我想编写 bash 脚本以递归地列出 sftp 上的所有文件(具有完整路径),然后在本地与路径交互(因此唯一需要 sftp 的是获取路径)。不幸的是,“ls -R”在那里不起作用。
任何关于如何使用一些基本的 POC 做到这一点的想法将不胜感激
Available commands:
bye Quit sftp
cd path Change remote directory to 'path'
chgrp grp path Change group of file 'path' to 'grp'
chmod mode path Change permissions of file 'path' to 'mode'
chown own path Change owner of file 'path' to 'own'
df [-hi] [path] Display statistics for current directory or
filesystem containing 'path'
exit Quit sftp
get [-Ppr] remote [local] Download file
help Display this help text
lcd path Change local directory to 'path'
lls [ls-options [path]] Display local directory listing
lmkdir path Create local directory
ln [-s] oldpath newpath Link remote file (-s for symlink)
lpwd Print local working directory
ls [-1afhlnrSt] [path] Display remote directory listing
lumask umask Set local umask to 'umask'
mkdir path Create remote directory
progress Toggle display of progress meter
put [-Ppr] local [remote] Upload file
pwd Display remote working directory
quit Quit sftp
rename oldpath newpath Rename remote file
rm path Delete remote file
rmdir path Remove remote directory
symlink oldpath newpath Symlink remote file
version Show SFTP version
!command Execute 'command' in local shell
! Escape to local shell
? Synonym for help
最佳答案
这个递归脚本完成了这项工作:
#!/bin/bash
#
URL=user@XXX.XXX.XXX.XXX
TMPFILE=/tmp/ls.sftp
echo 'ls -1l' > $TMPFILE
function handle_dir {
echo "====== $1 ========="
local dir=$1
sftp -b $TMPFILE "$URL:$dir" | tail -n +2 | while read info; do
echo "$info"
if egrep -q '^d' <<< $info; then
info=$(echo $info)
subdir=$(cut -d ' ' -f9- <<< $info)
handle_dir "$dir/$subdir"
fi
done
}
handle_dir "."
用 sftp 服务器数据填充 URL。
关于bash - 递归列出sftp上的所有文件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30504124/