Bash – Difference between * and ?* glob in Bash

bashwildcards

* is roughly a wild card character with a unlimited amount of length.
? is roughly a wild card character for one or zero length.

Is there a difference between using * vs ?* when searching for strings in bash?

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The difference is that in bash (as you tagged the question) * matches any string with length zero or more characters, while ?* matches a string with at least 1 character. Consider for example two files: file.txt and xfile.txt and try to list them with ls ?*file.txt or ls *file.txt.

One real case scenario when I use such construct is to list hidden files. Very often I just do

ls .??*

Double question marks are here to prevent listing the current directory . and the parent directory .., like it would be with a simpler form ls .*.

^{I need to point here that my .??* is not perfect; for example filenames with only two characters, like .f, don't match this pattern. More reliable solution is ls {..?,.[!.]}*, but usually that is too much to type for me.}

Related Solutions

Bash – select only one result from a bash glob

The following works in bash 4.2:

list=( /<root_path>/<process_two_path>/logs/* )
echo "${list[-1]}"

If your bash is an older version:

list=( /<root_path>/<process_two_path>/logs/* )
echo "${list[${#list[@]}-1]}"

Bash – Difference Between Single and Double Equal Signs (=) in Shell Comparisons

[[ $a == $b ]] is not comparison, it's pattern matching. You need [[ $a == "$b" ]] for byte-to-byte equality comparison. = is the same as == in any shell that supports [[...]] (introduced by ksh).

[[...]] is not standard sh syntax. The [ command is standard, and the standard comparison operator there is = (though some [ implementations also recognise ==).

Just like in any argument to any command, variable expansions must be quoted to prevent split+glob and empty removal (only the latter being performed in zsh), so:

[ "$a" = "$b" ]

In standard sh, pattern matching is done with case:

case $a in
  ($b) ...
esac

For completeness, other equality-like operators you may come across in shell scripts:

[ "$a" -eq "$b" ]: standard [ operator to compare decimal integer numbers. Some [ implementations allow blanks around the numbers, some allow arbitrary arithmetic expressions, but that's not portable. Portably, one can use [ "$(($a))" -eq "$(($b))" ] for that. See also [ "$((a == b))" -ne 0 ] which would be the standard equivalent (except that POSIXly, the behaviour is only specified if $a and $b contain integer constants) of:
((a == b)), from ksh and also found in zsh and bash, returns true if the evaluation of the arithmetic expression stored in $a yields the same number as that of $b. Typically, that's used for comparing numbers. Note that there are variations between shells as to how arithmetic expressions are evaluated and what numbers are supported (for instance bash and some implementation/versions of ksh don't support floating point or treat numbers with leading zeros as octal).
expr "$a" = "$b" does a number comparison if both operands are recognised as decimal integer numbers (some allowing blanks around the number), and otherwise checks if the two string operators have the same sorting order. It would also fail for values of $a or $b that are expr operators like (, substr...
awk 'BEGIN{exit !(ARGV[1] == ARGV[2])}' "$a" "$b": if $a and $b are recognised as numbers (at least decimal integer and floating point numbers like 1.2, -1.5e-4, leading trailing blanks ignored, some also recognising hexadecimal, octal or anything recognised by strtod()), then a numeric comparison is performed. Otherwise, depending on the implementation, it's either a byte-to-byte string comparison, or like for expr a strcoll() comparison, that is whether $a and $b sort the same.

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Related Solutions

Bash – select only one result from a bash glob

Bash – Difference Between Single and Double Equal Signs (=) in Shell Comparisons

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