Send command output in shell expect

Question

Following is the code I'm trying to execute with expect in shell script.

send "date +%d%m%y \n"  
set date1 "$expect_out(buffer)"  
puts "$date1"  
expect "#"

The output I'm getting is

date +%d%m%y  
111217

Is there any way to suppress the command(date +%d%m%y) and only store the output in variale date1?

yes it's obvious it's expect code. but there's no `expect` statement so i'm wondering how did the OP get the `$expect_out(buffer)`? — , Dec 11 '17 at 14:31
Possible duplicate of https://stackoverflow.com/questions/26803188/how-to-suppress-expect-send-output — Dinesh, Dec 12 '17 at 04:52

glenn jackman · Answer 1 · 2017-12-12T10:44:59.063

Try this: expect to see 6 digits as a whole word

send "date +%d%m%y\r"         ;# typically we use \r to "hit enter"
expect -re {\m(\d{6})\M}
set date1 "$expect_out(1,string)"  
puts "$date1"  
expect "#"

More generally, to select the text between the first line (the command) and the last line (the prompt), you can do:

lassign [regexp -inline {.*?\n(.*)\r\n#} $expect_out(buffer)] _ date1

where _ is a throw-away variable that will hold the entire matching text of the regex, and date1 will hold only the captured text.

See https://tcl.tk/man/tcl8.6/TclCmd/re_syntax.htm for Tcl regex syntax.

Without lassign, you could write

regexp {.*?\n(.*)\r\n#} $expect_out(buffer) _ date1

lassign is not working in my system. Is there any other way to get only the output of the command? — Kavya Jain, Dec 12 '17 at 06:57

Send command output in shell expect

1 Answers1