old_file.txt' > '/root/Desktop/my_new_file'
sed -n "$no_of_lines","$no_of_lines"p '/root/Desktop/old_file.txt'|sed 's/.$//g' >> '/root/Desktop/my_new_file'
opening the new_file I created, showed the output as follows:
sometext
moretext
lastlin
I apologize for my previous answer (wasn't reading carefully)
sed 's/.$//' filename | tee newFilename
This should do your job.
A couple perl solutions, for comparison/reference:
(echo 1a; echo 2b) | perl -e '$_=join("",<>); s/.$//; print'
(echo 1a; echo 2b) | perl -e 'while(<>){ if(eof) {s/.$//}; print }'
I find the first read-whole-file-into-memory approach can be generally quite useful (less so for this particular problem). You can now do regex's which span multiple lines, for example to combine every 3 lines of a certain format into 1 summary line.
For this problem, truncate would be faster and the sed version is shorter to type. Note that truncate requires a file to operate on, not a stream. Normally I find sed to lack the power of perl and I much prefer the extended-regex / perl-regex syntax. But this problem has a nice sed solution.

Replace whole line containing a string using Sed

I have a text file which has a particular line something like
sometext sometext sometext TEXT_TO_BE_REPLACED sometext sometext sometext
I need to replace the whole line above with
This line is removed by the admin.
The search keyword is TEXT_TO_BE_REPLACED
I need to write a shell script for this. How can I achieve this using sed?
You can use the change command to replace the entire line, and the -i flag to make the changes in-place. For example, using GNU sed:
sed -i '/TEXT_TO_BE_REPLACED/c\This line is removed by the admin.' /tmp/foo
You need to use wildcards (.*) before and after to replace the whole line:
sed 's/.*TEXT_TO_BE_REPLACED.*/This line is removed by the admin./'
The Answer above:
sed -i '/TEXT_TO_BE_REPLACED/c\This line is removed by the admin.' /tmp/foo
Works fine if the replacement string/line is not a variable.
The issue is that on Redhat 5 the \ after the c escapes the $. A double \\ did not work either (at least on Redhat 5).
Through hit and trial, I discovered that the \ after the c is redundant if your replacement string/line is only a single line. So I did not use \ after the c, used a variable as a single replacement line and it was joy.
The code would look something like:
sed -i "/TEXT_TO_BE_REPLACED/c $REPLACEMENT_TEXT_STRING" /tmp/foo
Note the use of double quotes instead of single quotes.
The accepted answer did not work for me for several reasons:
my version of sed does not like -i with a zero length extension
the syntax of the c\ command is weird and I couldn't get it to work
I didn't realize some of my issues are coming from unescaped slashes
So here is the solution I came up with which I think should work for most cases:
function escape_slashes {
sed 's/\//\\\//g'
}
function change_line {
local OLD_LINE_PATTERN=$1; shift
local NEW_LINE=$1; shift
local FILE=$1
local NEW=$(echo "${NEW_LINE}" | escape_slashes)
# FIX: No space after the option i.
sed -i.bak '/'"${OLD_LINE_PATTERN}"'/s/.*/'"${NEW}"'/' "${FILE}"
mv "${FILE}.bak" /tmp/
}
So the sample usage to fix the problem posed:
change_line "TEXT_TO_BE_REPLACED" "This line is removed by the admin." yourFile
All of the answers provided so far assume that you know something about the text to be replaced which makes sense, since that's what the OP asked. I'm providing an answer that assumes you know nothing about the text to be replaced and that there may be a separate line in the file with the same or similar content that you do not want to be replaced. Furthermore, I'm assuming you know the line number of the line to be replaced.
The following examples demonstrate the removing or changing of text by specific line numbers:
# replace line 17 with some replacement text and make changes in file (-i switch)
# the "-i" switch indicates that we want to change the file. Leave it out if you'd
# just like to see the potential changes output to the terminal window.
# "17s" indicates that we're searching line 17
# ".*" indicates that we want to change the text of the entire line
# "REPLACEMENT-TEXT" is the new text to put on that line
# "PATH-TO-FILE" tells us what file to operate on
sed -i '17s/.*/REPLACEMENT-TEXT/' PATH-TO-FILE
# replace specific text on line 3
sed -i '3s/TEXT-TO-REPLACE/REPLACEMENT-TEXT/'
for manipulation of config files
i came up with this solution inspired by skensell answer
configLine [searchPattern] [replaceLine] [filePath]
it will:
create the file if not exists
replace the whole line (all lines) where searchPattern matched
add replaceLine on the end of the file if pattern was not found
Function:
function configLine {
local OLD_LINE_PATTERN=$1; shift
local NEW_LINE=$1; shift
local FILE=$1
local NEW=$(echo "${NEW_LINE}" | sed 's/\//\\\//g')
touch "${FILE}"
sed -i '/'"${OLD_LINE_PATTERN}"'/{s/.*/'"${NEW}"'/;h};${x;/./{x;q100};x}' "${FILE}"
if [[ $? -ne 100 ]] && [[ ${NEW_LINE} != '' ]]
then
echo "${NEW_LINE}" >> "${FILE}"
fi
}
the crazy exit status magic comes from https://stackoverflow.com/a/12145797/1262663
In my makefile I use this:
#sed -i '/.*Revision:.*/c\'"`svn info -R main.cpp | awk '/^Rev/'`"'' README.md
PS: DO NOT forget that the -i changes actually the text in the file... so if the pattern you defined as "Revision" will change, you will also change the pattern to replace.
Example output:
Abc-Project written by John Doe
Revision: 1190
So if you set the pattern "Revision: 1190" it's obviously not the same as you defined them as "Revision:" only...
bash-4.1$ new_db_host="DB_HOSTNAME=good replaced with 122.334.567.90"
