I have a csv file full of lines like the following:
Aity Chel Jenni,Hendaland 229,2591 TE Amsterdam
I want to create a sed pattern for in an automated batch script that changes the info in this kind of formatting into the following formatting:
Aity Chel Jenni,Hendaland 30,2591 TE, Amsterdam
With a bit of research, I found out that I had to create a regex, then use an ampersand (&) character to have it change things around using the & to define the location of the regex.
I have tried the following:
sed 's/([1-9] [A-Z]{2}/&,/' file1 >file2
And have been trying variants of that trying to get the regexes down, but it doesn't seem to change anything.
Am I making a mistake in the usage of the ampersand or is my regex wrong?
Reading through the internet I can't seem to wrap my head around this function, can someone give me any examples/explain to me how to properly do this?
You are saying
sed 's/([1-9] [A-Z]{2}/&,/' file1 >file2
^
But you don't have to capture with () to use &. Instead, just say:
sed 's/[1-9] [A-Z]\{2\}/&,/' file
Note you need to escape the elements in the { } quantifier, unless you use -r:
sed -r 's/[1-9] [A-Z]{2}/&,/' file
Try the following:
sed -r 's:[0-9] [A-Z]{2}\b:&,:' file > out
About your own pattern, you're missing the closing parenthesis. And, iirc, you need to escape ( inside sed patterns to not match them literally.
The -r option enabled sed to use extended regex, which provides the {2} expansion.
Related
I have a string in text file where i want to replace the version number. Quotation marks can vary from ' to ". Also spaces around = can be there and can be not as well:
$data['MODULEXXX_VERSION'] = "1.0.0";
For testing i use
echo "_VERSION'] = \"1.1.1\"" | sed "s/\(_VERSION.*\)[1-9]\.[1-9]\.[1-9]/\11.1.2/"
which works perfectly.
When i change it to search in the file (the file has the same string):
sed "s/\(_VERSION.*\)[1-9]\.[1-9]\.[1-9]/\11.1.2/" -i test.php
, it does not find anything.
After after playing with the search part of regex, i found one more odd thing:
sed "s/\(_VERSION.*\)[1-9]\./\1***/" -i test.php
works and changes the string to $data['MODULEXXX_VERSION'] = "***0.0";, but
sed "s/\(_VERSION.*\)[1-9]\.[1-9]/\1***/" -i test.php
does not find anything anymore. Why?
I am using Ubuntu 17.04 desktop.
Anyone can explain what am I doing wrong? What would be the best command for replacing version numbers in the file for the string $data['MODULEXXX_VERSION'] = "***0.0";?
The main problem is that [1-9] doesn't match the 0s in the version number. You need to use [0-9].
Besides that, you may use the following sed command:
sed -r 's/(.*_VERSION['\''"]]\s*=\s*).*/\1"1.0.1";/' conf.php
This doesn't look at the current value, it simply replaces everything after the =.
I've used -r which enables extended posix regular expressions which makes it a bit simpler to formulate the pattern.
Another, probably cleaner attempt is to store the conf.php as a template like conf.php.tpl and then use a template engine to render the file. Or if you really want to use sed, the file may look like:
$data['FOO_VERSION'] = "FOO_VERSION_TPL";
Then just use:
sed 's/FOO_VERSION_TPL/1.0.1/' conf.php.tpl > conf.php
If there are multiple values to replace:
sed \
-e 's/FOO/BAR/' \
-e 's/HELLO/WORLD/' \
conf.php.tpl > conf.php
But I recommend a template engine instead of sed. That becomes more important when the content of the variables to replace may contain characters special to regular expressions.
I want to change some names in a file using sed. This is how the file looks like:
#! /bin/bash
SAMPLE="sample_name"
FULLSAMPLE="full_sample_name"
...
Now I only want to change sample_name & not full_sample_name using sed
I tried this
sed s/\<sample_name\>/sample_01/g ...
I thought \<> could be used to find an exact match, but when I use this, nothing is changed.
Adding '' helped to only change the sample_name. However there is another problem now: my situation was a bit more complicated than explained above since my sed command is embedded in a loop:
while read SAMPLE
do
name=$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 line like this
"RU_28" "CDM_279" "CDI_45"
"RU_567" "CDM_528" "CDI_10000"
I want to obtain the result below
"RU_28" "CDM_Unusued" "CDI_45"
"RU_567" "CDM_Unusued" "CDI_10000"
Do this for all the lines in the file
I'm using this commands:
sed 's/\"CDM_\w*\"/\"Unusued\"/g' File1.txt > File2.txt
It doesn't seem to works.
Thanks in advance!!!
You can use:
sed -i.bak 's/"\(CDM_\)[^"]*"/"\1Unused"/' file1.txt
You're not actually leaving "CDM_" on the right side. Your substitution says "Replace CDM_ and any number of words with Unusued." The common way to do what you actually want to do is to use parentheses around the section on the left you want to keep, and then use backreferences to indicate where they go on the right. In this case, you just need a single backreference, indicated by \1:
sed 's/"\(CDM_\)\w*"/"\1Unusued"/g' File1.txt > File2.txt
Note the backslashes before the parentheses on the left side; these can be omitted if using sed with -r (I think), for extended regexps, but as-is, they're necessary so sed knows they're not literal.
Edit: I've updated the command in response to the accurate comment by Birei, noting that the extraneous escapes for the double-quotes. (Note that the ones for the parentheses are still necessary).
I have a file that looks like this:
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
I want it to look like this
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
I thought I could use sed to do this but I can't figure out how to store something in a buffer and then modify it.
Am I even using the right tool?
Thanks
You don't have to get tricky with regular expressions and replacement strings: use sed's p command to print the line intact, then modify the line and let it print implicitly
sed 'p; s/\.png//'
Glenn jackman's response is OK, but it also doubles the rows which do not match the expression.
This one, instead, doubles only the rows which matched the expression:
sed -n 'p; s/\.png//p'
Here, -n stands for "print nothing unless explicitely printed", and the p in s/\.png//p forces the print if substitution was done, but does not force it otherwise
That is pretty easy to do with sed and you not even need to use the hold space (the sed auxiliary buffer). Given the input file below:
$ cat input
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
you should use this command:
sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
The result:
$ sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
This commands is just a replacement command (s///). It matches anything starting with #" followed by non-period chars ([^.]*) and then by .png",. Also, it matches all non-period chars before .png", using the group brackets \( and \), so we can get what was matched by this group. So, this is the to-be-replaced regular expression:
#"\([^.]*\)\.png",
So follows the replacement part of the command. The & command just inserts everything that was matched by #"\([^.]*\)\.png", in the changed content. If it was the only element of the replacement part, nothing would be changed in the output. However, following the & there is a newline character - represented by the backslash \ followed by an actual newline - and in the new line we add the #" string followed by the content of the first group (\1) and then the string ",.
This is just a brief explanation of the command. Hope this helps. Also, note that you can use the \n string to represent newlines in some versions of sed (such as GNU sed). It would render a more concise and readable command:
sed 's/#"\([^.]*\)\.png",/&\n#"\1",/' input
I prefer this over Carles Sala and Glenn Jackman's:
sed '/.png/p;s/.png//'
Could just say it's personal preference.
or one can combine both versions and apply the duplication only on lines matching the required pattern
sed -e '/^#".*\.png",/{p;s/\.png//;}' input