How to remove blank lines from a Unix file

How to remove blank lines from a Unix file - unix

I need to remove all the blank lines from an input file and write into an output file. Here is my data as below.
11216,33,1032747,64310,1,0,0,1.878,0,0,0,1,1,1.087,5,1,1,18-JAN-13,000603221321
11216,33,1033196,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,059762153003
11216,33,1033246,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,000603211032
11216,33,1033280,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,055111034001
11216,33,1033287,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,000378689701
11216,33,1033358,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,000093737301
11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802041926
11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802041954
11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802049326
11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802049383
11216,33,1036985,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000093415580
11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781202001
11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781261305
11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781603955
11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781615746

sed -i '/^$/d' foo
This tells sed to delete every line matching the regex ^$ i.e. every empty line. The -i flag edits the file in-place, if your sed doesn't support that you can write the output to a temporary file and replace the original:
sed '/^$/d' foo > foo.tmp
mv foo.tmp foo
If you also want to remove lines consisting only of whitespace (not just empty lines) then use:
sed -i '/^[[:space:]]*$/d' foo
Edit: also remove whitespace at the end of lines, because apparently you've decided you need that too:
sed -i '/^[[:space:]]*$/d;s/[[:space:]]*$//' foo

awk 'NF' filename
awk 'NF > 0' filename
sed -i '/^$/d' filename
awk '!/^$/' filename
awk '/./' filename
The NF also removes lines containing only blanks or tabs, the regex /^$/ does not.

Use grep to match any line that has nothing between the start anchor (^) and the end anchor ($):
grep -v '^$' infile.txt > outfile.txt
If you want to remove lines with only whitespace, you can still use grep. I am using Perl regular expressions in this example, but here are other ways:
grep -P -v '^\s*$' infile.txt > outfile.txt
or, without Perl regular expressions:
grep -v '^[[:space:]]*$' infile.txt > outfile.txt

sed -e '/^ *$/d' input > output
Deletes all lines which consist only of blanks (or is completely empty). You can change the blank to [ \t] where the \t is a representation for tab. Whether your shell or your sed will do the expansion varies, but you can probably type the tab character directly. And if you're using GNU or BSD sed, you can do the edit in-place, if that's what you want, with the -i option.
If I execute the above command still I have blank lines in my output file. What could be the reason?
There could be several reasons. It might be that you don't have blank lines but you have lots of spaces at the end of a line so it looks like you have blank lines when you cat the file to the screen. If that's the problem, then:
sed -e 's/ *$//' -e '/^ *$/d' input > output
The new regex removes repeated blanks at the end of the line; see previous discussion for blanks or tabs.
Another possibility is that your data file came from Windows and has CRLF line endings. Unix sees the carriage return at the end of the line; it isn't a blank, so the line is not removed. There are multiple ways to deal with that. A reliable one is tr to delete (-d) character code octal 15, aka control-M or \r or carriage return:
tr -d '\015' < input | sed -e 's/ *$//' -e '/^ *$/d' > output
If neither of those works, then you need to show a hex dump or octal dump (od -c) of the first two lines of the file, so we can see what we're up against:
head -n 2 input | od -c
Judging from the comments that sed -i does not work for you, you are not working on Linux or Mac OS X or BSD — which platform are you working on? (AIX, Solaris, HP-UX spring to mind as relatively plausible possibilities, but there are plenty of other less plausible ones too.)
You can try the POSIX named character classes such as sed -e '/^[[:space:]]*$/d'; it will probably work, but is not guaranteed. You can try it with:
echo "Hello World" | sed 's/[[:space:]][[:space:]]*/ /'
If it works, there'll be three spaces between the 'Hello' and the 'World'. If not, you'll probably get an error from sed. That might save you grief over getting tabs typed on the command line.

grep . file
grep looks at your file line-by-line; the dot . matches anything except a newline character. The output from grep is therefore all the lines that consist of something other than a single newline.

with awk
awk 'NF > 0' filename

To be thorough and remove lines even if they include spaces or tabs something like this in perl will do it:
cat file.txt | perl -lane "print if /\S/"
Of course there are the awk and sed equivalents. Best not to assume the lines are totally blank as ^$ would do.
Cheers

You can sed's -i option to edit in-place without using temporary file:
sed -i '/^$/d' file

Related

Removing comments from a datafile. What are the differences?

