I'm currently trying to find a line in a file
#define IMAX 8000
and replacing 8000 with another number.
Currently, stuck trying to pipe arguments from awk into sed.
grep '#define IMAX' 1d_Euler_mpi_test.c | awk '{print $3}' | sed
Not too sure how to proceed from here.
I would do something like:
sed -i '' '/^#define IMAX 8000$/s/8000/NEW_NUMBER/' 1d_Euler_mpi_test.c
Could you please try following. Place new number's value in place of new_number too.(tested this with GNU sed)
echo "#define IMAX 8000" | sed -E '/#define IMAX /s/[0-9]+$/new_number/'
In case you are reading input from an Input_file and want to save its output into Input_file itself use following then.
sed -E '/#define IMAX /s/[0-9]+$/new_number/' Input_file
Add -i flag in above code in case you want to save output into Input_file itself. Also my codes will catch any digits which are coming at the end of the line which has string #define IMAX so in case you only want to look for 8000 or any fixed number change [0-9]+$ to 8000 etc in above codes then.
You may use GNU sed.
sed -i -e 's/IMAX 8000/IMAX 9000/g' /tmp/file.txt
Which will invoke sed to do an in-place edit due to the -i option. This can be called from bash.
If you really really want to use just bash, then the following can work:
while read a ; do echo ${a//IMAX 8000/IMAX 9000} ; done < /tmp/file.txt > /tmp/file.txt.t ; mv /tmp/file.txt{.t,}
This loops over each line, doing a substitution, and writing to a temporary file (don't want to clobber the input). The move at the end just moves temporary to the original name.
Related
I have a file where some entries look like:
EMIG_BAD_ID = syscall.Errno( -0x12f)
I want to use sed to replace that negative number to make it positive,
EMIG_BAD_ID = syscall.Errno( 0x12f)
I've tried some ideas from web searches but not succeded.
E.g. this one exits with an error:
egrep EMIG_* _error.grep | \
sed -e 's/syscall.Errno(\1)/syscall.Errno(-\1)/g' _error.grep
sed: -e expression #1, char 40: Invalid back reference
What is wrong here?
If the format is exactly like you posted, you can use this fairly easy replacement:
sed 's/syscall.Errno( -/syscall.Errno( /g' _error.grep
To make the space between ( and - optional:
sed 's/syscall.Errno( \?-/syscall.Errno( /g' _error.grep
To answer your question / if you insist on using back references (and optional space), here's how to use back references:
sed 's/syscall.Errno( \?-\(.*\))/syscall.Errno(\1)/g' _error.grep
Some additional notes:
You don't need grep - if EMIG_BAD_ID is on the same line it's very easy to include that in the sed matching pattern.
You pipe from egrep to sed and let sed read from a file. That doesn't make sense. You should prefer reading files directly with sed; but if you need grep, just read from stdin (without the file argument). Specify -i with sed to perform an in-place edit.
Using sed
$ sed -E '/^EMIG/s/-([0-9]+)/\1/' input_file
EMIG_BAD_ID = syscall.Errno( 0x12f)
Thank you for your answers. Unfortunately I don't have a working answer yet.
Code below: (part of a bigger file)
cat _error.out
E2BIG = 0x40000007
EMIG_ARRAY_TOO_LARGE = -0x133
cat test.sh
cat _error.out | sed 's/=\(.*\)/= syscall.Errno(\1)/'
Result:
E2BIG = syscall.Errno( 0x40000007)
EMIG_ARRAY_TOO_LARGE = syscall.Errno( -0x133)
The problem is to use bash/grep/sed to make the negative number positive.
Thanks!
I'm trying to add the filename of a text file into the first line of a the same text file. for example if the file name is called test1.txt, then the first line when you open the file should be test1.
below is what I've done so for, the only problem i have is that the word "$file" is being written to the file not the file name. any help is appreciated.
for file in *.txt; do
sed -i '1 i\$file' $file;
awk 'sub("$", "\r")' "$file" > "$file"1;
mv "$file"1 "$file";
done
Without concise, testable sample input and expected output it's an untested guess but it SOUNDS like all you need is:
awk -i inplace -v ORS='\r\n' 'FNR==1{print FILENAME}1' *
No shell loop or multiple commands required.
The above uses GNU awk for inplace editing and I'm assuming the sub() in your code was intended to add a \r at the end of every line.
I've just started learning more about sed and awk and put this into a file called insert.sed and sourced it and passed it a file name:
sed -i '1s/^./'$1'\'$'\n/g' $1
In trying it, it seems to work okay:
rent$ cat x.txt
<<< Who are you?
rent$ source insert.sed x.txt
rent$ cat x.txt
x.txt
<< Who are you?
It is cutting off the first character of the first line so I'd have to fix that otherwise it does add the file name to first line.
I'm sure there's better ways of doing it.
If you want test1 on first line, with gnu sed
sed -i '1{x;s/.*/fich=$(ps -p $PPID -o args=);fich=${fich##*\\} };echo ${fich%%.*}/e;G}' test1.txt
I'm on Linux command line and I have file with
127.0.0.1
128.0.0.0
121.121.33.111
I want
127.0.0.1:80
128.0.0.0:80
121.121.33.111:80
I remember my colleagues were using sed for that, but after reading sed manual still not clear how to do it on command line?
You could try using something like:
sed -n 's/$/:80/' ips.txt > new-ips.txt
Provided that your file format is just as you have described in your question.
The s/// substitution command matches (finds) the end of each line in your file (using the $ character) and then appends (replaces) the :80 to the end of each line. The ips.txt file is your input file... and new-ips.txt is your newly-created file (the final result of your changes.)
Also, if you have a list of IP numbers that happen to have port numbers attached already, (as noted by Vlad and as given by aragaer,) you could try using something like:
sed '/:[0-9]*$/ ! s/$/:80/' ips.txt > new-ips.txt
So, for example, if your input file looked something like this (note the :80):
127.0.0.1
128.0.0.0:80
121.121.33.111
The final result would look something like this:
127.0.0.1:80
128.0.0.0:80
121.121.33.111:80
Concise version of the sed command:
sed -i s/$/:80/ file.txt
Explanation:
sed stream editor
-i in-place (edit file in place)
s substitution command
/replacement_from_reg_exp/replacement_to_text/ statement
$ matches the end of line (replacement_from_reg_exp)
:80 text you want to add at the end of every line (replacement_to_text)
file.txt the file name
How can this be achieved without modifying the original file?
If you want to leave the original file unchanged and have the results in another file, then give up -i option and add the redirection (>) to another file:
sed s/$/:80/ file.txt > another_file.txt
sed 's/.*/&:80/' abcd.txt >abcde.txt
If you'd like to add text at the end of each line in-place (in the same file), you can use -i parameter, for example:
sed -i'.bak' 's/$/:80/' foo.txt
However -i option is non-standard Unix extension and may not be available on all operating systems.
So you can consider using ex (which is equivalent to vi -e/vim -e):
ex +"%s/$/:80/g" -cwq foo.txt
which will add :80 to each line, but sometimes it can append it to blank lines.
So better method is to check if the line actually contain any number, and then append it, for example:
ex +"g/[0-9]/s/$/:80/g" -cwq foo.txt
If the file has more complex format, consider using proper regex, instead of [0-9].
You can also achieve this using the backreference technique
sed -i.bak 's/\(.*\)/\1:80/' foo.txt
You can also use with awk like this
awk '{print $0":80"}' foo.txt > tmp && mv tmp foo.txt
Using a text editor, check for ^M (control-M, or carriage return) at the end of each line. You will need to remove them first, then append the additional text at the end of the line.
sed -i 's|^M||g' ips.txt
sed -i 's|$|:80|g' ips.txt
sed -i 's/$/,/g' foo.txt
I do this quite often to add a comma to the end of an output so I can just easily copy and paste it into a Python(or your fav lang) array
We have a process which can use a file containing sed commands to alter piped input.
I need to replace a placeholder in the input with a variable value, e.g. in a single -e type of command I can run;
$ echo "Today is XX" | sed -e "s/XX/$(date +%F)/"
Today is 2012-10-11
However I can only specify the sed aspects in a file (and then point the process at the file), E.g. a file called replacements.sed might contain;
s/XX/Thursday/
So obviously;
$ echo "Today is XX" | sed -f replacements.sed
Today is Thursday
If I want to use an environment variable or shell value, though, I can't find a way to make it expand, e.g. if replacements.txt contains;
s/XX/$(date +%F)/
Then;
$ echo "Today is XX" | sed -f replacements.sed
Today is $(date +%F)
Including double quotes in the text of the file just prints the double quotes.
Does anyone know a way to be able to use variables in a sed file?
This might work for you (GNU sed):
cat <<\! > replacements.sed
/XX/{s//'"$(date +%F)"'/;s/.*/echo '&'/e}
!
echo "Today is XX" | sed -f replacements.sed
If you don't have GNU sed, try:
cat <<\! > replacements.sed
/XX/{
s//'"$(date +%F)"'/
s/.*/echo '&'/
}
!
echo "Today is XX" | sed -f replacements.sed | sh
AFAIK, it's not possible. Your best bet will be :
INPUT FILE
aaa
bbb
ccc
SH SCRIPT
#!/bin/sh
STRING="${1//\//\\/}" # using parameter expansion to prevent / collisions
shift
sed "
s/aaa/$STRING/
" "$#"
COMMAND LINE
./sed.sh "fo/obar" <file path>
OUTPUT
fo/obar
bbb
ccc
As others have said, you can't use variables in a sed script, but you might be able to "fake" it using extra leading input that gets added to your hold buffer. For example:
[ghoti#pc ~/tmp]$ cat scr.sed
1{;h;d;};/^--$/g
[ghoti#pc ~/tmp]$ sed -f scr.sed <(date '+%Y-%m-%d'; printf 'foo\n--\nbar\n')
foo
2012-10-10
bar
[ghoti#pc ~/tmp]$
In this example, I'm using process redirection to get input into sed. The "important" data is generated by printf. You could cat a file instead, or run some other program. The "variable" is produced by the date command, and becomes the first line of input to the script.
The sed script takes the first line, puts it in sed's hold buffer, then deletes the line. Then for any subsequent line, if it matches a double dash (our "macro replacement"), it substitutes the contents of the hold buffer. And prints, because that's sed's default action.
Hold buffers (g, G, h, H and x commands) represent "advanced" sed programming. But once you understand how they work, they open up new dimensions of sed fu.
Note: This solution only helps you replace entire lines. Replacing substrings within lines may be possible using the hold buffer, but I can't imagine a way to do it.
(Another note: I'm doing this in FreeBSD, which uses a different sed from what you'll find in Linux. This may work in GNU sed, or it may not; I haven't tested.)
I am in agreement with sputnick. I don't believe that sed would be able to complete that task.
However, you could generate that file on the fly.
You could change the date to a fixed string, like
__DAYOFWEEK__.
Create a temp file, use sed to replace __DAYOFWEEK__ with $(date +%Y).
Then parse your file with sed -f $TEMPFILE.
sed is great, but it might be time to use something like perl that can generate the date on the fly.
To add a newline in the replacement expression using a sed file, what finally worked for me is escaping a literal newline. Example: to append a newline after the string NewLineHere, then this worked for me:
#! /usr/bin/sed -f
s/NewLineHere/NewLineHere\
/g
Not sure it matters but I am on Solaris unix, so not GNU sed for sure.
