sed -n '$'!p abc.txt | tail +2 > def.txt
I have the above mentioned sed command in my code - I am unable to figure out what it does -I am going through sed tutorials to find it out but am not able to - Can some one please help me in figuring out what it does - Thanks
Taking this in stages:
sed -n abc.txt
"Run abc.txt through sed, but don't print anything out."
sed -n '$!p' abc.txt
(Note that I've corrected what I think was a misplaced quote mark.)
"Run abc.txt through sed; if a line isn't the last line, print it (i.e. print all but the last line)."
I guess you know the rest, but note that tail +2 is obsolete syntax-- tail -n 2 would be better.
EDIT:
To remove the last two lines, try
sed 'N;$d'
or if that doesn't work, crude but effective:
sed '$d' | sed '$d'
As far as the sed command '$'!p is concerned:
the $ matches only the last line of the input file.
the ! negates the sense of the match (so that it matches all but the last line).
the p prints out whatever was matched.
So basically this prints all but the last line of the file.
The -n option stops sed from performing its default action (to print the pattern space) - without that, you'd get one copy of the last line and two copies of all the other lines.
The quotes around $ are to stop the shell from trying to interpret it as a shell variable - I would have quoted the lot myself ('$!p') but that's a style issue, at least on bash. Other shells like csh (which uses ! for command history retrieval) may not be so forgiving.
Related
File Content
abab102
cdcd103
efef105
I want the username and id separated. Here, abab is user and 102 is id.
I use the command
sed 's/\([a-z]\)\{4\}\([0-9]\)\{3\}/Username:\1 ID:\2/' file.txt
Get this
Username:b ID:2
Username:d ID:3
Username:f ID:5
But I am expecting
Username:abab ID:102
Username:cdcd ID:103
Username:efef ID:105
But using the command
sed -e 's/\([a-z]\)\{4\}/Username:&/' -e 's/\([0-9]\)\{3\}/ID:&/' file2.txt
Output
Username:ababID:102
Username:cdcdID:103
Username:efefID:105
This output is close to what I need, but still I am expecting a blank space between Username:abab ID:102.
I want to know why \1 or \2 is not working here.
\([a-z]\) - between ( ) is one letter. If you repeat the \(\) group, back reference goes to the last matched expression. Put all the repetition inside.
's/\([a-z]\{4\}\)\([0-9]\{3\}\)/Username:\1 ID:\2/'
Ugh, simpler with extended:
sed -E 's/([a-z]{4})([0-9]{3})/Username:\1 ID:\2/'
Using sed
$ sed -E 's/[[:alpha:]]{4}/Username:& ID:/' input_file
Username:abab ID:102
Username:cdcd ID:103
Username:efef ID:105
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.
I would like to use sed to remove all occurances of this line if and only if it is this
<ab></ab>
If this line, I would not want to delete it
<ab>keyword</ab>
My attempt that's not working:
sed '/<ab></ab>/d'
Thanks for any insight. I'm not sure what's wrong as I should not have to escape anything?
I'm using a shell script named temp to execute this. My command is this:
cat foobar.html | ./temp
This is my temp shell script:
#!/bin/sh
sed -e '/td/!d' | sed '/<ab></ab>/d'
It looks like we have a couple of problems here. The first is with the / in the close-tag. sed uses this to delimit different parts of the command. Fortunately, all we have to do is escape it with \. Try:
sed '/<ab><\/ab>/d'
Here's an example on my machine:
$ cat test
<ab></ab>
<ab></ab>
<ab>test</ab>
$ sed '/<ab><\/ab>/d' test
<ab>test</ab>
$
The other problem is that I'm not sure what the purpose of sed -e '/td/!d' is. In it's default operating mode, you don't need to tell it not to delete something; just tell it exactly what you want to delete.
So, to do this on a file called input.html:
sed '/<ab><\/ab>/d' input.html
Or, to edit the file in-place, you can just do:
sed -i -e '/<ab><\/ab>/d' input.html
Additionally, sed lets you use any character you want as a delimiter; you don't have to use /. So if you'd prefer not to escape your input, you can do:
sed '\#<ab></ab>#d' input.html
Edit
In the comments, you mentioned wanting to delete lines that only contain </ab> and nothing else. To do that, you need to do what's called anchoring the match. The ^ character represents the beginning of the line for anchoring, and $ represents the end of the line.
sed '/^<\/ab>$/d' input.html
This will only match a line that contains (literally) </ab> and nothing else at all, and delete the line. If you want to match lines that contain whitespace too, but no text other than </ab>:
sed '/^[[:blank:]]*<\/ab>[[:blank:]]*$/d' input.html
[[:blank:]]* matches "0 or more whitespace characters" and is called a "POSIX bracket expression".
$ 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
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