Let's say that you would like to remove comments from a datafile using one of two methods:
cat file.dat | sed -e "s/\#.*//"
cat file.dat | grep -v "#"
How do these individual methods work, and what is the difference between them? Would it also be possible for a person to write the clean data to a new file, while avoiding any possible warnings or error messages to end up in that datafile? If so, how would you go about doing this?

How do these individual methods work, and what is the difference
between them?
Yes, they work same though sed and grep are 2 different commands. Your sed command simply substitutes all those lines which having # with NULL. On other hand grep will simply skip or ignore those lines which will skip lines which have # in it.
You could get more information on these by man page as follows:
man grep:
-v, --invert-match
Invert the sense of matching, to select non-matching lines. (-v is specified by POSIX.)
man sed:
s/regexp/replacement/
Attempt to match regexp against the pattern space. If successful, replace that portion matched with replacement. The
replacement may
contain the special character & to refer to that portion of the pattern space which matched, and the special escapes \1
through \9 to
refer to the corresponding matching sub-expressions in the regexp.
Would it also be possible for a person to write the clean data to a
new file, while avoiding any possible warnings or error messages to
end up in that datafile?
yes, we could re-direct the errors by using 2>/dev/null in both the commands.
If so, how would you go about doing this?
You could try like 2>/dev/null 1>output_file
Explanation of sed command: Adding explanation of sed command too now. This is only for understanding purposes and no need to use cat and then use sed you could use sed -e "s/\#.*//" Input_file instead.
sed -e " ##Initiating sed command here with adding the script to the commands to be executed
s/ ##using s for substitution of regexp following it.
\#.* ##telling sed to match a line if it has # till everything here.
//" ##If match found for above regexp then substitute it with NULL.

That grep -v will lose all the lines that have # on them, for example:
$ cat file
first
# second
thi # rd
so
$ grep -v "#" file
first
will drop off all lines with # on it which is not favorable. Rather you should:
$ grep -o "^[^#]*" file
first
thi
like that sed command does but this way you won't get empty lines. man grep:
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.

Delete empty lines from a file

I need to delete empty lines from a file (with spaces only - not null records).
The following command works only for null rows, but not in case of
spaces:
sed '/^$/d' filename
Can it be done using grep?

Use \s* for blank lines containing only whitespace:
sed '/^\s*$/d' file
To save the changes back to the file use the -i option:
sed -i '/^\s*$/d' file
Edit:
The regex ^\s*$ matches a line that only contains whitespace, grep -v print lines that don't match a given pattern so the following will print all none black lines:
grep -v '^\s*$' file

The POSIX portable way to do this is
sed -i '/^[[:blank:]]*$/d' file
or
grep -v '^[[:blank:]]*$' file

sed -i '/^[ \t]*$/d' file-name
it will delete all blank lines having any no. of white spaces (spaces or tabs) i.e. (0 or more)in the file..
NOte: there is a 'space' followed by '\t' inside the square bracket...
"-i" will force to write the updated contents back in the file... without this flag you can see the empty lines got deleted on the screen but the actual file will not be affected.

It happened that the file was copied from Windows machine a the sed command (sed '/^$/d' foo ) was not running correctly.
I ran following command and it worked.
$ dos2unix foo

Why sed removes last line?

$ cat file.txt
one
two
three
$ cat file.txt | sed "s/one/1/"
1
two
Where is the word "three"?
UPDATED:
There is no line after the word "three".

As Ivan suggested, your text file is missing the end of line (EOL) marker on the final line. Since that's not present, three is printed out by sed but then immediately over-written by your prompt. You can see it if you force an extra line to be printed.
sed 's/one/1/' file.txt && echo
This is a common problem since people incorrectly think of the EOL as an indication that there's a following line (which is why it's commonly called a "newline") and not as an indication that the current line has ended.

