clear stdin before place "read a"

[Ekkard Gerlach]

I want to clear stdin before 'read a':

#!/bin/bash
#
<.... my script ... (works for about 5 minutes!)>
echo "this is the result: "
[...]
echo "to go on press ENTER"
# here I want to clear stdin
read a
clear # clear the screen
[...]

because the user perhaps unfortunately has pressed some keys and when
read a is promted then a ist filled with the chars of these keys. So
the user doesn't see the result!

How can I clear stdin?
thx
Ekkard

[Glenn Jackman]

Perhaps: cat - >/dev/null

--
Glenn Jackman
Write a wise saying and your name will live forever. -- Anonymous

[pk]

On Wednesday 20 May 2009 21:52, Glenn Jackman wrote:

> At 2009-05-20 02:04PM, "Ekkard Gerlach" wrote:
>> I want to clear stdin before 'read a':
>>
>> #!/bin/bash
>> #
>> <.... my script ... (works for about 5 minutes!)>
>> echo "this is the result: "
>> [...]
>> echo "to go on press ENTER"
>> # here I want to clear stdin
>> read a
>> clear # clear the screen
>> [...]
>>
>> because the user perhaps unfortunately has pressed some keys and when
>> read a is promted then a ist filled with the chars of these keys. So
>> the user doesn't see the result!
>>
>> How can I clear stdin?
>
> Perhaps: cat - >/dev/null

But that will block if there's nothing in the stdin buffer. I can't think of
a way to do that in shell (which doesn't mean there isn't one of course); a
terrible hack could be using something like

read -t 1 -n <bignumber> dummyvar

before doing the actual read of user input, but it looks like a very poor
solution to me (and of course read must support -t; bash builtin read
does).

[Ekkard Gerlach]

pk schrieb:

> read -t 1 -n <bignumber> dummyvar
>

I'm using bash.

I tried:
#!/bin/bash
#
sleep 5
read -t 1 -n 323423423 dummyvar
read a
echo $a ready

the hack it works! Last input chars are displayed, but that doesn't
matter! The returns during sleep 5 are thrown away.

thx
Ekkard

[mop2]

If your buffer content always end with a newline ($'\n') you can try:
while read;do :;done

If not:
while read -n1;do :;done

[Ekkard Gerlach]

mop2 schrieb:

>
> If your buffer content always end with a newline ($'\n') you can try:
> while read;do :;done
>
> If not:
> while read -n1;do :;done

sorry, doesn't work in bash. In both examples "read" rests in a endless
loop.

Ekkard

[mop2]

I apologize, obviously "read" will wait forever in stdin (my mistake).
As already said by pk you can use "-t" for read's timeout. I think the
value 1 ("read" limit) is inconveniently high for that purpose, but
new version of bash seems accept lower values:
while read -n1 -t.1 a;do :;done

Bash here is:
$ bash --version|head -n1
GNU bash, version 4.0.24(1)-release (i686-pc-linux-gnu)

This new bash's resource will be very useful to me. :)

[mop2]

Ekkard Gerlach <ja...@aiai.de> wrote:

> the hack it works! Last input chars are displayed, but that doesn't
> matter! The returns during sleep 5 are thrown away.

Teste the key '-s' of the "read" while cleaning stdin buffer.

[pk]

On Friday 22 May 2009 12:34, mop2 wrote:

> As already said by pk you can use "-t" for read's timeout. I think the
> value 1 ("read" limit) is inconveniently high for that purpose, but
> new version of bash seems accept lower values:
> while read -n1 -t.1 a;do :;done
>
> Bash here is:
> $ bash --version|head -n1
> GNU bash, version 4.0.24(1)-release (i686-pc-linux-gnu)
>
> This new bash's resource will be very useful to me. :)

Ah thanks for the information, I tried with bash 3.2 which accepts only
integer values for -t. Then yes, 0.1 would be better, although that doesn't
make the solution less kludgy :)

[mop2]

Hi pk,
to me your solution seems very good, and in a simple test here, the loop
"while" ( suggested by me) seems not needed.
Then, your solution (using a very low timeout) seems very good to me,
perhaps better than a specif command (one more) in bash only for this purpose.