bash-4.1$
bash-4.1$ sed -i "/DB_HOST/c $new_db_host" test4sed
vim test4sed
'
'
'
DB_HOSTNAME=good replaced with 122.334.567.90
'
it works fine
To do this without relying on any GNUisms such as -i without a parameter or c without a linebreak:
sed '/TEXT_TO_BE_REPLACED/c\
This line is removed by the admin.
' infile > tmpfile && mv tmpfile infile
In this (POSIX compliant) form of the command
c\
text
text can consist of one or multiple lines, and linebreaks that should become part of the replacement have to be escaped:
c\
line1\
line2
s/x/y/
where s/x/y/ is a new sed command after the pattern space has been replaced by the two lines
line1
line2
cat find_replace | while read pattern replacement ; do
sed -i "/${pattern}/c ${replacement}" file
done
find_replace file contains 2 columns, c1 with pattern to match, c2 with replacement, the sed loop replaces each line conatining one of the pattern of variable 1
To replace whole line containing a specified string with the content of that line
Text file:
Row: 0 last_time_contacted=0, display_name=Mozart, _id=100, phonebook_bucket_alt=2
Row: 1 last_time_contacted=0, display_name=Bach, _id=101, phonebook_bucket_alt=2
Single string:
$ sed 's/.* display_name=\([[:alpha:]]\+\).*/\1/'
output:
100
101
Multiple strings delimited by white-space:
$ sed 's/.* display_name=\([[:alpha:]]\+\).* _id=\([[:digit:]]\+\).*/\1 \2/'
output:
Mozart 100
Bach 101
Adjust regex to meet your needs
[:alpha] and [:digit:]
are Character Classes and Bracket Expressions
This worked for me:
sed -i <extension> 's/.*<Line to be replaced>.*/<New line to be added>/'
An example is:
sed -i .bak -e '7s/.*version.*/ version = "4.33.0"/'
-i: The extension for the backup file after the replacement. In this case, it is .bak.
-e: The sed script. In this case, it is '7s/.*version.*/ version = "4.33.0"/'. If you want to use a sed file use the -f flag
s: The line number in the file to be replaced. In this case, it is 7s which means line 7.
Note:
If you want to do a recursive find and replace with sed then you can grep to the beginning of the command:
grep -rl --exclude-dir=<directory-to-exclude> --include=\*<Files to include> "<Line to be replaced>" ./ | sed -i <extension> 's/.*<Line to be replaced>.*/<New line to be added>/'
The question asks for solutions using sed, but if that's not a hard requirement then there is another option which might be a wiser choice.
The accepted answer suggests sed -i and describes it as replacing the file in-place, but -i doesn't really do that and instead does the equivalent of sed pattern file > tmp; mv tmp file, preserving ownership and modes. This is not ideal in many circumstances. In general I do not recommend running sed -i non-interactively as part of an automatic process--it's like setting a bomb with a fuse of an unknown length. Sooner or later it will blow up on someone.
To actually edit a file "in place" and replace a line matching a pattern with some other content you would be well served to use an actual text editor. This is how it's done with ed, the standard text editor.
printf '%s\n' '/TEXT_TO_BE_REPLACED/' d i 'This line is removed by the admin' . w q | \
ed -s /tmp/foo > /dev/null
Note that this only replaces the first matching line, which is what the question implied was wanted. This is a material difference from most of the other answers.
That disadvantage aside, there are some advantages to using ed over sed:
You can replace the match with one or multiple lines without any extra effort.
The replacement text can be arbitrarily complex without needing any escaping to protect it.
Most importantly, the original file is opened, modified, and saved. A copy is not made.
How it works
How it works:
printf will use its first argument as a format string and print each of its other arguments using that format, effectively meaning that each argument to printf becomes a line of output, which is all sent to ed on stdin.
The first line is a regex pattern match which causes ed to move its notion of "the current line" forward to the first line that matches (if there is no match the current line is set to the last line of the file).
The next is the d command which instructs ed to delete the entire current line.
After that is the i command which puts ed into insert mode;
after that all subsequent lines entered are written to the current line (or additional lines if there are any embedded newlines). This means you can expand a variable (e.g. "$foo") containing multiple lines here and it will insert all of them.
Insert mode ends when ed sees a line consisting of .
The w command writes the content of the file to disk, and
the q command quits.
The ed command is given the -s switch, putting it into silent mode so it doesn't echo any information as it runs,
the file to be edited is given as an argument to ed,
and, finally, stdout is thrown away to prevent the line matching the regex from being printed.
Some Unix-like systems may (inappropriately) ship without an ed installed, but may still ship with an ex; if so you can simply use it instead. If have vim but no ex or ed you can use vim -e instead. If you have only standard vi but no ex or ed, complain to your sysadmin.
It is as similar to above one..
sed 's/[A-Za-z0-9]*TEXT_TO_BE_REPLACED.[A-Za-z0-9]*/This line is removed by the admin./'
Below command is working for me. Which is working with variables
sed -i "/\<$E\>/c $D" "$B"
I very often use regex to extract data from files I just used that to replace the literal quote \" with // nothing :-)
cat file.csv | egrep '^\"([0-9]{1,3}\.[0-9]{1,3}\.)' | sed s/\"//g | cut -d, -f1 > list.txt

