bash script

[munna_dude]

#!/bin/bashk=" "
i=`rpm -ql mozilla | grep /usr/lib/mozilla- | tail -1 | cut -d/ -f4`
echo $i
path=`whereis $i |cut -d" " -f2`
pathname=$(path)
echo $pathname

how to use out put of "i" in "path" as input.
i wanna out put like this

for echo $i
mozilla-1.7.13

and
for echo $pathname
/usr/lib/mozilla-1.7.13

how to get this..
please help me

thank you in advance

[Franklin52]

Hi,

There are some possibilities, try it first with bash, you must strip the longest match of the substring from the front of the string.
Check this page of the abs guide how to extract substrings and help yourself :

http://tldp.org/LDP/abs/html/string-manipulation.html

Regards

[munna_dude]

Hi,

There are some possibilities, try it first with bash, you must strip the longest match of the substring from the front of the string.
Check this page of the abs guide how to extract substrings and help yourself :

http://tldp.org/LDP/abs/html/string-manipulation.html

Regards

i trie this in make file

k=" "
k=`mozilla -V | sed -e 's/\([^,]*\),.*/\L\1/' -e 's/ /-/'`
path=`whereis $k |cut -d" " -f2`
mozlib := $(shell $(path))

am including the lib path like this

gcc -g -O2 -o phone main.o support.o interface.o callbacks.o -L/root/.fedora -L${mozlib}

after making the terminal showing like this

gcc -g -O2 -o phone main.o support.o interface.o callbacks.o -L/root/.fedora -Lmozilla-1.7.13

here "path" contains the total path of the LIBdirectory...i.e
/usr/lib/mozilla-1.7.12
but am getting only
mozilla-1.7.12 in the make file

what i have to do to get /usr/lib/mozilla-1.7.12

thank you in advance