search and replace substring in string in bash

Question

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><a href="MeinKonto.do">Einstellungen verwalten</a></li>

to

<li><a href="<s:url action='MeinKonto'/>">Einstellungen verwalten</a></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.

Are your links always surrounded by double quotes?

Floris
– Floris

2014-06-10 23:33:19 +00:00
Commented Jun 10, 2014 at 23:33 — Floris
– Floris, Commented Jun 10, 2014 at 23:33

Nancy · Accepted Answer · 2014-06-11 03:51:11Z

1

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.

edited Jun 11, 2014 at 3:51

answered Jun 11, 2014 at 3:45

Nancy

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5 Comments

Qeole Over a year ago

If you have several links on a same line, and if one of it, other than the first one, is a .do to modify, your regex will delete links (sed is greedy, and your parenthesis catch as much as possible). No offense, but what's wrong with my solution, that made you post nearly the same?

Nancy Over a year ago

@Qeole Have you check your command? It just missed some double quotes or single quotes. And I think here you have \1 for href=" is a little redundant.

Qeole Over a year ago

The href=" thing does not not look that redundant too me, but OK. For the quotes, yes I checked my command, and for me with OP's example, it provides desired output. Could you please provide more details? Did you tried your command with input as described in my first comment?

Nancy Over a year ago

@Qeole yes, you're right in the first comment, I should add a [^\"] to avoid it. And for your command I just met error as Unmatched ".

Qeole Over a year ago

I didn't meet this error. If you still have the exact command you typed, or if you (or anyone else) ever reproduces it, I'm interested.

Qeole · Accepted Answer · 2014-06-10 23:21:11Z

1

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 /.

answered Jun 10, 2014 at 23:21

Qeole

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