Bash - integer comparison
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From: "Phil B" phil.remove.brady on 15 Jan 2010 15:32 "Bit Twister" <BitTwister(a)mouse-potato.com> wrote in message news:slrnhl1e3u.fjg.BitTwister(a)wb.home.test... > On Fri, 15 Jan 2010 17:51:43 -0000, Phil B wrote: >> Would someone help this newbie out please? I'm trying to detect 'after >> 6pm' but it is failing the if condition. >> In it's simplest form it is: >> >> #!/bin/bash >> hour=$(date +%H) >> if [ "$hour" -gt 18 ] ; then >> echo it is evening >> fi >> >> I think that $hour is treated as a string - how should I code this? > > It works without or without quotes for me. I cut/pasted your script > for testing from the command line and changed 18 to 11 for the test. > > > $ echo $SHELL > /bin/bash > > $ date > Fri Jan 15 12:41:27 CST 2010 > > $ hour=$(date +%H) > > $ if [ "$hour" -gt 11 ] ; then echo it is evening; fi > it is evening > > $ if [ $hour -gt 11 ] ; then echo it is evening; fi > it is evening > > > Feel free to use the set command to help debug your script. Example: > > #!/bin/bash > > set -x > > hour=$(date +%H) > if [ "$hour" -gt 18 ] ; then > echo it is evening > fi > > set - > #************** end test script **************** > > Other set args of interest -v -u > > Some light reading found here > http://tldp.org/LDP/abs/html/index.html > http://mywiki.wooledge.org/BashFAQ/050 > http://cfaj.freeshell.org/shell Thanks Bit Twister, You're quite right and thanks for restoring sanity. It was of course too late in the night when I was trying a more complex script - not thinking straight. thanks again Phil
From: Jasen Betts on 16 Jan 2010 04:25 On 2010-01-15, Phil B <phil.remove.brady(a)hotmail> wrote: > Would someone help this newbie out please? I'm trying to detect 'after > 6pm' but it is failing the if condition. > In it's simplest form it is: > > #!/bin/bash > hour=$(date +%H) > if [ "$hour" -gt 18 ] ; then > echo it is evening > fi > > I think that $hour is treated as a string - how should I code this? I would do it using an arithmetic expression. Not that there's anything wrong with using [ (or test) for this check, it's just that as a C programmer I find the arithmetic syntax more "natural", and the operators easier to remeber. #!/bin/bash hour=$(date +%H) if (( hour > 18 )) then echo it is evening fi --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: Martin on 16 Jan 2010 08:13 "Phil B" <phil.remove.brady(a)hotmail dot co dot united kingdom> wrote: > Would someone help this newbie out please? I'm trying to detect 'after > 6pm' but it is failing the if condition. > In it's simplest form it is: > > #!/bin/bash > hour=$(date +%H) > if [ "$hour" -gt 18 ] ; then > echo it is evening > fi > > I think that $hour is treated as a string - how should I code this? > > Regards > Phil This may not apply to your application, but when doing more sophisticated stuff (particularly with strings) it sometimes pays using Perl as scripting engine (from the point of view of performance and development efficiency). Your script would look like this: #!/usr/bin/perl my $hour = (localtime)[2]; if ($hour > 18) { printf("It is evening\n"); } Martin
From: Florian Diesch on 16 Jan 2010 09:20
"Phil B" <phil.remove.brady(a)hotmail dot co dot united kingdom> writes: > Would someone help this newbie out please? I'm trying to detect 'after > 6pm' but it is failing the if condition. > In it's simplest form it is: > > #!/bin/bash > hour=$(date +%H) > if [ "$hour" -gt 18 ] ; then > echo it is evening > fi Works for me - but of course it means "at least 7pm". Florian -- <http://www.florian-diesch.de/> |