bash - Linux/庆典 : remove text between two chars

Question

我有一个文件 file1它包含:

<     16 ./lnx/apps/vlc/tsconf_1.0-11_all.deb
<     16 ./lnx/apps/vlc/vlc
<   2000 ./lnx/apps/vlc/vlc-nox_2.0.8-1_i386.deb
<     16 ./lnx/apps/vlc/vlc-plugin-notify_2.0.8-1_i386.deb
<     32 ./lnx/apps/vlc/vlc-plugin-pulse_2.0.8-1_i386.deb
<     16 ./lnx/cmds/64bit_ubuntu_add_i386
<     16 ./lnx/cmds/acroread_dl
<     16 ./lnx/cmds/dl_from_gdrv
<     16 ./lnx/cmds/dpkg_install_list_in_txt_file
<     16 ./lnx/cmds/find_and_replace
<     16 ./lnx/cmd/pearl_script.ps1
<     16 ./lnx/cmds/rm_using_find
<     16 ./lnx/cmds/wget_dl_whl_ws

我想删除 < 之间的所有内容和 ./ (或介于 \n 和 ./ 之间)不包括 ./ ，所以输出如下:

./lnx/apps/vlc/tsconf_1.0-11_all.deb
./lnx/apps/vlc/vlc
./lnx/apps/vlc/vlc-nox_2.0.8-1_i386.deb
./lnx/apps/vlc/vlc-plugin-notify_2.0.8-1_i386.deb
./lnx/apps/vlc/vlc-plugin-pulse_2.0.8-1_i386.deb
./lnx/cmds/64bit_ubuntu_add_i386
./lnx/cmds/acroread_dl
./lnx/cmds/dl_from_gdrv
./lnx/cmds/dpkg_install_list_in_txt_file
./lnx/cmds/find_and_replace
./lnx/cmd/pearl_script.ps1
./lnx/cmds/rm_using_find
./lnx/cmds/wget_dl_whl_ws

我试过这些命令(它们的输出是准确的输入文件):

$ sed -n '/</,/ ./ p' file1
$ sed 's/< .* " ."/./' file1
$ sed -e 's/<\n[^ .]>/<\n.>/g' file1
$ sed -e 's/\(<\).*\(.\)/\1\2/' file1

可能这只是一件简单的事情，但我是 sed/awk/tr/grep/find 命令的新手。

Answer 1

$ sed 's/[^.]*//' file
./lnx/apps/vlc/tsconf_1.0-11_all.deb
./lnx/apps/vlc/vlc
./lnx/apps/vlc/vlc-nox_2.0.8-1_i386.deb
./lnx/apps/vlc/vlc-plugin-notify_2.0.8-1_i386.deb
./lnx/apps/vlc/vlc-plugin-pulse_2.0.8-1_i386.deb
./lnx/cmds/64bit_ubuntu_add_i386
./lnx/cmds/acroread_dl
./lnx/cmds/dl_from_gdrv
./lnx/cmds/dpkg_install_list_in_txt_file
./lnx/cmds/find_and_replace
./lnx/cmd/pearl_script.ps1
./lnx/cmds/rm_using_find
./lnx/cmds/wget_dl_whl_ws