I recently wrote a script that does a sed command, to replace all the occurrences of "string1" with "string2" in a file named "test.txt".
It looks like this:
sed -i 's/string1/string2/g' test.txt
The catch is, "string1" does not necessarily exist in test.txt.
I notice after executing a bunch of these sed commands, I get a number of empty files, left behind in the directory, with names that look like this:
"sed4l4DpD"
Does anyone know why this might be, and how I can correct it?
-i is the suffix given to the new/output file. Also, you need -e for the command.
Here's how you use it:
sed -i '2' -e 's/string1/string2/g' test.txt
This will create a file called test.txt2 that is the backup of test.txt
To replace the file (instead of creating a new copy - called an "in-place" substitution), change the -i value to '' (ie blank):
sed -i '' -e 's/string1/string2/g' test.txt
EDIT II
Here's actual command line output from a Mac (Snow Leopard) that show that my modified answer (removed space from between the -i and the suffix) is correct.
NOTE: On a linux server, there must be no space between it -i and the suffix.
> echo "this is a test" > test.txt
> cat test.txt
this is a test
> sed -i '2' -e 's/a/a good/' test.txt
> ls test*
test.txt test.txt2
> cat test.txt
this is a good test
> cat test.txt2
this is a test
> sed -i '' -e 's/a/a really/' test.txt
> ls test*
test.txt test.txt2
> cat test.txt
this is a really good test
I wasn't able to reproduce this with a quick test (using GNU sed 4.2.1) -- but strace did show sed creating a file called sedJd9Cuy and then renaming it to tmp (the file named on the command line).
It looks like something is going wrong after sed creates the temporary file and before it's able to rename it.
My best guess is that you've run out of room in the filesystem; you're able to create a new empty file, but unable to write to it.
What does df . say?
EDIT:
I still don't know what's causing the problem, but it shouldn't be too difficult to work around it.
Rather than
sed -i 's/string1/string2/g' test.txt
try something like this:
sed 's/string1/string2/g' test.txt > test.txt.$$ && mv -f test.txt.$$ test.txt
Something is going wrong with the way sed creates and then renames a text file to replace your original file. The above command uses sed as a simple input-output filter and creates and renames the temporary file separately.
So after much testing last night, it turns out that sed was creating these files when trying to operate on an empty string. The way i was getting the array of "$string1" arguments was through a grep command, which seems to be malformed. What I wanted from the grep was all lines containing something of the type "Text here '.'".
For example the string, "Text here 'ABC.DEF'" in a file, should have been caught by grep, then the ABC.DEF portion of the string, would be substituted by ABC_DEF. Unfortunately the grep I was using would catch lines of the type "Text here ''" (that is, nothing between the ''). When later on, the script attempted to perform a sed replacement using this empty string, the random file was created (probably because sed died).
Thanks for all your help in understanding how sed works.
Its better if you do it in this way:
cat large_file | sed 's/string1/string2/g' > file_filtred
Related
I've writted a sed script to replace all ^^ with NULL. It seems though that sed is only catching a pair, but not including the second in that pair as it continues to search.
echo "^^^^" | sed 's/\^\^/\^NULL\^/g'
produces
^NULL^^NULL^
when it should produce
^NULL^NULL^NULL^
Try with a loop to apply your command again to modified pattern space:
echo "^^^^" | sed ':a;s/\^\^/\^NULL\^/;t a;'
To edit a file in place on OSX, try the -i flag and multiline command:
sed -i '' ':a
s/\^\^/\^NULL\^/
t a' file
With GNU sed:
sed -i ':a;s/\^\^/\^NULL\^/;t a;' file
or simply redirect the command to a temporary file before renaming it:
sed ':a;s/\^\^/\^NULL\^/;t a;' file > tmp && mv tmp file
I really like SLePort solution, but since it is not working for you, you can try with (tested on Linux, not Mac):
echo "^^^^" | sed 's/\^\^/\^NULL\^/g; s//\^NULL\^/g'
It is doing the same as the former solution, but explicitly, not looping with tags.
You can omit the pattern in the second command and sed will use the previous pattern.
I succeeded in using a file descriptor with sed and giving the result on the standard output. Giving a file "file.txt" containing :
$ cat file.txt
foo
Foo
I open a file descriptor to file.txt, open a sub-shell, and give this file descriptor to sed :
$ (sed "/Foo/c\\bar" <&9 ) 9< file.txt
foo
bar
The result is correct.
Now, if I want to use the -i option of sed to change in place, I have troubles. I open the file descriptor in read and write mode, then give it to sed as input file :
$ (sed -i "/Foo/c\\bar" <&9 ) 9<> file.txt
sed: no input file
I do not understand why an input file is missing. Maybe sed needs a filename, and not a file descriptor when using the -i option ?
I tried a workaround which, of course, does not work as expected :
$ (sed "/Foo/c\\bar" <&9 >&9 ) 9<> file.txt
$ cat file.txt
foo
Foo
foo
bar
while I expected :
$ cat file.txt
foo
bar
Thanks in advance for your help !
Dunatotatos
You cannot edit anything "in place" with sed, and this is a great example of why -i is misnamed. gnu sed implements -i by creating a new file, writing the output to it, and then renaming the file. If you don't give sed the original filename, it doesn't know what to rename it.
sed -i expects a filename. You can't pass /dev/stdin (or similar), as sed will attempt to create a temporary file inside /dev.
You can't even save the output of sed into a temporary file and then write the output in the file descriptor again, as you can't rewind a file descriptor in Bash.
What you can do is figure out the original file name from the file descriptor. You can do this by using the link /proc/self/fd/9, like this:
sed -i "/Foo/c\\bar" "$(readlink /proc/self/fd/9)"
However, note that the original file may have been deleted or renamed, in which case this solution won't work. Also, this solution expects /proc to be available, which might not always be the case. /dev/fd/9 may be a good replacement.
