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up vote 1 down vote favorite

I want to get a string between two special characters. For example,

/home/oracle/ggs/text.ext

I want oracle/ggs in the above string.

How can I do this using bash?

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echo oracle/ggs will answer your question. Please clarify what the requirement are. Is that characters 7 to 16? From the 2nd to 4th /? –  Stéphane Chazelas Jun 11 '14 at 20:19

3 Answers 3

up vote 3 down vote 
$ expr /home/oracle/ggs/text.ext : '/home/\(.*\)/'
oracle/ggs

That returns the part after /home/ and before the last /.

Other possible approaches that will fulfil your requirements:

$ echo /home/oracle/ggs/text.ext | cut -d/ -f3,4
oracle/ggs
$ echo /home/oracle/ggs/text.ext | cut -c 7-16
oracle/ggs
$ echo oracle/ggs
oracle/ggs
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I got what I need thanks. >mysql_str=/home/oracle/ext/ggs/text.ext > path_str=echo ${mysql_str}| cut -f 2 -d '/' > expr ${mysql_str} : '/'${path_str}'/(.*)/' oracle/ext/ggs > echo $path_str home –  BOB Jun 11 '14 at 20:38
up vote 2 down vote

For this particular example, you can use awk:

echo "$string" | awk -F '/' '{printf "%s%s%s\n" ,$2,FS,$3}'

The same in Perl:

echo "$string" | perl -lanF '/' -E '$,="/";say @F[1,2]'
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oracle@ipt:> echo $x | awk -F '/' '{printf "%s%s%s\n" ,$3,FS,$4}' oracle/ggs I want to get the $4 dynamically any suggestions please? –  BOB Jun 11 '14 at 20:09
    
@StéphaneChazelas Can you please explain why the quotes are needed here? Won't echo be happy with any number of arguments? –  Joseph R. Jun 11 '14 at 20:23
1  
Leaving a variable unquoted in list context is the split+glob operator (the variable is split into words and each word is subject to globbing), that's independent of echo. Try with string='* *' for instance. Or with string=/home/oracle/ggs/text.ext where IFS is /. –  Stéphane Chazelas Jun 11 '14 at 20:25
up vote 0 down vote

Bash

IFS=/ read -ra a <<<'/home/oracle/ggs/text.ext' && 
(IFS=/; printf '%s\n' "${a[*]:2:2}") 
oracle/ggs
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