Expect: How to access variables and commands
0
I'm attempting to do a for loop so that i can read each file from the ls output inside my expect script.
/usr/bin/expect << 'EOF'
set timeout -1
spawn ssh root@server
expect {
"*assword: "
}
send "_t00rhs!r"
expect {
"*]# "
}
send 'for item in "$(ls)" do sort -k1 -n ${item}; done r'
expect {
"*]# "
}
The output shows that it hasnt got the value for item.
]0;root@atddpvm13:~[?1034h[root@atddpvm13 ~]# can't read "item": no such
variable
while executing
"send 'for item in "$(ls)" do sort -k1 -n ${item}"
bash expect spawn
asked Nov 20 '18 at 13:17
GillGill
64
add a comment |
0
I'm attempting to do a for loop so that i can read each file from the ls output inside my expect script.
/usr/bin/expect << 'EOF'
set timeout -1
spawn ssh root@server
expect {
"*assword: "
}
send "_t00rhs!r"
expect {
"*]# "
}
send 'for item in "$(ls)" do sort -k1 -n ${item}; done r'
expect {
"*]# "
}
The output shows that it hasnt got the value for item.
]0;root@atddpvm13:~[?1034h[root@atddpvm13 ~]# can't read "item": no such
variable
while executing
"send 'for item in "$(ls)" do sort -k1 -n ${item}"
bash expect spawn
asked Nov 20 '18 at 13:17
GillGill
64
Why did you escape the$before runninglsand not foritem?
– Inian
Nov 20 '18 at 13:20
If i escape both, i get this ]0;root@server:~[?1034h[root@server~]# usage: send [args] string while executing "send 'for item in "$(ls)" do sort -k1 -n ${item}"
– Gill
Nov 20 '18 at 13:29
take a look at sexpect with which you can write Expect scripts with shell code only.
– pynexj
Nov 20 '18 at 14:07
add a comment |
0
0
0
I'm attempting to do a for loop so that i can read each file from the ls output inside my expect script.
/usr/bin/expect << 'EOF'
set timeout -1
spawn ssh root@server
expect {
"*assword: "
}
send "_t00rhs!r"
expect {
"*]# "
}
send 'for item in "$(ls)" do sort -k1 -n ${item}; done r'
expect {
"*]# "
}
The output shows that it hasnt got the value for item.
]0;root@atddpvm13:~[?1034h[root@atddpvm13 ~]# can't read "item": no such
variable
while executing
"send 'for item in "$(ls)" do sort -k1 -n ${item}"
bash expect spawn
asked Nov 20 '18 at 13:17
GillGill
64
I'm attempting to do a for loop so that i can read each file from the ls output inside my expect script.
/usr/bin/expect << 'EOF'
set timeout -1
spawn ssh root@server
expect {
"*assword: "
}
send "_t00rhs!r"
expect {
"*]# "
}
send 'for item in "$(ls)" do sort -k1 -n ${item}; done r'
expect {
"*]# "
}
The output shows that it hasnt got the value for item.
]0;root@atddpvm13:~[?1034h[root@atddpvm13 ~]# can't read "item": no such
variable
while executing
"send 'for item in "$(ls)" do sort -k1 -n ${item}"
bash expect spawn
bash expect spawn
asked Nov 20 '18 at 13:17
GillGill
64
asked Nov 20 '18 at 13:17
GillGill
64
asked Nov 20 '18 at 13:17
GillGill
64
asked Nov 20 '18 at 13:17
GillGill
64
asked Nov 20 '18 at 13:17
GillGill
64
64
Why did you escape the$before runninglsand not foritem?
– Inian
Nov 20 '18 at 13:20
If i escape both, i get this ]0;root@server:~[?1034h[root@server~]# usage: send [args] string while executing "send 'for item in "$(ls)" do sort -k1 -n ${item}"
– Gill
Nov 20 '18 at 13:29
take a look at sexpect with which you can write Expect scripts with shell code only.
– pynexj
Nov 20 '18 at 14:07
add a comment |
Why did you escape the$before runninglsand not foritem?
– Inian
Nov 20 '18 at 13:20
If i escape both, i get this ]0;root@server:~[?1034h[root@server~]# usage: send [args] string while executing "send 'for item in "$(ls)" do sort -k1 -n ${item}"
– Gill
Nov 20 '18 at 13:29
take a look at sexpect with which you can write Expect scripts with shell code only.