As reference to others without bash4, behaviour of some versions of bash
that I have here, for the parameter "-t":

PROMPT$ f(){ date +%s.%N;}
PROMPT$ export -f f

PROMPT$ $0 --version|head -n1

GNU bash, version 4.0.24(1)-release (i686-pc-linux-gnu)

PROMPT$ f;read -t1;f
1242991418.271353662
1242991419.276413729
PROMPT$ f;read -t.1;f
1242991423.486857764
1242991423.592517631
PROMPT$ f;read -t.01;f
1242991426.867548100
1242991426.880439021
PROMPT$ f;read -t.001;f
1242991429.878183736
1242991429.884461932

PROMPT$ ls -ldtr /bin/bash*
-rwxr-xr-x 1 root root 679620 2006-05-13 20:55 /bin/bash31*
-rwxr-xr-x 1 root root 674752 2008-01-13 00:59 /bin/bash32*
-rwxr-xr-x 1 root root 676288 2008-05-10 12:28 /bin/bash3239*
lrwxrwxrwx 1 root root 8 2008-05-10 15:32 /bin/bashold -> bash3239*
-rwxr-xr-x 1 root root 674720 2008-11-27 04:55 /bin/bash3248*
-rwxr-xr-x 1 root root 748228 2009-03-13 04:57 /bin/bash4010*
-rwxr-xr-x 1 root root 774180 2009-04-13 21:52 /bin/bash4017*
-r-xr-xr-x 1 root root 6778 2009-05-17 02:07 /bin/bashbug*
-rwxr-xr-x 1 root root 774564 2009-05-17 02:07 /bin/bash4024*
lrwxrwxrwx 1 root root 8 2009-05-17 02:14 /bin/bash -> bash4024*
lrwxrwxrwx 1 root root 8 2009-05-17 02:18 /bin/bash.17 -> bash4017*

PROMPT$ bash3239
PROMPT$ $0 --version|head -n1
GNU bash, version 3.2.39(1)-release (i686-pc-linux-gnu)
PROMPT$ f;read -t.1;f
1242991466.067515682
bash3239: read: .1: invalid timeout specification
1242991466.070565025

PROMPT$ bash4010
PROMPT$ $0 --version|head -n1
GNU bash, version 4.0.10(1)-release (i686-pc-linux-gnu)
PROMPT$ f;read -t.1;f
1242991526.170081626
1242991526.273839251
PROMPT$

[ysxiao]

read function with '-t' has some problems in lower bash, I spent a long time to find a solution. Here is the one, call this before question user please try it on:

fflush_stdin() {
local dummy=""
local originalSettings=""
originalSettings=`stty -g`
stty -icanon min 0 time 0
while read dummy ; do : ; done
stty "$originalSettings"
}

--http://compgroups.net/comp.unix.shell/clear-stdin-before-place-read-a/507380

[Rajib Bandopadhyay]

Dear friends,
Sorry to intrude. I am new to scripting, directed from Advanced Bash-Scripting Guide.html#SHA-BANG by Mendel Cooper. Would you please tell me the role of the code /dev/null? YOu could also email your reply to my gmailbox, if you desire privacy.
Regards,
Rajib Bandopadhyay

[Kees Nuyt]

On Sun, 26 Aug 2012 04:52:34 -0700 (PDT), Rajib Bandopadhyay <bkpsu...@gmail.com> wrote:

>On Wednesday, May 20, 2009 11:34:23 PM UTC+5:30, Ekkard Gerlach wrote:
>> I want to clear stdin before 'read a':
>>

[snipped]

>> Ekkard
>
> I am very sorry to intrude, Ekkard. I am new to
> this user group, directed from the advanced
> Bash-scripting guide by Mendel Cooper. I am
> new to scripting, interested in learning the
> nitty-gritty.
> I was stuck at a script which uses /dev/null .
> What is the role of /dev/null?

http://en.wikipedia.org/wiki//dev/null

> I am sure you could help me, since you have used it here.
> ===========================
> The script is as follows:
> cleanup: A script to clean up log files in /var/log
> #!/bin/bash
> # Cleanup
> # Run as root, of course.
>
> cd /var/log
> cat /dev/null > messages
> cat /dev/null > wtmp
> echo "Log files cleaned up."
> ===========================

In this case, /dev/null is used as input for cat.
It yields an EOF (end of file) immediately, so the
output files (messages and wtmp) will be emptied.

> And BTW, does this group have rules regarding posting?
> Please guide me on the group rules.

Well, one rule is not to hijack a thread, that is to follow-up on a message about some other subject. Instead, post to a new thread.

Also, don't ask the same question repeatedly.

Your favourite search engine will help you find general rules for usenet, search for "usenet posting etiquette".

For example:

http://tgos.org/newbie/rules.html


Enjoy.

Best regards,
--
( Kees Nuyt
)
c[_]