AWK Print Line If Specific Character Is Matched

Shell Programming and Scripting
1

Hello,

I have a file as such:

FFFFFFF6C000000     225280     225240          -          - rwxs-    [ shmid=0x7000181 ]
FFFFFFFF79C00000       3240       3240          -          - rwxs-    [ shmid=0x7000181 ]
FFFFFFFF7A000000       4096       4096          -          - rwxs-    [ shmid=0x7000186 ]
FFFFFFFF7A400000         64         64          -          - rwxs-    [ shmid=0x7000186 ]
FFFFFFFF7A700000        288        248          -          - r-x--  libresolv.so.2
FFFFFFFF7A848000         24         24          -          - rwx--  libresolv.so.2
FFFFFFFF7A84E000          8          8          -          - rwx--  libresolv.so.2
FFFFFFFF7A900000         16         16          -          - r-x--  nss_dns.so.1

How do I get awk to only print lines which have the 's' character in position 65. (field 6)

Thanks,

PW

2 
awk '$6 ~ "s"' file
3

Thanks Franklin52, that didn't work, but it did get me in the right direction. I'm using:

awk '$6 ~ /...s./' file

which seems to do what I need.

Thanks.

PW

4

awk '{if (substr($0,65,1)=="s")print $0}' filename