I want to do some simple string replace in Bash with sed. I am Ubuntu 10.10.
Just see the following code, it is self-explanatory:
name="A%20Google.."
echo $name|sed 's/\%20/_/'|sed 's/\.+/_/'
I want to get A_Google_ but I get A_Google..
The sed 's/\.+/_/' part is obviously wrong.
BTW, sed 's/\%20/_/' and sed 's/%20/_/' both work. Which is better?
sed speaks POSIX basic regular expressions, which don't include + as a metacharacter. Portably, rewrite to use *:
sed 's/\.\.*/_/'
or if all you will ever care about is Linux, you can use various GNU-isms:
sed -r 's/\.\.*/_/' # turn on POSIX EREs (use -E instead of -r on OS X)
sed 's/\.\+/_/' # GNU regexes invert behavior when backslash added/removed
That last example answers your other question: a character which is literal when used as is may take on a special meaning when backslashed, and even though at the moment % doesn't have a special meaning when backslashed, future-proofing means not assuming that \% is safe.
Additional note: you don't need two separate sed commands in the pipeline there.
echo $name | sed -e 's/\%20/_/' -e 's/\.+/_/'
(Also, do you only need to do that once per line, or for all occurrences? You may want the /g modifier.)
The sed command doesn't understand + so you'll have to expand it by hand:
sed 's/\.\.*/_/'
Or tell sed that you want to use extended regexes:
sed -r 's/\.+/_/' # GNU
sed -E 's/\.+/_/' # OSX
Which switch, -r or -E, depends on your sed and it might not even support extended regexes so the portable solution is to use \.\.* in place of \.+. But, since you're on Linux, you should have GNU sed so sed -r should do the trick.
Related
I have two sed command which includes in my cook.sh script. One command is
sed -E -i "s/^(\\\$mainDomain=\")[^\"]+(\";)$/\1$MainDomain\2/" /var/config.php
This is working fine.
But the below command which is almost same. But it is not working.
sed -E -i "s/^(\\\$authURI=\")[^\"]+(\";)$/\1$duo_auth\2/" /var/config.php
That give the below error message
sed: -e expression #1, char 36: unknown option to `s'
Any idea on this ?
The issue is likely due to your replacement variable $duo_auth having a un-escaped /, change the default sed separator from / to ~ as
sed -E -i "s~^(\\\$authURI=\")[^\"]+(\";)$~\1$duo_auth\2~" /var/config.php
Try it without -i for seeing if the replacement is as expected and put it back after confirmation.
Example:-
cat /var/config.php
<?php
$authURI="dev.digin.io";
now setting the variable
duo_auth="http://auth.uri.digin.io:3048/"
Now the replacement, without -i
sed -E "s~^(\\\$authURI=\")[^\"]+(\";)$~\1$duo_auth\2~" /var/config.php
<?php
$authURI="http://auth.uri.digin.io:3048/";
The problem is probably due to $duo_auth containing an unescaped /. This means that the sed editing script will have a syntax error.
You may pick any other character to use as a delimiter in the s/.../.../ command, for example #:
sed "s#....#....#"
Just make sure that you pick a character that is not ever going to turn up in either $duo_auth or $authURI.
While testing, I'd recommend that you avoid using in-place-editing (-i) with sed. Also, the -i switch is horribly non-portable between sed implementations (some requires an argument).
Instead, do the slightly more cumbersome
sed -e "s#...#...#" data.in >data.in.tmp && mv -f data.in.tmp data.in
While testing, check the data.in.tmp file before moving it.
I am trying to convert 2015-06-03_18-05-30 to 20150603180530 using sed.
I have this:
$ var='2015-06-03_18-05-30'
$ echo $var | sed 's/\-\|\_//g'
$ echo $var | sed 's/-|_//g'
None of these are working. Why is the alternation not working?
As long as your script has a #!/bin/bash (or ksh, or zsh) shebang, don't use sed or tr: Your shell can do this built-in without the (comparatively large) overhead of launching any external tool:
var='2015-06-03_18-05-30'
echo "${var//[-_]/}"
That said, if you really want to use sed, the GNU extension -r enables ERE syntax:
$ sed -r -e 's/-|_//g' <<<'2015-06-03_18-05-30'
20150603180530
See http://www.regular-expressions.info/posix.html for a discussion of differences between BRE (default for sed) and ERE. That page notes, in discussing ERE extensions:
Alternation is supported through the usual vertical bar |.
If you want to work on POSIX platforms -- with /bin/sh rather than bash, and no GNU extensions -- then reformulate your regex to use a character class (and, to avoid platform-dependent compatibility issues with echo[1], use printf instead):
printf '%s\n' "$var" | sed 's/[-_]//g'
[1] - See the "APPLICATION USAGE" section of that link, in particular.
Something like this ought to do.
sed 's/[-_]//g'
This reads as:
s: Search
/[-_]/: for any single character matching - or _
//: replace it with nothing
g: and do that for every character in the line
Sed operates on every line by default, so this covers every instance in the file/string.
I know you asked for a solution using sed, but I offer an alternative in tr:
$ var='2015-06-03_18-05-30'
$ echo $var | tr -d '_-'
20150603180530
tr should be a little faster.
Explained:
tr stands for translate and it can be used to replace certain characters with another ones.
-d option stands for delete and it removes the specified characters instead of replacing them.
'_-' specifies the set of characters to be removed (can also be specified as '\-_' but you need to escape the - there because it's considered another option otherwise).
