Replacing text in sed with escape characters - unix

I have a very large file where I need to replace the characters \x01\n with the character \n. How would this be done with sed? So far I have:
$ sed -i 's/\x01\n/\n' file
extra characters at the end of d command
But perhaps I'm missing a few escape characters.


Don't write \n in the line oriented sed:
sed -i 's/\x01$//' file


I think you are just missing the closing '/':
$ sed -i 's/\x01\n/\n/' file


This might work for you (GNU sed):
sed -zi 's/\x01\n/\n/g' file
This slurps in the whole (or parts delimited by \x00) file and replaces \x01\n by \n globally.

Related

Remove all special characters from a file.

I have a file with special characters and I only want alpha/numbers back.
File:
!12asdfasdf#$!#$%$##123123%##$%123
Return:
12asdfasdf123123123

You could either use sed or awk for this:
Sed
sed "s/[^a-z|0-9]//g;" file
Awk
awk '{gsub(/[[:punct:]]/,"")}1' file

Sed replace only exact match

I wan't to replace a string like Europe12 with Europe12_yesturday in a file. Without changing the Europe12-36 strings that also exists in the file.
I tried:
$basename=Europe12
sed -i 's/\b$basename\b/${basename}_yesterday/g' file.txt
but this also changed the Europe12-36 strings.

Require a space or end of line character:
sed 's/Europe12\([ ]|$\)/Europe12_yesturday\1/g' input

Manually construct the delimiter list you want instead of using \b, \W or \<. - is not part of the word characters (alphanumericals), so that's why this also matches your other string. So try something like this, expanding the list as needed: [-a-zA-Z0-9].

You can do it in 2 times:
sed -e 's/Europe12/Europe12_yesturday/g' -e 's/Europe12_yesturday-36/Europe12-36/g' file.txt

sed 's/\(Europe12[[:blank:]]\)/\1_yesturday/g;s/Europe12$/&_yesturday/' YourFile
[[:blank:]] could be completeted with any boundary you accept also like .,;:/]) etc (be carrefull of regex meaning of this char in this case)

It is little late to reply..
It can be achieved easily by "word boundary" notation (\<..\>)
sed -i 's/\<$basename\>/${basename}_yesterday/g' file.txt

How to remove blank lines from a Unix file

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

Replacing all commas with spaces - S/R with a Unix Command?

I'm looking for a Unix command that will allow me to search/replace in a file - I need to replace all commas in a certain file with spaces. I need to do this in a script and I'm looking to avoid parsing/reading the file line by line. Is there a simple unix command that will allow me to do this?

sed 's/,/ /g' filename >resultfile

You can use awk, sed, vi, ex or even Perl, PHP etc ... depends what you are proficient with.
sed example:
sed -i 's/,/ /g' filename_here

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

Resources