– pynexj
Nov 20 '18 at 14:07
Why did you escape the
$ before running ls and not for item?– Inian
Nov 20 '18 at 13:20
Why did you escape the
$ before running ls and not for item?– Inian
Nov 20 '18 at 13:20
If i escape both, i get this ]0;root@server:~[?1034h[root@server~]# usage: send [args] string while executing "send 'for item in "$(ls)" do sort -k1 -n ${item}"
– Gill
Nov 20 '18 at 13:29
If i escape both, i get this ]0;root@server:~[?1034h[root@server~]# usage: send [args] string while executing "send 'for item in "$(ls)" do sort -k1 -n ${item}"
– Gill
Nov 20 '18 at 13:29
take a look at sexpect with which you can write Expect scripts with shell code only.
– pynexj
Nov 20 '18 at 14:07
take a look at sexpect with which you can write Expect scripts with shell code only.
– pynexj
Nov 20 '18 at 14:07
add a comment |
1 Answer
1
active
oldest
votes
0
From the expect side:
- single quotes have no special meaning in expect
- use braces for non-interpolating quoting
- however the
rsequence will not be interpreted in braces
So you get
send {for item in $(ls); do sort -k1 -n "$item"; done}
send "r"
From the shell side
- don't parse
ls
- don't use
forto iterate lines
You want
send "printf '%sn' * | sort -k1 -nr"
# or
send "ls -1vr"
answered Nov 20 '18 at 14:14
glenn jackmanglenn jackman
168k26147239
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
From the expect side:
- single quotes have no special meaning in expect
- use braces for non-interpolating quoting
- however the
rsequence will not be interpreted in braces
So you get
send {for item in $(ls); do sort -k1 -n "$item"; done}
send "r"
From the shell side
- don't parse
ls
- don't use
forto iterate lines
You want
send "printf '%sn' * | sort -k1 -nr"
# or
send "ls -1vr"
answered Nov 20 '18 at 14:14
glenn jackmanglenn jackman
168k26147239
add a comment |
0
From the expect side:
- single quotes have no special meaning in expect
- use braces for non-interpolating quoting
- however the
rsequence will not be interpreted in braces
So you get
send {for item in $(ls); do sort -k1 -n "$item"; done}
send "r"
From the shell side
- don't parse
ls
- don't use
forto iterate lines
You want
send "printf '%sn' * | sort -k1 -nr"
# or
send "ls -1vr"
answered Nov 20 '18 at 14:14
glenn jackmanglenn jackman
168k26147239
add a comment |
0
0
0
From the expect side:
- single quotes have no special meaning in expect
- use braces for non-interpolating quoting
- however the
rsequence will not be interpreted in braces
So you get
send {for item in $(ls); do sort -k1 -n "$item"; done}
send "r"
From the shell side
- don't parse
ls
- don't use
forto iterate lines
You want
send "printf '%sn' * | sort -k1 -nr"
# or
send "ls -1vr"
answered Nov 20 '18 at 14:14
glenn jackmanglenn jackman
168k26147239
From the expect side:
- single quotes have no special meaning in expect
- use braces for non-interpolating quoting
- however the
rsequence will not be interpreted in braces
So you get
send {for item in $(ls); do sort -k1 -n "$item"; done}
send "r"
From the shell side
- don't parse
ls
- don't use
forto iterate lines
You want
send "printf '%sn' * | sort -k1 -nr"
# or
send "ls -1vr"
answered Nov 20 '18 at 14:14
glenn jackmanglenn jackman
168k26147239
answered Nov 20 '18 at 14:14
glenn jackmanglenn jackman
168k26147239
answered Nov 20 '18 at 14:14
glenn jackmanglenn jackman
168k26147239
answered Nov 20 '18 at 14:14
glenn jackmanglenn jackman
168k26147239
168k26147239
add a comment |
add a comment |
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Why did you escape the
$before runninglsand not foritem?– Inian
Nov 20 '18 at 13:20
If i escape both, i get this ]0;root@server:~[?1034h[root@server~]# usage: send [args] string while executing "send 'for item in "$(ls)" do sort -k1 -n ${item}"
– Gill
Nov 20 '18 at 13:29
take a look at sexpect with which you can write Expect scripts with shell code only.
– pynexj
Nov 20 '18 at 14:07