Easy:
sed 's/[-_]//g'
The character class [-_] matches of the characters from the set.
sed 's/[^[:digit:]]//g' YourFile
Could you tell me what failed on echo $var | sed 's/\-\|\_//g', it works here (even if escapping - and _ are not needed and assuming you use a GNU sed due to \| that only work in this enhanced version of sed)
Having
DEST_PATH='/var/www/clones'
site='xyz.com'
sed -i -e "s/\$log_path\s=\s'\(.*\)'/\$log_path = '$DEST_PATH\/$site\/logs'/" $DEST_PATH/$site/configuration.php
The problem is the forward slashes in first variable, because this is what is being processed and returns error:
sed -i -e "s/\$log_path\s=\s'\(.*\)'/\$log_path = '/var/www/clones\/xyz.com\/logs'/" configuration.php
When this is what actually should be run:
sed -i -e "s/\$log_path\s=\s'\(.*\)'/\$log_path = '\/var\/www\/clones\/xyz.com\/logs'/" configuration.php
So I know, I could replace all the / inside $DEST_PATH, and run the sed again, but I was wondering if you know or can think of any other/better way of doing so. Ideally, maybe having sed automatically escape the '$DEST_PATH/$site/logs' if possible.
Are you using a modern enough version of sed (e.g., GNU sed)? Then you are not required to use / to separate pattern and substitution. Any character will do.
E.g., you can use
s,pattern,substitution,
instead of
s/pattern/substitution/
I'm trying to transform this 3.11.0.17.16 into 3.11.0-17-generic using only bash and unix tools. The 16 in the original string can be anything. I feel like sed is the answer, but I'm not comfortable with its flavor of regex. How would you do this?
Version using awk instead of sed:
echo "3.11.0.17.16" | awk -F. '{printf "%s.%s.%s-%s-generic\n",$1,$2,$3,$4}'
echo "3.11.0.17.16" | sed 's/\.\([0-9][0-9]*\)\.[0-9][0-9]*$/-\1-generic/'
3.11.0-17-generic
This only accepts digits in the final component. If you want to accept arbitrary characters other than . there (you can't allow . or the match will become ambiguous) then write instead
echo "3.11.0.17.gr#wl1x" | sed 's/\.\([0-9][0-9]*\)\.[^.][^.]*$/-\1-generic/'
In a portable sed invocation you are limited to POSIX basic regular expressions, which most importantly means you cannot use +, ?, or |, and ( ) { } are ordinary characters unless \-escaped. Many sed implementations now accept an -E option that brings their regex syntax in line with egrep, but that is not a feature even of the very latest revision of POSIX so you cannot rely on it.
Substring removal using bash parameter expansion and extended globs
shopt -s extglob
version=3.11.0.17.16
version=${version%.+(!(.))}
printf "%s-%s-generic\n" ${version%.+(!(.))} ${version##*.}
3.11.0-17-generic
If you anchor the regex you are trying to match onto the last 3 sets of digits you would get
echo "3.11.0.17.16" | sed 's!\([0-9]*\)\.\([0-9]*\)\.\([0-9]*\)$!\1-\2-generic!'
i have a file which contains several instances of \n.
i would like to replace them with actual newlines, but sed doesn't recognize the \n.
i tried
sed -r -e 's/\n/\n/'
sed -r -e 's/\\n/\n/'
sed -r -e 's/[\n]/\n/'
and many other ways of escaping it.
is sed able to recognize a literal \n? if so, how?
is there another program that can read the file interpreting the \n's as real newlines?
Can you please try this
sed -i 's/\\n/\n/g' input_filename
What exactly works depends on your sed implementation. This is poorly specified in POSIX so you see all kinds of behaviors.
The -r option is also not part of the POSIX standard; but your script doesn't use any of the -r features, so let's just take it out. (For what it's worth, it changes the regex dialect supported in the match expression from POSIX "basic" to "extended" regular expressions; some sed variants have an -E option which does the same thing. In brief, things like capturing parentheses and repeating braces are "extended" features.)
On BSD platforms (including MacOS), you will generally want to backslash the literal newline, like this:
sed 's/\\n/\
/g' file
On some other systems, like Linux (also depending on the precise sed version installed -- some distros use GNU sed, others favor something more traditional, still others let you choose) you might be able to use a literal \n in the replacement string to represent an actual newline character; but again, this is nonstandard and thus not portable.
If you need a properly portable solution, probably go with Awk or (gasp) Perl.
perl -pe 's/\\n/\n/g' file
In case you don't have access to the manuals, the /g flag says to replace every occurrence on a line; the default behavior of the s/// command is to only replace the first match on every line.
awk seems to handle this fine:
echo "test \n more data" | awk '{sub(/\\n/,"**")}1'
test ** more data
Here you need to escape the \ using \\
$ echo "\n" | sed -e 's/[\\][n]/hello/'
sed works one line at a time, so no \n on 1 line only (it's removed by sed at read time into buffer). You should use N, n or H,h to fill the buffer with more than one line, and then \n appears inside. Be careful, ^ and $ are no more end of line but end of string/buffer because of the \n inside.
\n is recognized in the search pattern, not in the replace pattern. Two ways for using it (sample):
sed s/\(\n\)bla/\1blabla\1/
sed s/\nbla/\
blabla\
/
The first uses a \n already inside as back reference (shorter code in replace pattern);
the second use a real newline.
So basically
sed "N
$ s/\(\n\)/\1/g
"
works (but is a bit useless). I imagine that s/\(\n\)\n/\1/g is more like what you want.