Bash - integer comparison
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From: "Phil B" phil.remove.brady on 15 Jan 2010 12:51 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 $hour is treated as a string - how should I code this? Regards Phil
From: Lew Pitcher on 15 Jan 2010 13:40 On January 15, 2010 12:51, in alt.os.linux, phil.remove.brady(a)hotmail dot co dot united kingdom wrote: > Would someone help this newbie out please? I'm trying to detect 'after > 6pm' but it is failing the if condition. > In it's simplest form it is: > > #!/bin/bash > hour=$(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? In this case, $hour (which /is/ a string) is re-evaluated as an integer, and compared to (the integer) 18 First off, remember that Bash doesn't have a builtin [ test operator; the [ in your if statement is actually /bin/[, which is link to /bin/test. The builtin test operator in bash is [[ (with it's corresponding ]] terminator). So, your Bash if statement is actually running /bin/[ (/bin/test) to evaluate the test condition. The manpage for /bin/[ and /bin/test says: INTEGER1 -gt INTEGER2 INTEGER1 is greater than INTEGER2 and that matches what you've coded. -- Lew Pitcher Master Codewright & JOAT-in-training | Registered Linux User #112576 Me: http://pitcher.digitalfreehold.ca/ | Just Linux: http://justlinux.ca/ ---------- Slackware - Because I know what I'm doing. ------
From: Bit Twister on 15 Jan 2010 13:48 On Fri, 15 Jan 2010 17:51:43 -0000, Phil B wrote: > Would someone help this newbie out please? I'm trying to detect 'after > 6pm' but it is failing the if condition. > In it's simplest form it is: > > #!/bin/bash > hour=$(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 [ "$hour" -gt 18 ] ; then echo it is evening fi set - #************** end test script **************** Other set args of interest -v -u Some light reading found here http://tldp.org/LDP/abs/html/index.html http://mywiki.wooledge.org/BashFAQ/050 http://cfaj.freeshell.org/shell
From: Aragorn on 15 Jan 2010 13:51 On Friday 15 January 2010 19:40 in alt.os.linux, somebody identifying as Lew Pitcher wrote... > On January 15, 2010 12:51, in alt.os.linux, phil.remove.brady(a)hotmail > dot co dot united kingdom wrote: > >> Would someone help this newbie out please? I'm trying to detect >> 'after 6pm' but it is failing the if condition. >> In it's simplest form it is: >> >> #!/bin/bash >> hour=$(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? > > In this case, $hour (which /is/ a string) is re-evaluated as an > integer, and compared to (the integer) 18 > > First off, remember that Bash doesn't have a builtin [ test operator; I'm afraid that's not true, Lew. ;-) > the [ in your if statement is actually /bin/[, which is link to > /bin/test. The builtin test operator in bash is [[ (with it's > corresponding ]] terminator). > > So, your Bash if statement is actually running /bin/[ (/bin/test) to > evaluate the test condition. [aragorn] $> type [ [ is a shell builtin :-) -- *Aragorn* (registered GNU/Linux user #223157)
From: Lew Pitcher on 15 Jan 2010 13:54
On January 15, 2010 13:51, in alt.os.linux, aragorn(a)chatfactory.invalid wrote: > On Friday 15 January 2010 19:40 in alt.os.linux, somebody identifying as > Lew Pitcher wrote... > >> On January 15, 2010 12:51, in alt.os.linux, phil.remove.brady(a)hotmail >> dot co dot united kingdom wrote: >> >>> Would someone help this newbie out please? I'm trying to detect >>> 'after 6pm' but it is failing the if condition. >>> In it's simplest form it is: >>> >>> #!/bin/bash >>> hour=$(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? >> >> In this case, $hour (which /is/ a string) is re-evaluated as an >> integer, and compared to (the integer) 18 >> >> First off, remember that Bash doesn't have a builtin [ test operator; > > I'm afraid that's not true, Lew. ;-) [snip] > [aragorn] $> type [ > [ is a shell builtin > > :-) Oops... OK. Then I missed it in the manpage. OTOH, the shell builtin [ seems to work the same way that /bin/[ works, and the manpage for /bin/[ is good documentation for it. -- Lew Pitcher Master Codewright & JOAT-in-training | Registered Linux User #112576 Me: http://pitcher.digitalfreehold.ca/ | Just Linux: http://justlinux.ca/ ---------- Slackware - Because I know what I'm doing. ------ |