Shell script how to list file name in ascending order

Pathname expansion (glob expansion) returns a list of filenames which is alphabetically sorted according to your current locale. If you have something simple like UTF-8 or C, your sorting order will be ASCII sorted. This is visible in the result of the OP. The file with name 19.jpg is sorted before 1.jpg because the lt;dot>-character has a higher lexicographical order than the character 9.

If you want to traverse your files in a different sorting order, then a different approach needs to be taken.

Under the bold assumption that the OP requests to traverse the files in a numeric sorted way, i.e. order the names according to a number at the beginning of the file-name, you can do the following:

while IFS= read  -r -d '' file; do 
   echo "filename: $file"
done < <(find .  -maxdepth 1 -type f -name '*.jpg' -print0 | sort -z -n)

Here we use find to list all files in the current directory (depth==1) we print them with a \0 as a separator, and use sort to ask for the requested sorting, indicating that we have \0 as the field separator. Instead of using a for-loop, we use a while-loop to read the information.

See BashPitFall001 For some details

note: sort -z is a GNU extension

not quite sure if this is what you're asking, but you have the echo inside your loop which will cause it to be printed in a different row each time.

you can do:

list "" 
for i in ./*.jpg
do
filename=$(basename "$i")
list="$list $filename"
done    
echo  "files: $list"

which would output

files: 1.jpg 2.jpg

Nevertheless, You should clarify your question.

From your requirement to "process them in the sequence of names 1.jpg, 2.jpg etc", this will accomplish that. The sort specifies a numeric key obtained by defining the first field as the string before a "." delimiter.

#!/usr/bin/env bash

shopt -s nullglob
allfiles=(*.jpg);

for f in "${allfiles[@]}"
do
    echo "$f"
done | sort -t"." -k1n