使用 find 从 bash 脚本中递归获取文件名
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Get the filename from bash script recursively using find
提问by RameshVel
I am trying retrieve the filename from the find command recursively. This command prints all the filenames with full path
我正在尝试从 find 命令中递归检索文件名。此命令打印所有带有完整路径的文件名
> for f in $(find -name '*.png'); do echo "$f";done
> ./x.png
> ./bg.png
> ./s/bg.png
But when i try to just get the name of the file using these commands, it prints
但是当我尝试使用这些命令获取文件名时，它会打印
for f in $(find -name '*.png'); do echo "${f##*/}";done
bg.png
and
和
for f in $(find -name '*.png'); do echo $(basename $f);done
bg.png
It omits other 2 files. I am new to shell scripting. I couln't figure out whats wrong with this one.
它省略了其他 2 个文件。我是 shell 脚本的新手。我想不通这个有什么问题。
EDIT:
编辑：
THis is what i have actually wanted.
这就是我真正想要的。
- I want to loop through a directory recursively and find all png images
- send it to pngnq for RGBA compression
- It outputs the new file with
orgfilename-nq8.png - send it to pngcrush and rename and generate a new file (org file will be overwritten )
- remove new file
- 我想递归遍历目录并找到所有 png 图像
- 将其发送到 pngnq 进行 RGBA 压缩
- 它输出新文件
orgfilename-nq8.png - 将它发送到 pngcrush 并重命名并生成一个新文件（组织文件将被覆盖）
- 删除新文件
i have a code which works on a single directory
我有一个适用于单个目录的代码
for f in *.png; do pngnq -f -n 256 "$f" && pngcrush "${f%.*}"-nq8.png "$f";rm "${f%.*}"-nq8.png; done
I want to do this recursively
我想递归地做到这一点
回答by Gilles Quenot
Simply do :
简单地做：
find -name '*.png' -printf '%f\n'
If you want to run something for each files :
如果你想为每个文件运行一些东西：
find -name '*.png' -printf '%f\n' |
while read file; do
# do something with "$file"
done
Or with xargs:
或与xargs：
find -name '*.png' -printf '%f\n' | xargs -n1 command
Be sure you cannot use finddirectly like this :
确保您不能find像这样直接使用：
find -name '*.png' -exec command {} +
or
或者
find -name '*.png' -exec bash -c 'do_something with ${1##*/}' -- {} \;
Search -printfon http://unixhelp.ed.ac.uk/CGI/man-cgi?findor in
-printf在http://unixhelp.ed.ac.uk/CGI/man-cgi?find 上搜索或在
man find | less +/^' *-printf'
