As I read,
test -n $string ==> Exit status is 0 if $string is not null, 1 otherwise
and
test -z $string ==> Exit status is 0 if $string is null, 1 otherwise
But in this particular example, (I tried to create a null string) it seems I am missing some thing.
#!/bin/sh str="" test -n $str echo $? test -z $str echo $?
Output of this is:
0 0
Can anyone give an explanation for this strange behavior??
Answer
Put $str inside double quotes!
The -n test requires that the string be quoted within the test brackets. Using an unquoted string with ! -z, or even just the unquoted string alone within test brackets (see Example 7-6) normally works, however, this is an unsafe practice. Always quote a tested string. Other Comparison Operators