我不明白 Bash 脚本中“$x*”中的 * 是什么

Question

Here's some code for some context.

for x in {a..z};
do
    for y in $x*;do
        if  [ "$y" != "$x*" ]; then
        let "count++"

If someone can try to explain what that 2nd for loop does I would appreciate it. Thank you

Answer 1

It's a really inefficient and ugly way of counting how many files in the current directory start with the letters a through z. The outer loop executes once for each letter, and for each one, the inner loop executes once for each file starting with that letter - $x* expands to a* through z* and then to files matching that pattern (Or just the pattern if none do, assuming nullglob is off. The if catches those cases though it has issues if the filename is literally a* etc.).

A better way to count the number of files matching a pattern in bash is to store all the filenames in an array using a single more efficient glob pattern, and then get the array's length:

shopt -s nullglob
files=([a-z]*)
count=${#files[@]}

Answer 2

By default ( shopt -u nullglob ), for the glob pattern foo* , if there are no files matching it, the pattern will expand to the literal string foo* . See the following example which you can try.

$ cat ../foo.sh
count=0
for x in {a..z}; do
    for y in $x*; do
        if  [ "$y" != "$x*" ]; then
            let "count++"
        fi
    done
done
echo $count
$ ls
a1 a2 b3 c4 z5
$ bash ../foo.sh
5
$ x=a
$ for file in $x*; do echo "$file"; done
a1
a2
$ x=b
$ for file in $x*; do echo "$file"; done
b3
$ x=f
$ for file in $x*; do echo "$file"; done
f*
$

Search for nullglob in the bash manual for more details.

As others pointed out the code is not technically correct unless you are sure there are no filenames including a literal * char. Or you can change [ "$y" != "$x*" ] to [ -e "$y" ] .

$ ls -1
a1
a2
b3
c4
z5
z6*
$ cat ../foo.sh
count=0
for x in {a..z}; do
    for y in $x*; do
        if  [ -e "$y" ]; then
        #   ^^^^^^^^^^^
            let "count++"
        fi
    done
done
echo $count
$ bash ../foo.sh
6
$