bash - 在没有解压缩的情况下获取 rar 中的文件名

Question

试图将一个 bash 脚本放在一起，但是我坚持这一点。rar 被分成 x 个文件，rar 内是 1 个单个文件。我正在做的如下：

cd $dir
for rarfile in $(find -iname "*.part1.rar")
        do
                echo "Rar file: " $rarfile >> $dir/execute.log
                name = $(unrar lb "$rarfile")
                echo "Name of file inside rar container: " $name >> $dir/execute.log
                extension ="${name##*.}"
                echo "Extension: " $extension >> $dir/execute.log
                filename = ${name%.*}
                echo "Name: " $filename >> $dir/execute.log
#               unrar x -y -o- $rarfile $uprar_dir
        done

execute.log 如下：

Rar file:  ./file.part1.rar
Name of file inside rar container:
Extension:
Name:

似乎无法让 $name 工作。然而，unrar 工作正常。请帮忙。

Answer 1

in bash to assign value to a variable you cannot have spaces ie:

name=$(unrar lb "$rarfile")

instead of:

name = $(unrar lb "$rarfile")