I need to replace within a little bash script a string inside a file but... I am getting weird results.
Let's say I want to replace:
<tag><![CDATA[text]]></tag>
With:
<tag><![CDATA[replaced_text]]></tag>
Should I use sed? I think due to / and [ ] I am getting weird results...
What would be the best way of approaching this?
Perl with -p option works almost as sed and it has \Q (quote) switch for its regexes:
perl -pe 's{\Q<tag><![CDATA[text]]></tag>}
{<tag><![CDATA[replaced_text]]></tag>}' YOUR_FILE
And in Perl you can use different punctuation to delimiter your expressions (s{...}{...} in my example).
Yes, you need to escape the brackets, and either escape slashes or use different delimiters.
sed 's,<tag><!\[CDATA\[text\]\]></tag>,<tag><!\[CDATA\[replaced)text\]\]></tag>,'
That said, SGML and XML are not actually any better than HTML when it comes to using regexes; don't expect this to generalize.
This should be enough:
$ echo '<tag><![CDATA[text]]></tag>' | sed 's/\[text\]/\[replaced_text\]/'
<tag><![CDATA[replaced_text]]></tag>
You can also change your / separator inside sed to a different character like ,, | or %.
Just use a delimiter other than /, here I use #:
sed -i 's#<tag><!\[CDATA\[text\]\]></tag>#<tag><![CDATA[replaced_text]]></tag>#g' filename
-i to have sed change the file instead of printing it out.
g is for matching more than once (global).
But do you know the exact string you want to match, both the tag and the text?
For instance, if you want to replace the text in all with your replaced_text:
perl -i -pe 's#(<tag><!\[CDATA\[)(.*?)(\]\]></tag>)#\1replaced_text\3#g' filename
Switched to perl because sed doesn't support non-greedy multipliers (the *?).
Related
I have an input stream of many lines which look like this:
path/to/file: example: 'extract_me.proto'
path/to/other-file: example: 'me_too.proto'
path/to/something/else: example: 'and_me_2.proto'
...
I'd like to just extract the *.proto filenames from these lines, and I have tried:
[INPUT] | sed 's/^.*\([a-zA-Z0-9_]+\.proto\).*$/\1/'
I know that part of my problem is that .* is greedy and I'm going to get things like e.proto and o.proto and 2.proto, but I can't even get that far... it just outputs with the same lines as the input. Any help would be greatly appreciated.
I find it helpful to use extended regex for this purpose (-r) in which case you need not escape your brackets.
sed -r 's/^.*[^a-zA-Z0-9_]([a-zA-Z0-9_]+\.proto).*$/\1/'
The addition of [^a-zA-Z0-9_] forces the .* to not be greedy.
Since you tag your command with linux, I'll assume you have GNU grep. Pick one of
grep -oP '\w+\.proto' file
grep -o "[^']+\\.proto" file
one way to do it:
sed 's/^.*[^a-zA-Z0-9_]\([a-zA-Z0-9_]\+\.proto\).*$/\1/'
escaped the + char
put a negation before the alphanum+underscore to delimit the leading chars
another way: use single quote delimitation, after all it's here for that:
sed "s/^.*'\([a-zA-Z0-9_]\+\.proto\)'.*\$/\1/"
Use this sed:
sed "s/^.*'\([a-zA-Z0-9_]\+\.proto\).*$/\1/"
+ - Extended-RegEx. So, you need to escape to get special meaning. The preceding item will be matched one or more times.
Another way:
sed "s/^.*'\([^']\+\.proto\)'.*$/\1/"
With GNU sed:
sed -E "s/.*'([^']+)'$/\1/"
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.
I've read lots of posts to understand how to correctly escape white spaces and special characters inside strings using sed, but still i can't make it, here's what i'm trying to achieve.
I have a file containing the some strings like this one:
JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=some_value"
and i'm trying to replace 'some_value' using the following:
sed -i "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1$DORG_APACHE_CATALINA_JSESSIONID/" $JBOSS_CONFIGURATION/jboss.configuration
$JBOSS_CONFIGURATION is a variable containing an absolute Linux path.
jboss.configuration is a file i'm pointing as the target for replace
operations.
$DORG_APACHE_CATALINA_JSESSIONID contains the value i want instead
of 'some_value'.
Please note that the pattern:
JAVA_OPTS="$JAVA_OPTS -D
Is always present, and org.apache.catalina.jsessionid is an example of a variable value i'm trying to replace with this script.
What's missing/wrong ? i tried also escaping whitespaces using \s without success,
and echoing the whole gives me the following:
echo "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1$DORG_APACHE_CATALINA_JSESSIONID/"
s/^\(JAVA_OPTS="$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*$/\1/
is echo interpreting the search pattern as sed does ?
any info/help/alternative ways of doing it are highly welcome,
thank you all
echo 'JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=some_value"' | (export DORG_APACHE_CATALINA_JSESSIONID=FOO/BAR/FOOBAR; sed "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1${DORG_APACHE_CATALINA_JSESSIONID////\/}\"/")
Note the bash expansion (in order to escape any / that may trip up sed) and the extra \" after $DORG_APACHE_CATALINA_JSESSIONID in order to properly close the double quote. Other than that your sed expression works for me and the above command outputs the follwoing result:
JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=FOO/BAR/FOOBAR"
You can use sed like this:
sed -r '/\$JAVA_OPTS -D/{s/^(.+=).*$/\1'"$DORG_APACHE_CATALINA_JSESSIONID"'/;}' $JBOSS_CONFIGURATION/jboss.configuration
You can specify a pattern that'll match the desired string rather than trying to specify it exactly.
The following should work for you:
sed -i 's#^\(JAVA_OPTS.*Dorg.apache.catalina.jsessionid\)=\([^"]*\)"#\1='"$DORG_APACHE_CATALINA_JSESSIONID"'"#' $JBOSS_CONFIGURATION/jboss.configuration
sed 's/=\w.*$/='"$DORG_APACHE_CATALINA_JSESSIONID"'/' $JBOSS_CONFIGURATION/jboss.configuration
I want to locate all of the files that contain self.vars['stuff'] and replace that text in-line with self.v.stuff.
I have tried using: sed -e "s/self.vars\['/self.v./" -e "s/'\]//", but that affects other lines that happen to contain ] as well...
Any ideas?
Use a backreference. This works in sed, assuming you only use single-quotes for the keys:
s/self\.vars\['\([^]]\+\)'\]/self.v.\1/g
I am using sed -e "s/foo/$bar/" -e "s/some/$text/" file.whatever to replace a phrase in a certain file. The problem is that the $bar string contains multiple special characters like /. So when I try to replace something in a text file using the following code...
#!/bin/bash
bar="http://stackoverflow.com/"
sed -e "s/foo/$bar/" -e "s/some/$text/ file.whatever
...then I get an error saying : sed: unknown option to s is there anything I can do about it?
You can use any delimiter. s#some#SOME# for example. Another good delimiter is vertical-bar. Other chars can work but have special significance for some contexts such as regular expressions.
You can get this difficulty in sed regardless of what delimiters you use, especially if you don't know what the string contains. I'd pick a different method for passing the shell variables into the helper interpreter.
awk -v rep1="$bar" -v rep2="$text" '{sub(/foo/, rep1); sub(/some/, rep2); print}'
or
perl -spe 's/foo/$rep1/; s/some/$rep2/' -- -rep1="$bar" -rep2="$text"
Correctness trumps brevity in this case.
(reference for Perl example)