如何使用awk将文件名打印到具有特定格式的新文件？

Question

I'm trying to print a list of all jpg files in a directory to a new .txt file.

The output format in that .txt file should look like that:

<img src="filename.jpg">

Currently, I have this command:

ls -al *.jpg | awk ‘{print”<img src=”$9">"}' > list_of_files.txt

But it doesn't work. What would be the right command to get the right formatting?

Answer 1

One in awk:

$ awk '
BEGIN {
    for(i=1;i<ARGC;i++)                       # loop all argument files
        printf "<img src=\"%s\">\n", ARGV[i]  # output as requested
    exit                                      # never touch any files
}' *.jpg

Output sample:

<img src="foo.jpg">
<img src="bar.jpg">

Answer 2

您不需要awk或任何其他外部工具，shell可以自己完成：

printf '<img src="%s">\n' *.jpg > list_of_files.txt

Answer 3

ls -A1 *.jpg | awk '{print "<img src=\""$0"\">"}' > list_of_files.txt

Safe version (in case to have something like rm -rf .. .jpg in your files:

shopt -s nullglob
for f in *; do
  if [[ "$f" == *.jpg ]]; then
    echo "<img src=\""$f"\">" >> list_of_files.txt
  fi
done