How to pull out specific part of a variable and set to another variable?

Ruby
1

I have a variable that is a long string. The variable looks like this:

“this sentence is my variable and the important information is {inside
these brackets}”

The information I need is the information inside the brackets including
the brackets themselves, as the brackets are required for when I need to
reuse this information. How do I pull out just that piece? It will
ALWAYS be in brackets and be the only brackets in that variable.

2

On Mon, Apr 9, 2012 at 7:28 PM, Charlie B. [email protected] wrote:

I have a variable that is a long string. The variable looks like this:

“this sentence is my variable and the important information is {inside
these brackets}”

The information I need is the information inside the brackets including
the brackets themselves, as the brackets are required for when I need to
reuse this information. How do I pull out just that piece? It will
ALWAYS be in brackets and be the only brackets in that variable.

You can try the String#[] method and a regular expression to look for
the brackets and what’s inside. For example:

s = "asdlkmlsdmflkmflk elfm lwemf le {asdfsdf sdfdsf} asdf df "
=> "asdlkmlsdmflkmflk elfm lwemf le {asdfsdf sdfdsf} asdf df "

s[/{[^}]+}/]
=> “{asdfsdf sdfdsf}”

Jesus.

3

Here’s a possible approach

If there should be just one match:

pry(main)> var = “this sentence is my variable and the important
information is {inside these brackets}”
=> “this sentence is my variable and the important information is
{inside
these brackets}”
pry(main)> new_var = var.match(/({.})/)[1]*
=> “{inside these brackets}”

If there could be more than one match:

pry(main)> new_var2 = var.match(/({.})/).captures*
=> ["{inside these brackets}"]
pry(main)>

new_var has only a match, while new_var2 has an array of all the
occurrencies

Hope it helped

Federico M. Iachetti

4

Wow … nice one
Federico M. Iachetti

2012/4/9 Jess Gabriel y Galn [email protected]

5

2012/4/9 Iachetti Federico M. [email protected]:

If there could be more than one match:

pry(main)> new_var2 = var.match(/({.*})/).captures
=> [“{inside these brackets}”]
pry(main)>

new_var has only a match, while new_var2 has an array of all the
occurrencies

Careful, the * is greedy, so it will consume as many characters as it
can:

s = “adf {first one} seomthing {second one} dfdsf”
=> “adf {first one} seomthing {second one} dfdsf”
s[/({.*})/]
=> “{first one} seomthing {second one}”

That’s why I used the negative character set, to match everything that
is not a closed bracket. You can also use the ? modifier:

s[/({.*?})/]
=> “{first one}”

Jesus.

6

sorry … my mistake
Federico M. Iachetti

2012/4/9 Iachetti Federico M. [email protected]

7

With match I didn’t have that problem. It just puts every match on an
array

Federico M. Iachetti

2012/4/9 Jess Gabriel y Galn [email protected]

8

The regular expression by Jesus works perfect. I just tested and it
works great. Thanks for the quick turn around!

9

“Iachetti Federico Martín” <iachetti.federico@gmai wrote in post
#1055670:

If there could be more than one match:

pry(main)> new_var2 = var.match(/({.})/).captures*
=> ["{inside these brackets}"]
pry(main)>

This is wrong in several important ways:

var = “abc {foo} def {bar}”
=> “abc {foo} def {bar}”

var.match(/({.})/).captures
SyntaxError: compile error
(irb):2: invalid regular expression; there’s no previous pattern, to
which ‘{’ would define cardinality at 2: /({.})/
from (irb):2

var.match(/({.*})/).captures
=> ["{foo} def {bar}"]

var.match(/({.*?})/).captures
=> ["{foo}"]

What I think you’re looking for is:

var.scan(/{.*?}/)
=> ["{foo}", “{bar}”]