On Jun 9, 2014 10:41 AM, "Thibault, Daniel" <daniel.thiba...@drdc-rddc.gc.ca> wrote: > > Configuration Information [Automatically generated, do not change]: > Machine: x86_64 > OS: linux-gnu > Compiler: gcc > Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='x86_64' -DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='x86_64-pc-linux-gnu' -DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' -DSHELL -DHAVE_CONFIG_H -I. -I../bash -I../bash/include -I../bash/lib -D_FORTIFY_SOURCE=2 -g -O2 -fstack-protector --param=ssp-buffer-size=4 -Wformat -Wformat-security -Werror=format-security -Wall > uname output: Linux sds-dut-vb 3.9.3 #1 SMP Mon Mar 24 18:48:39 EDT 2014 x86_64 x86_64 x86_64 GNU/Linux > Machine Type: x86_64-pc-linux-gnu > > Bash Version: 4.2 > Patch Level: 25 > Release Status: release > > Description: > The string comparison operators -n and -z are designed to be mutually > complementary. ! -z should be interchangeable with -n and ! -n should be > interchangeable with -z. But such is not the case. Consider these lines: > > $ if [ -z `pgrep pname` ]; then echo "not r" ; else echo "r" ; fi > $ if [ ! -z `pgrep pname` ]; then echo "r" ; else echo "not r" ; fi > $ if [ -n `pgrep pname` ]; then echo "r" ; else echo "not r" ; fi > $ if [ ! -n `pgrep pname` ]; then echo "not r" ; else echo "r" ; fi > > They should be equivalent but are not: -z correctly detects the process's > presence or absence, while -n returns true even when the process is not > running. > > Turns out this is how the script needs to be written to work correctly: > > $ if [ -z "`pgrep pname`" ]; then echo "not r" ; else echo "r" ; fi > $ if [ ! -z "`pgrep pname`" ]; then echo "r" ; else echo "not r" ; fi > $ if [ -n "`pgrep pname`" ]; then echo "r" ; else echo "not r" ; fi > $ if [ ! -n "`pgrep pname`" ]; then echo "not r" ; else echo "r" ; fi > > Repeat-By: > See the examples above. >
If you use single square brackets, you must quote the argument. For example: $ if [ -z "$(pgrep pname)" ]; then echo "not r" ; else echo "r" ; fi etc. And you should use $() instead of backticks. In Bash, you should use double square brackets. Please see http://stackoverflow.com/a/3870055/26428
