Sed find and replace

Question

I need to amended the following text in a few hundred documents ...

{% url project.app.views.chart %}

so it reads

{% url "project.app.views.chart" %}

Do you know the next way to do this based on the text between the quotes being different each time ?

Thanks,

Answer 1

Solution

This works as soons as you don't have any % char in your input strings:

sed -r 's/url ([^%]+) %/url "\1" %/g'

Test

$ echo '{% url test %}{% url test 2 %}' | sed -r 's/url ([^%]+) %/url "\1" %/g'
{% url "test" %}{% url "test 2" %}

Explanations

Let's explain this answer. The -r option is here because extended regexes are the one I prefer to use.

Now, the substitution part.

sed 's/pattern/replacement/g'

Is used to replace text, matching pattern as a regular expression and replacing it by replacement .

The g is a substitution option that means you want to replace the pattern as many times as it is present in the text.

So here the pattern is url ([^%]+) % . url and % part are pretty logical and easy to get (if not, tell me). The ([^%]+) means "capture everything that isn't a % and that is composed by at least a character" . The parenthesis allow to save this part, to reuse it after with \\x where x is the number of the parenthesis group (Here it will be \\1 ).

So when it receives {% url test %} , the pattern matches url test % where test is saved under \\1 , and it rewrites it with url "\\1" % so url "test" % . The parts before and after the pattern aren't modified, so it will output {% url "test" %}

Other possibilities

To avoid adding another set of quotes if there are already there, just add [^"] after url that way : url [^"] .

Here is the full command :

sed -r 's/url ([^"][^%]+) %/url "\1" %/g'

Answer 2

Assuming you want to wrap string between url and %} .

Something like this may work:

sed -e 's/url /url "/' -e 's/ %}/" %}/'