Question

我是新手来编写脚本并需要帮助：

我需要从目录中删除特定文件。我的目标是在每个子目录中找到一个名为＆＃34; filename.A＆＃34;的文件。并删除所有以＆＃34; filename＆＃34;开头的文件延伸B，那就是：＆＃34; filename01.B＆＃34; ，＆＃34; filename02.B＆＃34;等。

我试过了：

B_folders="$(find /someparentdirectory -type d -name "*.B" | sed 's#  (.*\)/.*#\1#'|uniq)"
A_folders="$(find "$B_folders" -type f -name "*.A")"

for FILE in "$A_folders" ; do
   A="${file%.A}"
   find "$FILE" -name "$A*.B" -exec rm -f {}\;
done

当目录名称包含空格时，开始出现问题。

有关正确方法的建议吗？

编辑：

我的目标是在每个子目录中找到（名称中可能有空格），文件格式为：＆＃34; filename.A＆＃34;

如果存在此类文件：

检查＆＃34; filename * .B＆＃34;存在并删除它，即：删除：＆＃34; filename01.B＆＃34; ，＆＃34; filename02.B＆＃34;等。

Answer 1

在bash 4中，它只是

shopt -s globstar nullglob
for f in some_parent_directory/**/filename.A; do
    rm -f "${f%.A}"*.B
done

Answer 2

如果空格是唯一的问题，您可以修改for内的find，如下所示：

find "$FILE" -name "$A*.B" -print0 | xargs -0 rm

男人找到节目：

  -print0
          True; print the full file name on the standard output, followed by a null character (instead of the newline character that  -print  uses).   This  allows
          file  names that contain newlines or other types of white space to be correctly interpreted by programs that process the find output.  This option corre-
          sponds to the -0 option of xargs.

和xarg的手册

  -0     Input  items are terminated by a null character instead of by whitespace, and the quotes and backslash are not special (every character is taken literal-
          ly).  Disables the end of file string, which is treated like any other argument.  Useful when input items might contain  white  space,  quote  marks,  or
          backslashes.  The GNU find -print0 option produces input suitable for this mode.

在bash中使用find with variables

2 个答案: