I have an F5 bigip.conf text file in which I want to change the route domain from 701 to 703 for all lines showing "10.166.201." The route domain is represented by %701
10.166.201.10%701
10.166.201.15%701
10.166.201.117%701
I am able to do this with bash but the problem is that the "else printf" command (I've also tried echo), which is supposed to print out all other lines, incorrectly parses things like "\r\n" and leaves them as "rn"
#!/bin/bash
while read line
do
if [[ $line = *"10.166.201"* ]];
then
printf '%s\n' "$line" | sed -e 's/701/703/'
else printf '%s\n' "$line"
fi
done < bigip.conf > bigip.conf_updated
Is there a way to stop printf and echo from modifying the "\r\n"?
Is there a better way to do this in sed/awk?
Thanks.
Use a regexp address:
sed '/10\.166\.201\./s/%701/%703/' bigip.conf
Once you made sure that the command works for you, you can change the file in place using -i:
sed -i'' '/10\.166\.201\./s/%701/%703/' bigip.conf
With GNU sed you can omit the option value for -i:
sed -i '/10\.166\.201\./s/%701/%703/' bigip.conf
sed:
sed -E 's/(10\.166\.201\.[[:digit:]]+%70)1/\13/'
The captured group, (10\.166\.201\.[[:digit:]]+%70) matches 10.166.201. literally, then one or more digits, then %70 literally
Outside the captured group, 1 matches literally; in the replacement, the the captured group is used and 1 is replaced by 3
Example:
% cat file.txt
10.166.201.10%701
10.166.201.15%701
10.166.201.117%701
% sed -E 's/(10\.166\.201\.[[:digit:]]+%70)1/\13/' file.txt
10.166.201.10%703
10.166.201.15%703
10.166.201.117%703
Following awk may help you on same.
awk '{gsub(/\r/,"")} /your_string/{sub(/701/,"703")} 1' Input_file
Also in case you want to save output into same Input_file itself then do following:
awk '{gsub(/\r/,"")} /your_string/{sub(/701/,"703")} 1' Input_file > temp_file && mv temp_file Input_file
EDIT: In case your Input_file has \r in them then I added {gsub(/\r/,"")} in my above codes, in case you don't have them you could remove them from codes.
EDIT2: Changing string to your_string also change . to \. too in your address.
The clear, simple, robust, efficient way is:
awk 'BEGIN{RS=ORS="\r\n"; FS=OFS="%"} index($1,"10.166.201.")==1{ $2="703" } 1' file
Note that you don't need to escape the .s or anchor to avoid partial matches because the above simply treats the IP address as a string appearing at the start of the line. The above uses GNU awk for multi-char RS to preserve your \r\n line endings.
Related
Given a file, for example:
potato: 1234
apple: 5678
potato: 5432
grape: 4567
banana: 5432
sushi: 56789
I'd like to grep for all lines that start with potato: but only pipe the numbers that follow potato:. So in the above example, the output would be:
1234
5432
How can I do that?
grep 'potato:' file.txt | sed 's/^.*: //'
grep looks for any line that contains the string potato:, then, for each of these lines, sed replaces (s/// - substitute) any character (.*) from the beginning of the line (^) until the last occurrence of the sequence : (colon followed by space) with the empty string (s/...// - substitute the first part with the second part, which is empty).
or
grep 'potato:' file.txt | cut -d\ -f2
For each line that contains potato:, cut will split the line into multiple fields delimited by space (-d\ - d = delimiter, \ = escaped space character, something like -d" " would have also worked) and print the second field of each such line (-f2).
or
grep 'potato:' file.txt | awk '{print $2}'
For each line that contains potato:, awk will print the second field (print $2) which is delimited by default by spaces.
or
grep 'potato:' file.txt | perl -e 'for(<>){s/^.*: //;print}'
All lines that contain potato: are sent to an inline (-e) Perl script that takes all lines from stdin, then, for each of these lines, does the same substitution as in the first example above, then prints it.
or
awk '{if(/potato:/) print $2}' < file.txt
The file is sent via stdin (< file.txt sends the contents of the file via stdin to the command on the left) to an awk script that, for each line that contains potato: (if(/potato:/) returns true if the regular expression /potato:/ matches the current line), prints the second field, as described above.
or
perl -e 'for(<>){/potato:/ && s/^.*: // && print}' < file.txt
The file is sent via stdin (< file.txt, see above) to a Perl script that works similarly to the one above, but this time it also makes sure each line contains the string potato: (/potato:/ is a regular expression that matches if the current line contains potato:, and, if it does (&&), then proceeds to apply the regular expression described above and prints the result).
Or use regex assertions: grep -oP '(?<=potato: ).*' file.txt
grep -Po 'potato:\s\K.*' file
-P to use Perl regular expression
-o to output only the match
\s to match the space after potato:
\K to omit the match
.* to match rest of the string(s)
sed -n 's/^potato:[[:space:]]*//p' file.txt
One can think of Grep as a restricted Sed, or of Sed as a generalized Grep. In this case, Sed is one good, lightweight tool that does what you want -- though, of course, there exist several other reasonable ways to do it, too.
This will print everything after each match, on that same line only:
perl -lne 'print $1 if /^potato:\s*(.*)/' file.txt
This will do the same, except it will also print all subsequent lines:
perl -lne 'if ($found){print} elsif (/^potato:\s*(.*)/){print $1; $found++}' file.txt
These command-line options are used:
-n loop around each line of the input file
-l removes newlines before processing, and adds them back in afterwards
-e execute the perl code
You can use grep, as the other answers state. But you don't need grep, awk, sed, perl, cut, or any external tool. You can do it with pure bash.
Try this (semicolons are there to allow you to put it all on one line):
$ while read line;
do
if [[ "${line%%:\ *}" == "potato" ]];
then
echo ${line##*:\ };
fi;
done< file.txt
## tells bash to delete the longest match of ": " in $line from the front.
$ while read line; do echo ${line##*:\ }; done< file.txt
1234
5678
5432
4567
5432
56789
or if you wanted the key rather than the value, %% tells bash to delete the longest match of ": " in $line from the end.
$ while read line; do echo ${line%%:\ *}; done< file.txt
potato
apple
potato
grape
banana
sushi
The substring to split on is ":\ " because the space character must be escaped with the backslash.
You can find more like these at the linux documentation project.
Modern BASH has support for regular expressions:
while read -r line; do
if [[ $line =~ ^potato:\ ([0-9]+) ]]; then
echo "${BASH_REMATCH[1]}"
fi
done
grep potato file | grep -o "[0-9].*"
Given a file, for example:
potato: 1234
apple: 5678
potato: 5432
grape: 4567
banana: 5432
sushi: 56789
I'd like to grep for all lines that start with potato: but only pipe the numbers that follow potato:. So in the above example, the output would be:
1234
5432
How can I do that?
grep 'potato:' file.txt | sed 's/^.*: //'
grep looks for any line that contains the string potato:, then, for each of these lines, sed replaces (s/// - substitute) any character (.*) from the beginning of the line (^) until the last occurrence of the sequence : (colon followed by space) with the empty string (s/...// - substitute the first part with the second part, which is empty).
or
grep 'potato:' file.txt | cut -d\ -f2
For each line that contains potato:, cut will split the line into multiple fields delimited by space (-d\ - d = delimiter, \ = escaped space character, something like -d" " would have also worked) and print the second field of each such line (-f2).
or
grep 'potato:' file.txt | awk '{print $2}'
For each line that contains potato:, awk will print the second field (print $2) which is delimited by default by spaces.
or
grep 'potato:' file.txt | perl -e 'for(<>){s/^.*: //;print}'
All lines that contain potato: are sent to an inline (-e) Perl script that takes all lines from stdin, then, for each of these lines, does the same substitution as in the first example above, then prints it.
or
awk '{if(/potato:/) print $2}' < file.txt
The file is sent via stdin (< file.txt sends the contents of the file via stdin to the command on the left) to an awk script that, for each line that contains potato: (if(/potato:/) returns true if the regular expression /potato:/ matches the current line), prints the second field, as described above.
or
perl -e 'for(<>){/potato:/ && s/^.*: // && print}' < file.txt
The file is sent via stdin (< file.txt, see above) to a Perl script that works similarly to the one above, but this time it also makes sure each line contains the string potato: (/potato:/ is a regular expression that matches if the current line contains potato:, and, if it does (&&), then proceeds to apply the regular expression described above and prints the result).
Or use regex assertions: grep -oP '(?<=potato: ).*' file.txt
grep -Po 'potato:\s\K.*' file
-P to use Perl regular expression
-o to output only the match
\s to match the space after potato:
\K to omit the match
.* to match rest of the string(s)
sed -n 's/^potato:[[:space:]]*//p' file.txt
One can think of Grep as a restricted Sed, or of Sed as a generalized Grep. In this case, Sed is one good, lightweight tool that does what you want -- though, of course, there exist several other reasonable ways to do it, too.
This will print everything after each match, on that same line only:
perl -lne 'print $1 if /^potato:\s*(.*)/' file.txt
This will do the same, except it will also print all subsequent lines:
perl -lne 'if ($found){print} elsif (/^potato:\s*(.*)/){print $1; $found++}' file.txt
These command-line options are used:
-n loop around each line of the input file
-l removes newlines before processing, and adds them back in afterwards
-e execute the perl code
You can use grep, as the other answers state. But you don't need grep, awk, sed, perl, cut, or any external tool. You can do it with pure bash.
Try this (semicolons are there to allow you to put it all on one line):
$ while read line;
do
if [[ "${line%%:\ *}" == "potato" ]];
then
echo ${line##*:\ };
fi;
done< file.txt
## tells bash to delete the longest match of ": " in $line from the front.
$ while read line; do echo ${line##*:\ }; done< file.txt
1234
5678
5432
4567
5432
56789
or if you wanted the key rather than the value, %% tells bash to delete the longest match of ": " in $line from the end.
$ while read line; do echo ${line%%:\ *}; done< file.txt
potato
apple
potato
grape
banana
sushi
The substring to split on is ":\ " because the space character must be escaped with the backslash.
You can find more like these at the linux documentation project.
Modern BASH has support for regular expressions:
while read -r line; do
if [[ $line =~ ^potato:\ ([0-9]+) ]]; then
echo "${BASH_REMATCH[1]}"
fi
done
grep potato file | grep -o "[0-9].*"
I am having trouble figuring out how to grep the characters between two single quotes .
I have this in a file
version: '8.x-1.0-alpha1'
and I like to have the output like this (the version numbers can be various):
8.x-1.0-alpha1
I wrote the following but it does not work:
cat myfile.txt | grep -e 'version' | sed 's/.*\?'\(.*?\)'.*//g'
Thank you for your help.
Addition:
I used the sed command sed -n "s#version:\s*'\(.*\)'#\1#p"
I also like to remove 8.x- which I edited to sed -n "s#version:\s*'8.x-\(.*\)'#\1#p".
This command only works on linux and it does not work on MAC. How to change this command to make it works on MAC?
sed -n "s#version:\s*'8.x-\(.*\)'#\1#p"
If you just want to have that information from the file, and only that you can quickly do:
awk -F"'" '/version/{print $2}' file
Example:
$ echo "version: '8.x-1.0-alpha1'" | awk -F"'" '/version/{print $2}'
8.x-1.0-alpha1
How does this work?
An awk program is a series of pattern-action pairs, written as:
condition { action }
condition { action }
...
where condition is typically an expression and action a series of commands.
-F "'": Here we tell awk to define the field separator FS to be a <single quote> '. This means the all lines will be split in fields $1, $2, ... ,$NF and between each field there is a '. We can now reference these fields by using $1 for the first field, $2 for the second ... etc and this till $NF where NF is the total number of fields per line.
/version/{print $2}: This is the condition-action pair.
condition: /version/:: The condition reads: If a substring in the current record/line matches the regular expression /version/ then do action. Here, this is simply translated as if the current line contains a substring version
action: {print $2}:: If the previous condition is satisfied, then print the second field. In this case, the second field would be what the OP requests.
There are now several things that can be done.
Improve the condition to be /^version :/ && NF==3 which reads _If the current line starts with the substring version : and the current line has 3 fields then do action
If you only want the first occurance, you can tell the system to exit immediately after the find by updating the action to {print $2; exit}
I'd use GNU grep with pcre regexes:
grep -oP "version: '\\K.*(?=')" file
where we are looking for "version: '" and then the \K directive will forget what it just saw, leaving .*(?=') to match up to the last single quote.
Try something like this: sed -n "s#version:\s*'\(.*\)'#\1#p" myfile.txt. This avoids the redundant cat and grep by finding the "version" line and extracting the contents between the single quotes.
Explanation:
the -n flag tells sed not to print lines automatically. We then use the p command at the end of our sed pattern to explicitly print when we've found the version line.
Search for pattern: version:\s*'\(.*\)'
version:\s* Match "version:" followed by any amount of whitespace
'\(.*\)' Match a single ', then capture everything until the next '
Replace with: \1; This is the first (and only) capture group above, containing contents between single quotes.
When your only want to look at he quotes, you can use cut.
grep -e 'version' myfile.txt | cut -d "'" -f2
grep can almost do this alone:
grep -o "'.*'" file.txt
But this may also print lines you don't want to: it will print all lines with 2 single quotes (') in them. And the output still has the single quotes (') around it:
'8.x-1.0-alpha1'
But sed alone can do it properly:
sed -rn "s/^version: +'([^']+)'.*/\1/p" file.txt
How can I replace/delete characters in a file while leaving comment lines unchanged? I'm looking for a something to the effect of the following lines (where 'X' is replaced for 'Y' in file.txt), just substantially faster:
while read line
do
if [[ ${line:0:1} = "#" ]]
then
echo "$line"
else
echo "$line" | tr "X" "Y"
fi
done < file.txt
Thank you!
Equivalent, more accurate (and faster) will be this sed command as compared to your script:
sed '/^ *#/!{s/X/Y/g;}' file.txt
This means match any line that doesn't have 0 or more spaces followed by # at the start of line and replace X with Y globally.
i am willing to bet perl will be faster than all above :
perl -i -pe 's/X/Y/g unless /^#/' file.txt
for fast replacement, use sed, and only replace in lines not starting with "#":
cat foo.txt | sed -e '/^#/! s/X/Y/g'
sed -i '/^#/! s/{what_to_replace}/{to_what_to_replace}/g' file.txt
awk version:
awk '!/^ *#/{gsub(/X/,"Y")}1' file.txt
Do look for word boundaries to prevent sub strings of your substitution from getting replaced. For example, with gawk you can use \< and \>
All of the lines with comments in a file begin with #. How can I delete all of the lines (and only those lines) which begin with #? Other lines containing #, but not at the beginning of the line should be ignored.
This can be done with a sed one-liner:
sed '/^#/d'
This says, "find all lines that start with # and delete them, leaving everything else."
I'm a little surprised nobody has suggested the most obvious solution:
grep -v '^#' filename
This solves the problem as stated.
But note that a common convention is for everything from a # to the end of a line to be treated as a comment:
sed 's/#.*$//' filename
though that treats, for example, a # character within a string literal as the beginning of a comment (which may or may not be relevant for your case) (and it leaves empty lines).
A line starting with arbitrary whitespace followed by # might also be treated as a comment:
grep -v '^ *#' filename
if whitespace is only spaces, or
grep -v '^[ ]#' filename
where the two spaces are actually a space followed by a literal tab character (type "control-v tab").
For all these commands, omit the filename argument to read from standard input (e.g., as part of a pipe).
The opposite of Raymond's solution:
sed -n '/^#/!p'
"don't print anything, except for lines that DON'T start with #"
you can directly edit your file with
sed -i '/^#/ d'
If you want also delete comment lines that start with some whitespace use
sed -i '/^\s*#/ d'
Usually, you want to keep the first line of your script, if it is a sha-bang, so sed should not delete lines starting with #!. also it should delete lines, that just contain only a hash but no text. put it all together:
sed -i '/^\s*\(#[^!].*\|#$\)/d'
To be conform with all sed variants you need to add a backup extension to the -i option:
sed -i.bak '/^\s*#/ d' $file
rm -Rf $file.bak
You can use the following for an awk solution -
awk '/^#/ {sub(/#.*/,"");getline;}1' inputfile
This answer builds upon the earlier answer by Keith.
egrep -v "^[[:blank:]]*#" should filter out comment lines.
egrep -v "^[[:blank:]]*(#|$)" should filter out both comments and empty lines, as is frequently useful.
For information about [:blank:] and other character classes, refer to https://en.wikipedia.org/wiki/Regular_expression#Character_classes.
If you want to delete from the file starting with a specific word, then do this:
grep -v '^pattern' currentFileName > newFileName && mv newFileName currentFileName
So we have removed all the lines starting with a pattern, writing the content into a new file, and then copy the content back into the source/current file.
You also might want to remove empty lines as well
sed -E '/(^$|^#)/d' inputfile
Delete all empty lines and also all lines starting with a # after any spaces:
sed -E '/^$|^\s*#/d' inputfile
For example, see the following 3 deleted lines (including just line numbers!):
1. # first comment
2.
3. # second comment
After testing the command above, you can use option -i to edit the input file in place.
Just this!
Here is it with a loop for all files with some extension:
ll -ltr *.filename_extension > list.lst
for i in $(cat list.lst | awk '{ print $8 }') # validate if it is the 8 column on ls
do
echo $i
sed -i '/^#/d' $i
done