Extract string from brackets - bash

Question

I'm pretty new at bash so this is a pretty noob question..
Suppose I have a string:
string1 [string2] string3 string4
I would like to extract string2 from the square brackets; but the brackets may be surrounding any other string at any other time.
How would I use sed, etc, to do this? Thanks!

Answer 1

Try this:
echo $str | cut -d "[" -f2 | cut -d "]" -f1

Answer 2

Here's one way using awk:
echo "string1 [string2] string3 string4" | awk -F'[][]' '{print $2}'
This sed option also works:
echo "string1 [string2] string3 string4" | sed 's/.*\[\([^]]*\)\].*/\1/g'
Here's a breakdown of the sed command:
s/ <-- this means it should perform a substitution
.* <-- this means match zero or more characters
\[ <-- this means match a literal [ character
\( <-- this starts saving the pattern for later use
[^]]* <-- this means match any character that is not a [ character
the outer [ and ] signify that this is a character class
having the ^ character as the first character in the class means "not"
\) <-- this closes the saving of the pattern match for later use
\] <-- this means match a literal ] character
.* <-- this means match zero or more characters
/\1 <-- this means replace everything matched with the first saved pattern
(the match between "\(" and "\)" )
/g <-- this means the substitution is global (all occurrences on the line)

Answer 3

In pure bash:
STR="string1 [string2] string3 string4"
STR=${STR#*[}
STR=${STR%]*}
echo $STR

Answer 4

Specify awk multiple delimiters with -F '[delimiters]'
If the delimiters are square brackets, put them back to back like this ][
awk -F '[][]' '{print $2}'
otherwise you will have to escape them
awk -F '[\\[\\]]' '{print $2}'
Other examples to get the value between the brackets:
echo "string1 (string2) string3" | awk -F '[()]' '{print $2}'
echo "string1 {string2} string3" | awk -F '[{}]' '{print $2}'

Answer 5

Here's another one , but it takes care of multiple occurrences, eg
$ echo "string1 [string2] string3 [string4 string5]" | awk -vRS="]" -vFS="[" '{print $2}'
string2
string4 string5
The simple logic is this, you split on "]" and go through the split words finding a "[", then split on "[" to get the first field. In Python
for item in "string1 [string2] string3 [string4 string5]".split("]"):
if "[" in item:
print item.split("]")[-1]

Answer 6

Here is an awk example, but I'm matching on parenthesis which also makes it more obvious of how the -F works.
echo 'test (lskdjf)' | awk -F'[()]' '{print $2}'

Answer 7

Another awk:
$ echo "string1 [string2] string3 [string4]" |
awk -v RS=[ -v FS=] 'NR>1{print $1}'
string2
string4

Answer 8

Read file in which the delimiter is square brackets:
$ cat file
123;abc[202];124
125;abc[203];124
127;abc[204];124
To print the value present within the brackets:
$ awk -F '[][]' '{print $2}' file
202
203
204
At the first sight, the delimiter used in the above command might be confusing. Its simple. 2 delimiters are to be used in this case: One is [ and the other is ]. Since the delimiters itself is square brackets which is to be placed within the square brackets, it looks tricky at the first instance.
Note: If square brackets are delimiters, it should be put in this way only, meaning first ] followed by [. Using the delimiter like -F '[[]]' will give a different interpretation altogether.
Refer this link: http://www.theunixschool.com/2012/07/awk-10-examples-to-read-files-with.html

Answer 9

Inline solution could be:
a="first \"Foo1\" and second \"Foo2\""
echo ${a#*\"} | { read b; echo ${b%%\"*}; }
You can test in single line:
a="first \"Foo1\" and second \"Foo2\""; echo ${a#*\"} | { read b; echo ${b%%\"*}; }
Output: Foo1
Example with brackets:
a="first [Foo1] and second [Foo2]"
echo ${a#*[} | { read b; echo ${b%%]*}; }
That in one line:
a="first [Foo1] and second [Foo2]"; echo ${a#*[} | { read b; echo ${b%%]*}; }
Output: Foo1