end_date=$(date +"%m/%d/%Y")
/usr/bin/perl -pi -e "s/_end_date_/${end_date}/g" filename
I want to replace string '_end_date_' with the current date. Since the current date has slashes in it (yes, I want the slashes), I need to escape them. How can I do this?
I've tried several ways, like replacing slashes with "/" using sed and Perl itself, but it didn't work. Finally I used 'cut' to break date in 3 parts and escaped slashes, but this solution doesn't look good. Is there a better solution?
In Perl you can choose which character to use to separate parts of a regular expression. The following code will work fine.
end_date = $(date +"%m/%d/%Y")
/usr/bin/perl -pi -e "s#_end_date_#${end_date}#g" filename
This is to avoid the 'leaning toothpick' syndrome with / alternating.
Use a different s delimiter: s{all/the/slashes/you/want/}{replacement}.
I would recommend changing the delimiter, but you can almost always get by with quotemeta:
/usr/bin/perl -pi -e "my \$ed=quotemeta('${end_date}');s/_end_date_/\$ed/g" filename
But you also have this route:
/usr/bin/perl -pi -e 'BEGIN { use POSIX qw<strftime>; $ed=quotemeta(strftime( q[%m/%d/%Y], localtime())); } s/_end_date_/$ed/'
which does the same thing as your two lines.
Building on Axeman's answer, the following works for me :
perl -MPOSIX=strftime -p -e'$ed=strftime( q[%m/%d/%Y], localtime()) ;s/_end_date_/$ed/'
A few things to note
The quotemeta isn't needed because the compiler isn't looking for a / in the variable $ed.
I have used single quotes ' rather than " as otherwise you end up having to quote $
I prefer using -MPOSIX=strftime to BEGIN { use POSIX qw<strftime> }
Related
I am looking to translate this regular expression into grep flavour:
I am trying to filter all lines that contain refs/changes/\d+/$VAR/
Example of line that should match, assuming that VAR=285900
b3fb1e501749b98c69c623b8345a512b8e01c611 refs/changes/00/285900/9
Current code:
VAR=285900
grep 'refs/changes/\d+/$VAR/' sample.txt
I am trying to filter all lines that contain refs/changes/\d+/$VAR/
That would be
grep "refs/changes/[[:digit:]]\{1,\}/$VAR/"
or
grep -E "refs/changes/[[:digit:]]+/$VAR/"
Note that the \d+ notation is a perl thing. Some overfeatured greps might support it with an option, but I don't recommend it for portability reasons.
inside simple quotes I cannot use variable expansion
You can mix and match quotes:
foo=not; echo 'single quotes '"$foo"' here'
with double quotes it does match anything.
It's not clear what you're doing, so we can't say why it doesn't work. It should work. There is no need to escape forward slashes for grep, they don't have any special meaning.
I want to change some names in a file using sed. This is how the file looks like:
#! /bin/bash
SAMPLE="sample_name"
FULLSAMPLE="full_sample_name"
...
Now I only want to change sample_name & not full_sample_name using sed
I tried this
sed s/\<sample_name\>/sample_01/g ...
I thought \<> could be used to find an exact match, but when I use this, nothing is changed.
Adding '' helped to only change the sample_name. However there is another problem now: my situation was a bit more complicated than explained above since my sed command is embedded in a loop:
while read SAMPLE
do
name=$SAMPLE
sed -e 's/\<sample_name\>/$SAMPLE/g' /path/coverage.sh > path/new_coverage.sh
done < $1
So sample_name should be changed with the value attached to $SAMPLE. However when running the command sample_name is changed to $SAMPLE and not to the value attached to $SAMPLE.
I believe \< and \> work with gnu sed, you just need to quote the sed command:
sed -i.bak 's/\<sample_name\>/sample_01/g' file
In GNU sed, the following command works:
sed 's/\<sample_name\>/sample_01/' file
The only difference here is that I've enclosed the command in single quotes. Even when it is not necessary to quote a sed command, I see very little disadvantage to doing so (and it helps avoid these kinds of problems).
Another way of achieving what you want more portably is by adding the quotes to the pattern and replacement:
sed 's/"sample_name"/"sample_01"/' script.sh
Alternatively, the syntax you have proposed also works in GNU awk:
awk '{sub(/\<sample_name\>/, "sample_01")}1' file
If you want to use a variable in the replacement string, you will have to use double quotes instead of single, for example:
sed "s/\<sample_name\>/$var/" file
Variables are not expanded within single quotes, which is why you are getting the the name of your variable rather than its contents.
#user1987607
You can do this the following way:
sed s/"sample_name">/sample_01/g
where having "sample_name" in quotes " " matches the exact string value.
/g is for global replacement.
If "sample_name" occurs like this ifsample_name and you want to replace that as well
then you should use the following:
sed s/"sample_name ">/"sample_01 "/g
So that it replaces only the desired word. For example the above syntax will replace word "the" from a text file and not from words like thereby.
If you are interested in replacing only first occurence, then this would work fine
sed s/"sample_name"/sample_01/
Hope it helps
I have the following task:
I have to replace several links, but only the links which ends with .do
Important: the files have also other links within, but they should stay untouched.
<li>Einstellungen verwalten</li>
to
<li>Einstellungen verwalten</li>
So I have to search for links with .do, take the part before and remember it for example as $a , replace the whole link with
<s:url action=' '/>
and past $a between the quotes.
I thought about sed, but sed as I know does only search a whole string and replace it complete.
I also tried bash Parameter Expansions in combination with sed but got severel problems with the quotes and the variables.
cat ./src/main/webapp/include/stoBox2.jsp | grep -e '<a href=".*\.do">' | while read a;
do
b=${a#*href=\"};
c=${b%.do*};
sed -i 's/href=\"$a.do\"/href=\"<s:url action=\'$a\'/>\"/g' ./src/main/webapp/include/stoBox2.jsp;
done;
any ideas ?
Thanks a lot.
sed -i sed 's#href="\(.*\)\.do"#href="<s:url action='"'\1'"'/>"#g' ./src/main/webapp/include/stoBox2.jsp
Use patterns with parentheses to get the link without .do, and here single and double quotes separate the sed command with 3 parts (but in fact join with one command) to escape the quotes in your text.
's#href="\(.*\)\.do"#href="<s:url action='
"'\1'"
'/>"#g'
parameters -i is used for modify your file derectly. If you don't want to do this just remove it. and save results to a tmp file with > tmp.
Try this one:
sed -i "s%\(href=\"\)\([^\"]\+\)\.do%\1<s:url action='\2'/>%g" \
./src/main/webapp/include/stoBox2.jsp;
You can capture patterns with parenthesis (\(,\)) and use it in the replacement pattern.
Here I catch a string without any " but preceding .do (\([^\"]\+\)\.do), and insert it without the .do suffix (\2).
There is a / in the second pattern, so I used %s to delimit expressions instead of traditional /.
I have a string ( e.g. 3122323123123) and want to replace any 1->ax, 2->by and 3->cz.
How do I do that in bash?
I started with the character set [123] and tried with "sed", but didn't know how to write the replacement expression ?
Regex is not the tool for you here. There's nothing in your question that requires any regex.
You didn't specify your language, but if you're working in PHP, you could use the function strtr() which does exactly what you are looking for.
And good old str_replace() can probably also do what you want too, as it can accept arrays for the search/replacement arguments.
Most other languages should have similar capabilities that mean you shouldn't need regex for this.
Look at standard tr utility.
% echo "3122323123123" | tr "123" "abc"
cabbcbcabcabc
If you want to replace a character with multiple characters, you can use sed for every replacement:
% echo "3122323123123" | sed -e "s/1/ax/g" -e "s/2/by/g" -e "s/3/cz/g"
czaxbybyczbyczaxbyczaxbycz
In c#
string input = "3122323123123";
string output = intput.Replace('1','a').Replace('2','b').Replace('3','c');
Using Perl tr/// for example:
$ echo "3122323123123" | perl -pe "tr/123/abc/"
cabbcbcabcabc
I'm trying to change strings like this:
<a href='../Example/case23.html'><img src='Blablabla.jpg'
To this:
<a href='../Example/case23.html'><img src='<?php imgname('case23'); ?>'
And I've got this monster of a regular expression:
find . -type f | xargs perl -pi -e \
's/<a href=\'(.\.\.\/Example\/)(case\d\d)(.\.html\'><img src=\')*\'/\1\2\3<\?php imgname\(\'\2\'); \?>\'/'
But it isn't working. In fact, I think it's a problem with Bash, which could probably be pointed out rather quickly.
r: line 4: syntax error near unexpected token `('
r: line 4: ` 's/<a href=\'(.\.\.\/Example\/)(case\d\d)(.\.html\'><img src=\')*\'/\1\2\3<\?php imgname\(\'\2\'); \?>\'/''
But if you want to help me with the regular expression that'd be cool, too!
Teaching you how to fish:
s/…/…/
Use a separator other than / for the s operator because / already occurs in the expression.
s{…}{…}
Cut down on backslash quoting, prefer [.] over \. because we'll shellquote later. Let's keep backslashes only for the necessary or important parts, namely here the digits character class.
s{<a href='[.][.]/Example/case(\d\d)[.]html'>…
Capture only the variable part. No need to reassemble the string later if the most part is static.
s{<a href='[.][.]/Example/case(\d\d)[.]html'><img src='[^']*'}{<a href='../Example/case$1.html'><img src='<?php imgname('case$1'); ?>'}
Use $1 instead of \1 to denote backreferences. [^']* means everything until the next '.
To serve now as the argument for the Perl -e option, this program needs to be shellquoted. Employ the following helper program, you can also use an alias or shell function instead:
> cat `which shellquote`
#!/usr/bin/env perl
use String::ShellQuote qw(shell_quote); undef $/; print shell_quote <>
Run it and paste the program body, terminate input with Ctrl+d, you receive:
's{<a href='\''[.][.]/Example/case(\d\d)[.]html'\''><img src='\''[^'\'']*'\''}{<a href='\''../Example/case$1.html'\''><img src='\''<?php imgname('\''case$1'\''); ?>'\''}'
Put this together with shell pipeline.
find . -type f | xargs perl -pi -e 's{<a href='\''[.][.]/Example/case(\d\d)[.]html'\''><img src='\''[^'\'']*'\''}{<a href='\''../Example/case$1.html'\''><img src='\''<?php imgname('\''case$1'\''); ?>'\''}'
Bash single-quotes do not permit any escapes.
Try this at a bash prompt and you'll see what I mean:
FOO='\'foo'
will cause it to prompt you looking for the fourth single-quote. If you satisfy it, you'll find FOO's value is
\foo
You'll need to use double-quotes around your expression. Although in truth, your HTML should be using double-quotes in the first place.
Single quotes within single quotes in Bash:
set -xv
echo ''"'"''
echo $'\''
I wouldn't use a one-liner. Put your Perl code in a script, which makes it much easier to get the regex right without wondering about escaping quotes and such.
I'd use a script like this:
#!/usr/bin/perl -pi
use strict;
use warnings;
s{
( <a \b [^>]* \b href=['"] [^'"]*/case(\d+)\.html ['"] [^>]* > \s*
<img \b [^>]* \b src=['"] ) [^'"<] [^'"]*
}{$1<?php imgname('case$2'); ?>}gix;
and then do something like:
find . -type f | xargs fiximgs
– Michael
if you install the package mysql, it comes with a command called replace.
With the replace command you can:
while read line
do
X=`echo $line| replace "<a href='../Example/" ""|replace ".html'><" " "|awk '{print $1}'`
echo "<a href='../Example/$X.html'><img src='<?php imgname('$X'); ?>'">NewFile
done < myfile
same can be done with sed. sed s/'my string'/'replace string'/g.. replace is just easier to work with special characters.