Hi to all,

i've never programmed a shell skriptbefore and i need your help please!

i have to search the first occurence of a string in a file -> then read the next 4 characters!

fileexample:

lsap state: LSAP_DISCONNECTES, slsap_sel: 0x10, dlsap_sel: 0xff, (IrCOMM)
lsap state: LSAP_DISCONNECTES, slsap_sel: 0x0, dlsap_sel: 0xff, (IrIAS srv)


what i need ist the 4 characters after the first occurence of 'slsap_sel: ' which is 0x10


I know that that's no problem for you!

Thanks for any help
tina

This should do it...

awk -F"," ' # -F flag sets the field delimiter to a ","
$2~/slsap/ # If the 2nd field matches "sslap"...
{
# print the the substring of 2nd field,
# starting from 2 characters positions after the ":" in the second field
print substr($2,index($2,": ")+2);
# once you've found a match, you don't need to process anymore
exit }' yourFile

Without comments...

awk -F"," '$2~/slsap/ {print substr($2,index($2,": ")+2); exit}' yourFile

HTH

Hi Damian,

thanks for your answer!
i already got it, and my script is very close to yours ;-)

awk '/slsap_sel: /{
if ( substr($0,index($0,"slsap_sel: ")+11,4 )=="0x6a" )
... do somethin ...
exit
}' irlmp

bye tina