Replace whitespace with a comma in a text file in Linux

I need to edit a few text files (an output from sar) and convert them into CSV files.
I need to change every whitespace (maybe it's a tab between the numbers in the output) using sed or awk functions (an easy shell script in Linux).
Can anyone help me? Every command I used didn't change the file at all; I tried gsub.
tr ' ' ',' <input >output
Substitutes each space with a comma, if you need you can make a pass with the -s flag (squeeze repeats), that replaces each input sequence of a repeated character that is listed in SET1 (the blank space) with a single occurrence of that character.
Use of squeeze repeats used to after substitute tabs:
tr -s '\t' <input | tr '\t' ',' >output
Try something like:
sed 's/[:space:]+/,/g' orig.txt > modified.txt
The character class [:space:] will match all whitespace (spaces, tabs, etc.). If you just want to replace a single character, eg. just space, use that only.
EDIT: Actually [:space:] includes carriage return, so this may not do what you want. The following will replace tabs and spaces.
sed 's/[:blank:]+/,/g' orig.txt > modified.txt
as will
sed 's/[\t ]+/,/g' orig.txt > modified.txt
In all of this, you need to be careful that the items in your file that are separated by whitespace don't contain their own whitespace that you want to keep, eg. two words.
without looking at your input file, only a guess
awk '{$1=$1}1' OFS=","
redirect to another file and rename as needed
What about something like this :
cat texte.txt | sed -e 's/\s/,/g' > texte-new.txt
(Yes, with some useless catting and piping ; could also use < to read from the file directly, I suppose -- used cat first to output the content of the file, and only after, I added sed to my command-line)
EDIT : as #ghostdog74 pointed out in a comment, there's definitly no need for thet cat/pipe ; you can give the name of the file to sed :
sed -e 's/\s/,/g' texte.txt > texte-new.txt
If "texte.txt" is this way :
$ cat texte.txt
this is a text
in which I want to replace
spaces by commas
You'll get a "texte-new.txt" that'll look like this :
$ cat texte-new.txt
this,is,a,text
in,which,I,want,to,replace
spaces,by,commas
I wouldn't go just replacing the old file by the new one (could be done with sed -i, if I remember correctly ; and as #ghostdog74 said, this one would accept creating the backup on the fly) : keeping might be wise, as a security measure (even if it means having to rename it to something like "texte-backup.txt")
This command should work:
sed "s/\s/,/g" < infile.txt > outfile.txt
Note that you have to redirect the output to a new file. The input file is not changed in place.
sed can do this:
sed 's/[\t ]/,/g' input.file
That will send to the console,
sed -i 's/[\t ]/,/g' input.file
will edit the file in-place
Here's a Perl script which will edit the files in-place:
perl -i.bak -lpe 's/\s+/,/g' files*
Consecutive whitespace is converted to a single comma.
Each input file is moved to .bak
These command-line options are used:
-i.bak edit in-place and make .bak copies
-p loop around every line of the input file, automatically print the line
-l removes newlines before processing, and adds them back in afterwards
-e execute the perl code
If you want to replace an arbitrary sequence of blank characters (tab, space) with one comma, use the following:
sed 's/[\t ]+/,/g' input_file > output_file
or
sed -r 's/[[:blank:]]+/,/g' input_file > output_file
If some of your input lines include leading space characters which are redundant and don't need to be converted to commas, then first you need to get rid of them, and then convert the remaining blank characters to commas. For such case, use the following:
sed 's/ +//' input_file | sed 's/[\t ]+/,/g' > output_file
This worked for me.
sed -e 's/\s\+/,/g' input.txt >> output.csv

Resources