How to make $! to work inside `sh -c "$!"`

Question

Please help to see if there is someway to make below scrip workable?

name='bob'
# below script reports 'bash: !: event not found' error
sh -c "echo $name; sleep 20 & pid1=$!; sleep 10 & pid2=$!; echo \"pid1: $pid1, pid2: $pid2\"" 
# below script $name will not become bob
sh -c 'echo $name; sleep 20 & pid1=$!; sleep 10 & pid2=$!; echo \"pid1: $pid1, pid2: $pid2\"'

[Add 01/03]

This question is somewhat duplicate of this one, it's my fault that the description of the original question is not clear and accurate enough, and I try to create a new one to make it more specific and clear.

try `sh -c 'echo '"$name"'; sleep 20 & pid1=$!; sleep 10 & pid2=$!; echo \"pid1: $pid1, pid2: $pid2\"'` — tomix86, Jan 02 '18 at 13:44
@hek2mgl Then `$name` won't be expanded (unless you separately `export` it). — tripleee, Jan 02 '18 at 13:46
[*Like I already told you,*](https://stackoverflow.com/a/47900893/874188) use `set +H` to prevent Bash from producing the `event not found` error. — tripleee, Jan 02 '18 at 13:47
@Cyrus But `sh` does not have this issue. The Csh-style history expansion (mis)feature is a Bash extension. — tripleee, Jan 02 '18 at 13:51
Please post a [mcve]. At least the sleep calls are irrelevant to this question and make it more difficult to read. — Melebius, Jan 02 '18 at 13:59
One background process of some sort is probably required for a MCVE since the code needs to want to use `$!` but I'm really, really hoping the OP will simply accept the proposed duplicate. — tripleee, Jan 02 '18 at 14:13

tripleee · Accepted Answer · 2018-01-03T04:27:03.770

1

You seem to be trying to solve the event not found problem.

It happens when you have set -H in Bash and use an exclamation mark in an interactive command (outside of single quotes or a backslash escape).

I have posted an answer and I think half a dozen comments explaining what is happening and how to fix it.

set +H

This command disables the Bash history mechanism which uses the exclamation mark in interactive sessions.

In addition, within double quotes, you have to escape the dollar sign to pass it verbatim to the subshell.

sh -c "echo \"$name\"
    sleep 20 & pid1=\$!
    sleep 10 & pid2=\$!
    echo \"pid1: \$pid1, pid2: \$pid2\""

edited Jan 03 '18 at 04:27

answered Jan 02 '18 at 13:54

tripleee

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This isn't a proper answer but I'm hoping at least this will prompt you to clarify if this precise problem is not the one you are still trying to solve. Otherwise please accept the duplicate I proposed. – tripleee Jan 02 '18 at 13:55
I am afraid maybe I didn't catch your point, actually I want to store `$!` as PID number to a variable and use it later, the `event not found` is a side-error I encountered for my script: `sh -c "echo $name; sleep 20 & pid1=$!; sleep 10 & pid2=$!; echo \"pid1: $pid1, pid2: $pid2\""`, and if I use `set +H; sh -c "echo $name; sleep 20 & pid1=$!; sleep 10 & pid2=$!; echo \"pid1: $pid1, pid2: $pid2\""`, there is no `event not found` error, but `$!` doesn't convert to PID number what I want to get. – Y.Huang Jan 03 '18 at 02:27
Uplated answer. Though this is basically a duplicate of https://stackoverflow.com/a/37746186/874188 – tripleee Jan 03 '18 at 04:32

William Pursell · Answer 2 · 2018-01-02T18:26:23.200

1

The comments propose mixing quotes and disabling history expansion, but a better solution (IMO) is to pass the name as a parameter:

sh -c 'echo "$1"; sleep 20 & pid1=$!; ...' sh "$name"

edited Jan 02 '18 at 18:26

answered Jan 02 '18 at 14:11

William Pursell

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Why the second `sh`? Just to make the parameter be `$1` instead of `$0`? – Melebius Jan 02 '18 at 14:23
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@Melebius Yes. Custom dictates that `$0` should be the name of the shell. It's not necessary, but is best practice. – William Pursell Jan 02 '18 at 14:25

How to make $! to work inside `sh -c "$!"`

2 Answers2