Using comments from other posts:
older versions of sed do not process the last line of a file if no EOL or "new line" is present.
echo can be used to add a new line
Then, to solve the problem you can re-order the commands:
( cat file.txt && echo ) | sed 's/one/1/'

I guess there is no new line character after last line. sed didn't find line separator after last line and ignore it.
Update
I suggest you to rewrite this in perl (if you have it installed):
cat file.txt | perl -pe 's/one/1/'

Instead of cat'ing the file and piping into sed, run sed with the file name as an argument after the substitution string, like so:
sed "s/one/1/" file.txt
When I did it this way, I got the "three" immediately following by the prompt:
1
two
three$

A google search shows that the man page for some versions of sed (not the GNU or BSD versions, which work as you'd expect) indicate that it won't process an incomplete line (one that's not newline-terminated) at the end of a file. The solution is to ensure your files end with a newline, install GNU sed, or use awk or perl instead.

here's an awk solution
awk '{gsub("one","1")}1' file.txt

find and replace from command line unix

I have a multi line text file where each line has the format
..... Game #29832: ......
I want to append the character '1' to each number on each line (which is different on every line), does anyone know of a way to do this from the command line?
Thanks

sed -i -e 's/Game #[0-9]*/&1/' file
-i is for in-place editing, and & means whatever matched from the pattern. If you don't want to overwrite the file, omit the -i flag.

Using sed:
cat file | sed -e 's/\(Game #[0-9]*\)/\11/'

sed 's/ Game #\([0-9]*\):/ Game #1\1:/' yourfile.txt

GNU awk
awk '{b=gensub(/(Game #[0-9]+)/ ,"\\11","g",$0); print b }' file

Replace \n with \r\n in Unix file

I'm trying to do the opposite of this question, replacing Unix line endings with Windows line endings, so that I can use SQL Server bcp over samba to import the file. I have sed installed but not dos2unix. I tried reversing the examples but to no avail.
Here's the command I'm using.
sed -e 's/\n/\r\n/g' myfile
I executed this and then ran od -c myfile, expecting to see \r\n where there used to be \n. But there all still \n. (Or at least they appear to be. The output of od overflows my screen buffer, so I don't get to see the beginning of the file).
I haven't been able to figure out what I'm doing wrong. Any suggestions?

When faced with this, I use a simple perl one-liner:
perl -pi -e 's/\n/\r\n/' filename
because sed behavior varies, and I know this works.

What is the problem with getting dos2unix onto the machine?
What is the platform you are working with?
Do you have GNU sed or regular non-GNU sed?
On Solaris, /usr/bin/sed requires:
sed 's/$/^M/'
where I entered the '^M' by typing controlV controlM. The '$' matches at the end of the line, and replaces the end of line with the control-M. You can script that, too.
Mechanisms expecting sed to expand '\r' or '\\r' to control-M are going to be platform-specific, at best.

You don't need the -e option.
$ matches the endline character. This sed command will insert a \r character before the end of line:
sed 's/$/\r/' myfile

Just adding a \r (aka ^M, see Jonathan Leffler's answer) in front of \n is not safe because the file might have mixed mode EOL, so then you risk ending up with some lines becomming \r\r\n. The safe thing to do is first remove all '\r' characters, and then insert (a single) \r before \n.
#!/bin/sh
sed 's/^M//g' ${1+"$#"} | sed 's/$/^M/'
Updated to use ^M.

sed 's/\([^^M]\)$/\0^M/' your_file
This makes sure you only insert a \r when there is no \r before \n. This worked for me.

Try using:
echo " this is output" > input
sed 's/$/\r/g' input |od -c

Maybe if you try it this way
cat myfile | sed 's/\n/\r\n/g' > myfile.win
will work, from my understanding your just making the replacements to the console output, you need to redirect output to a file, in this case myfile.win, then you could just rename it to whatever you want. The whole script would be (running inside a directory full of this kind of files):
#!/bin/bash
for file in $(find . -type f -name '*')
do
cat $file | sed 's/\n/\r\n/g' > $file.new
mv -f $file.new $file
done

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