shell - 检查外壳脚本中是否存在目录(Check if a directory exists in a shell script)

Question

什么命令可用于检查Shell脚本中是否存在目录？

  ask by Grundlefleck translate from so

Answer 1

To check if a directory exists in a shell script you can use the following:

(要检查shell脚本中是否存在目录，可以使用以下命令：)

if [ -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY exists.
fi

Or to check if a directory doesn't exist:

(或检查目录是否不存在：)

if [ ! -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY doesn't exist.
fi

---

However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check.

(但是，正如乔恩·埃里克森 （ Jon Ericson）所指出的，如果您不考虑到目录的符号链接也将通过此检查，则后续命令可能无法按预期运行。)

Eg running this:

(例如运行此：)

ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then 
  rmdir "$SYMLINK" 
fi

Will produce the error message:

(会产生错误信息：)

rmdir: failed to remove `symlink': Not a directory

So symbolic links may have to be treated differently, if subsequent commands expect directories:

(因此，如果后续命令需要目录，则可能必须区别对待符号链接：)

if [ -d "$LINK_OR_DIR" ]; then 
  if [ -L "$LINK_OR_DIR" ]; then
    # It is a symlink!
    # Symbolic link specific commands go here.
    rm "$LINK_OR_DIR"
  else
    # It's a directory!
    # Directory command goes here.
    rmdir "$LINK_OR_DIR"
  fi
fi

---

Take particular note of the double-quotes used to wrap the variables, the reason for this is explained by 8jean in another answer .

(请特别注意用于包装变量的双引号，其原因由8jean 在另一个答案中解释。)

If the variables contain spaces or other unusual characters it will probably cause the script to fail.

(如果变量包含空格或其他异常字符，则可能会导致脚本失败。)