变量分配之间的差异

Question

I had a question that was answered here: https://unix.stackexchange.com/questions/264925/awk-fs-with-back-slashes

After that I tried adding it to a .sh script that starts with:

#!/bin/bash

My Line:

category1Filenames=(\`find . -maxdepth 1 -type f |
gawk 'BEGIN { FS = "(\\\\./)|(\\\\.)" } ; { print $2 }' | sort -u\`)

And then I print the output using:

for genericFilename in ${category1Filenames[@]}; do  
  printf "%s\n" $genericFilename  
done 

But this gives me the error:

gawk: cmd. line:1: warning: escape sequence `\.' treated as plain `.'

If I change the line to be:

category1Filenames=$(find . -maxdepth 1 -type f |
gawk 'BEGIN { FS = "(\\\\./)|(\\\\.)" } ; { print $2 }' | sort -u)

Then it works as expected.

I'm assuming that the problem was with using the back tick characters, but I don't understand why. Probably because I don't understand the behavior of ` vs ' vs " . If anyone can point me to documentation that can explain this behavior (maybe tldp) then it would be greatly appreciated.

Answer 1

The backquote (`) is used in the old-style command substitution, The foo=$(command) syntax is recommended instead. Backslash handling inside $() is less surprising, and $() is easier to nest. See BashFAQ82

Using backquotes, you'll have to escape required sequence for gawk :

category1Filenames=(`find . -maxdepth 1 -type f | gawk 'BEGIN { FS = "(\\\./)|(\\\.)" } ; { print $2 }' | sort -u`)

Aside: Feed categoryFilenames array in a loop instead:

while read -r line; do
    categoryFilenames+=("$line")
done < <(find . -maxdepth 1 -type f | gawk 'BEGIN { FS = "(\./)|(\.)" } ; { print $2 }' | sort -u)

Now it is safer to iterate over categoryFilenames like you did, but with double-quotes:

for genericFilename in "${category1Filenames[@]}"; do  
  printf "%s\n" "$genericFilename"  
done  

Don't forget to "Double quote" every literal that contains spaces/metacharacters and every expansion: "$var" , "$(command "$var")" , "${array[@]}" , "a & b" .

Answer 2

Instead of find , just iterate over all the files in the current directory, and use the shell to process only regular files and extract the file name.

for name in *; do
    [[ -f $name ]] || continue
    name_only=${name%%.*}
    printf "%s\n" "$name_only"
done