Shell Programming and Scripting

Hi

Can somebody please help me know how do i match the basename using a regular expression using posix standard in shell script

suppose i want to match

/u01/Sybase/data/master.dbf the result should be master.dbf as i want to match everything after the last /

regards

Code:

 echo /u01/Sybase/data/master.dbf | grep -o '[^/]*$'

If u make a little search, you could have found the answer yourself...

Code:

 
>str="/u01/Sybase/data/master.dbf "
>echo ${str##*/}
>master.dbf

>str="/u01/Sybase/data/master.dbf"
>echo $str
/u01/Sybase/data/master.dbf
>basename $str
master.dbf

Hi

Thanks to all responded and the responses are also excellent at this point of time i am very keen on the regex solution

Code:

 echo /u01/Sybase/data/master.dbf | grep -o '[^/]*$'

due to my inquisitiveness in regex what does this regex do

Code:

grep -o '[^/]*$'

I know individually
^ matches begining of the line
$ matches end of the line
* matches anything
/ is a escape sequence

i dont understand what a [] does and what does the combined expression all put together mean including the greps -o switch

I'll try to explain it:

Code:

[^/]*$

[] is called a bracket expression (see man grep).

^ as first character within the bracket expression means: don't match characters within the brackets

[^/] means select all characters except a slash

*$ means match the characters untill the last character

Thus the whole regex means: select all the characters till the end without a slash.


Regards

Hi Franklin

Thank you for the excellent explanation.

However what does the -o do it doesnt sem to work on AIX

when i do

Code:

echo /u01/Sybase/data/master.dbf | grep '[^/]'

shoudlnt it return everything minus the / slash in my case it returns

Code:

u01 Sybase data master.dbf

regards