我有一个ZIP文件目录(在Windows机器上创建的)。我可以手动解压缩他们使用unzip文件名,但我怎么能解压缩所有的ZIP文件在当前文件夹通过shell?

使用Ubuntu Linux服务器。


当前回答

for i in `ls *.zip`; do unzip $i; done

其他回答

如果你所说的“当前目录”是指zip文件所在的目录,那么我将使用以下命令:

find . -name '*.zip' -execdir unzip {} \; 

节选自find的手册页

-execdir command ;
-execdir command {} +

Like -exec, but the specified command is run from the subdirectory containing the matched file, which is not normally the directory in which you started find. This a much more secure method for invoking commands, as it avoids race conditions during resolution of the paths to the matched files. As with the -exec option, the '+' form of -execdir will build a command line to process more than one matched file, but any given invocation of command will only list files that exist in the same subdirectory. If you use this option, you must ensure that your $PATH environment variable does not reference the current directory; otherwise, an attacker can run any commands they like by leaving an appropriately-named file in a directory in which you will run -execdir.

这是Pedro Lobito回答的一个变体,使用如何递归地遍历目录以删除具有某些扩展名的文件:

shopt -s globstar
root_directory="."

for zip_file_name in **/*.{zip,sublime\-package}; do
    directory_name=`echo $zip_file_name | sed 's/\.\(zip\|sublime\-package\)$//'`
    printf "Unpacking zip file \`$root_directory/$zip_file_name\`...\n"

    if [ -f "$root_directory/$zip_file_name" ]; then
        mkdir -p "$root_directory/$directory_name"
        unzip -o -q "$root_directory/$zip_file_name" -d "$directory_name"

        # Some files have the executable flag and were not being deleted because of it.
        # chmod -x "$root_directory/$zip_file_name"
        # rm -f "$root_directory/$zip_file_name"
    fi
done

unpack -e *.zip,安装atool。 具有智能地处理错误的优点,并且总是解包到子目录中,除非zip只包含一个文件。因此,不存在使用大量文件污染当前目录的危险,而在没有目录结构的zip文件上进行unzip则存在这种危险。

要解压缩目录中的所有文件,只需在终端中键入以下命令:

unzip '*.zip'

下面是一个没有使用ls的一行程序,它为文件创建了带有zip名称的文件夹。它适用于当前目录中的任何zip。

for z in *.zip; do unzip "$z" -d "${z%".zip"}"; done

你可以把它添加到你的。bashrc

alias unzip_all='for z in *.zip; do unzip "$z" -d "${z%".zip"}"; done'

灵感源自:

方法#2:在https://www.cyberciti.biz/faq/linux-unix-shell-unzipping-many-zip-files/中使用Shell For循环(长版本)从Linux命令行解压多个文件