I have a CSV. I want to edit the 35th field of the CSV and write the change back to the 35th field. This is what I am doing on bash:
awk -F "," '{print $35}' test.csv | sed -i 's/^0/+91/g'
so, I am pulling the 35th entry using awk and then replacing the "0" in the starting position in the string with "+91". This one works perfet and I get desired output on the console.
Now I want this new entry to get written in the file. I am thinking of sed's "in -place" replacement feature but this fetuare needs and input file. In above command, I cannot provide input file because my primary command is awk and sed is taking the input from awk.
Thanks.
You should choose one of the two tools. As for sed, it can be done as follows:
sed -ri 's/^(([^,]*,){34})0([^,]*)/\1+91\3/' test.csv
Not sure about awk, but #shellter's comment might help with that.
The in-place feature of sed is misnamed, as it does not edit the file in place. Instead, it creates a new file with the same name. eg:
$ echo foo > foo
$ ln -f foo bar
$ ls -i foo bar # These are the same file
797325 bar 797325 foo
$ echo new-text > foo # Changes bar
$ cat bar
new-text
$ printf '/new/s//newer\nw\nq\n' | ed foo # Edit foo "in-place"; changes bar
9
newer-text
11
$ cat bar
newer-text
$ ls -i foo bar # Still the same file
797325 bar 797325 foo
$ sed -i s/new/newer/ foo # Does not edit in-place; creates a new file
$ ls -i foo bar
797325 bar 792722 foo
Since sed is not actually editing the file in place, but writing a new file and then renaming it to the old file, you might as well do the same.
awk ... test.csv | sed ... > test.csv.1 && mv test.csv.1 test.csv
There is the misperception that using sed -i somehow avoids the creation of the temporary file. It does not. It just hides the fact from you. Sometimes abstraction is a good thing, but other times it is unnecessary obfuscation. In the case of sed -i, it is the latter. The shell is really good at file manipulation. Use it as intended. If you do need to edit a file in place, don't use the streaming version of ed; just use ed
So, it turned out there are numerous ways to do it. I got it working with sed as below:
sed -i 's/0\([0-9]\{10\}\)/\+91\1/g' test.csv
But this is little tricky as it will edit any entry which matches the criteria. however in my case, It is working fine.
Similar implementation of above logic in perl:
perl -p -i -e 's/\b0(\d{10})\b/\+91$1/g;' test.csv
Again, same caveat as mentioned above.
More precise way of doing it as shown by Lev Levitsky because it will operate specifically on the 35th field
sed -ri 's/^(([^,]*,){34})0([^,]*)/\1+91\3/g' test.csv
For more complex situations, I will have to consider using any of the csv modules of perl.
Thanks everyone for your time and input. I surely know more about sed/awk after reading your replies.
This might work for you:
sed -i 's/[^,]*/+91/35' test.csv
EDIT:
To replace the leading zero in the 35th field:
sed 'h;s/[^,]*/\n&/35;/\n0/!{x;b};s//+91/' test.csv
or more simply:
|sed 's/^\(\([^,]*,\)\{34\}\)0/\1+91/' test.csv
If you have moreutils installed, you can simply use the sponge tool:
awk -F "," '{print $35}' test.csv | sed -i 's/^0/+91/g' | sponge test.csv
sponge soaks up the input, closes the input pipe (stdin) and, only then, opens and writes to the test.csv file.
As of 2015, moreutils is available in package repositories of several major Linux distributions, such as Arch Linux, Debian and Ubuntu.
Another perl solution to edit the 35th field in-place:
perl -i -F, -lane '$F[34] =~ s/^0/+91/; print join ",",#F' test.csv
These command-line options are used:
-i edit the file in-place
-n loop around every line of the input file
-l removes newlines before processing, and adds them back in afterwards
-a autosplit mode – split input lines into the #F array. Defaults to splitting on whitespace.
-e execute the perl code
-F autosplit modifier, in this case splits on ,
#F is the array of words in each line, indexed starting with 0
$F[34] is the 35 element of the array
s/^0/+91/ does the substitution