Date: Mon, 16 May 2005 07:42:53 +1000 (EST) From: Ian Smith <smithi@nimnet.asn.au> To: =?ISO-8859-1?Q?Mikko_Ty=F6l=E4j=E4rvi?= <mbsd@pacbell.net> Cc: freebsd-questions@freebsd.org Subject: Re: simple? sh problen Message-ID: <Pine.BSF.3.96.1050516073341.22570A-100000@gaia.nimnet.asn.au> In-Reply-To: <20050515133928.F10135@sotec.home>
next in thread | previous in thread | raw e-mail | index | archive | help
On Sun, 15 May 2005, Mikko Työläjärvi wrote: > On Mon, 16 May 2005, Ian Smith wrote: > > How do I test whether a sh argument is an integer or not, so as to avoid > > failing on a syntax error from otherwise working code such as: > > > > [ $3 -lt 10 -o $3 -gt 600 ] && echo "$0 $1 $2: $3 invalid" && exit 1 > > > > when $3 is a non-integer argument? Do I need to delve into awk and REs, > > or is there something more simple I've missed in mans test, expr, etc? > > Here are some suggestions for functions to do the test: > > isnum() { > expr "$1" : '^[0-9][0-9]*$' >/dev/null > } > > isnum() { > case "$1" in > *[^0-9]*|'') return 1;; > esac > return 0 > } > > The second one is likely to be faster unless "expr" is a shell builtin > (typically it it not). Thanks Mikko; I liked the look of the case version, and it works fine: if ! isnum $3; then echo "$0 $1 $2: $3 is not an integer"; exit 1; fi > $.02, Worth every penny. Cheers, Ian
Want to link to this message? Use this URL: <https://mail-archive.FreeBSD.org/cgi/mid.cgi?Pine.BSF.3.96.1050516073341.22570A-100000>