Another thing to be aware of is that sed -i works by replacing the the file with a new one: after running sed -i, your fd 9 won't refer the newly created file. To workaround this problem:
name="$(readlink /proc/self/fd/9)"
cp "$name" "$name.tmp"
sed "/Foo/c\\bar" "$name.tmp" > "$name"
This way, your fd 9 will still refer the same file before and after running sed. You might want to use mktemp to create the temporary file, and atexit to ensure that it gets deleted.
So, disclaimer: I am pretty new to using bash and zsh, so there is a chance the answer is really simple. Nonetheless. I checked previous postings and couldn't find anything. (edit: I have tried this in both bash and zsh shells- same problem.)
I have a directory with many files and am trying to remove the first line from each file.
So say the directory contains: file1.txt file2.txt file3.txt ... etc.
I am using the sed command (non-GNU):
sed -i -e "1d" *.txt
For some reason, this is only removing the first line of the first file. I thought that the *.txt would affect all files matching the pattern in directory. Strangely, it is creating the file duplicates with -e appended, but both the duplicate and original are the same.
I tried this with other commands (e.g. ls *.txt) and it works fine. Is there something about sed I am missing?
Thank you in advance.
Different versions of sed in differing operating systems support various parameters.
OpenBSD (5.4) sed
The -i flag is unavailable. You can use the following /bin/sh syntax:
for i in *.txt
do
f=`mktemp -p .`
sed -e "1d" "${i}" > "${f}" && mv -- "${f}" "${i}"
done
FreeBSD (11-CURRENT) sed
The -i flag requires an extension, even if it's empty. Thus must be written as sed -i "" -e "1d" *.txt
GNU sed
This looks to see if the argument following -i is another option (or possibly a command). If so, it assumes an in-place modification. If it appears to be a file extension such as ".bak", it will rename the original with the ".bak" and then modify it into the original file's name.
There might be other variations on other platforms, but those are the three I have at hand.
use it without -e !
for one file use:
sed -i '1d' filename
for all files use :
sed -i '1d' *.txt
or
files=/path/to/files/*.extension ; for var in $files ; do sed -i '1d' $var ; done
.for me i use ubuntu and debian based systems , this method is working for me 100% , but for other platformes i'm not sure , so this is other method :
replace first line with emty pattern , and remove empty lines , (double commands):
for files in $(ls /path/to/files/*.txt); do sed -i "s/$(head -1 "$files")//g" "$files" ; sed -i '/^$/d' "$files" ; done
Note: if your files contain splash '/' , then it will give error , so in this case sed command should look like this ( sed -i "s[$(head -1 "$files")[[g" )
hope that's what you're looking for :)
The issue here is that the line number isn't reset when sed opens a new file, so 1 only matches the first line of the first file.
One solution is to use a shell loop, calling sed once for each file. Gumnos' answer shows how to do this in the most widely compatible way, although if you have a version of sed supporting the -i flag, you could do this instead:
for i in *.txt; do
sed -i.bak '1d' "$i"
done
It is possible to avoid creating the backup file by passing an empty suffix but personally, I don't think it's such a bad thing. One day you'll be grateful for it!
It appears that you're not working with GNU tools but if you were, I would recommend using GNU awk for this task. The variable FNR is useful here, as it keeps track of the record number for each file individually, allowing you to do this:
gawk -i inplace 'FNR>1' *.txt
Using the inplace extension, this allows you to remove the first line from each of your files, by only printing the lines where FNR is greater than 1.
Testing it out:
$ seq 5 > file1
$ seq 5 > file2
$ gawk -i inplace 'FNR>1' file1 file2
$ cat file1
2
3
4
5
$ cat file2
2
3
4
5
The last argument you are passing to the Sed is the problem
try something like this.
var=(`find *txt`)
for file in "${var[#]}"
do
sed -i -e 1d $file
done
This did the trick for me.
I have hundreds of text files in one directory. For all files, I want to delete all the lines that begin with HETATM. I would need a csh or bash code.
I would think you would use grep, but I'm not sure.
Use sed like this:
sed -i -e '/^HETATM/d' *.txt
to process all files in place.
-i means "in place".
-e means to execute the command that follows.
/^HETATM/ means "find lines starting with HETATM", and the following d means "delete".
Make a backup first!
If you really want to do it with grep, you could do this:
#!/bin/bash
for f in *.txt
do
grep -v "^HETATM" "%f" > $$.tmp && mv $$.tmp "$f"
done
It makes a temporary file of the output from grep (in file $$.tmp) and only overwrites your original file if the command executes successfully.
Using the -v option of grep to get all the lines that do not match:
grep -v '^HETATM' input.txt > output.txt
I'm trying to extract lines from certain files that do not begin with # (commented out). How would I run through a file, ignore everything with a # in front of it, but copy each line that does not start with a # into a different file.
Thanks
Simpler: grep -v '^[[:space:]]*#' input.txt > output.txt
This assumes that you're using Unix/Linux shell and the available Unix toolkit of commands AND that you want to keep a copy of the original file.
cp file file.orig
mv file file.fix
sed '/^[ ]*#/d' file.fix > file
rm file.fix
Or if you've got a nice shiny new GNU sed that all be summarized as
cp file file.orig
sed -i '/^[ ]*#/d' file
In both cases, the regexp in the sed command is meant to be [spaceCharTabChar]
So you saying, delete any line that begins with an (optional space or tab chars) #, but print everything else.
I hope this helps.
grep -v ^\# file > newfile
grep -v ^\# file | grep -v ^$ > newfile
Not fancy regex, but I provide this method to Jr. Admins as it helps with understanding of pipes and redirection.