bash 如何检查目录是否有子目录？
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How do I check if a directory has child directories?
提问by Dave
Using bash, how do I write an if statement that checks if a certain directory, stored in the a script variable named "$DIR", contains child directories that are not "." or ".."?
使用 bash，我如何编写 if 语句来检查存储在名为“$DIR”的脚本变量中的某个目录是否包含不是“.”的子目录。或者 ”..”？
Thanks, - Dave
谢谢， - 戴夫
回答by AlG
Here's one way:
这是一种方法：
#!/usr/bin/bash
subdircount=`find /d/temp/ -maxdepth 1 -type d | wc -l`
if [ $subdircount -eq 2 ]
then
echo "none of interest"
else
echo "something is in there"
fi
回答by robert
Here's a more minimalist solution that will perform the test in a single line..
这是一个更简约的解决方案，它将在一行中执行测试..
ls $DIR/*/ >/dev/null 2>&1 ;
if [ $? == 0 ];
then
echo Subdirs
else
echo No-subdirs
fi
By putting /after the *wildcard you select only directories, so if there is no directories then lsreturns error-status 2and prints the message ls: cannot access <dir>/*/: No such file or directory. The 2>&1captures stderrand pipes it into stdoutand then the whole lot gets piped to null(which gets rid of the regular ls output too, when there is files).
通过将/后*通配符只选择目录，因此如果没有目录，那么ls将返回错误状态2和打印消息ls: cannot access <dir>/*/: No such file or directory。在2>&1捕获标准错误和管道入标准输出，然后一大堆得到管道输送到零（这摆脱了常规的LS输出过，当有文件）。
回答by sbtkd85
I'm not really sure what you trying to do here, but you can use find:
我不太确定你想在这里做什么，但你可以使用find：
find /path/to/root/directory -type d
If you want to script it:
如果要编写脚本：
find $DIR/* -type d
should do the trick.
应该做的伎俩。
回答by jfg956
A solution in pure bash, without needing any other program execution. This is not the most compact solution, but if run in a loop, it could be more efficient because no process creation is needed. If there is a lot of files in '$dir', the filename expansion could break though.
纯 bash 中的解决方案，无需任何其他程序执行。这不是最紧凑的解决方案，但如果在循环中运行，它可能会更高效，因为不需要创建进程。如果 中有很多文件'$dir'，文件名扩展可能会中断。
shopt -s dotglob # To include directories beginning by '.' in file expansion.
nbdir=0
for f in $dir/*
do
if [ -d $f ]
then
nbdir=$((nbdir+1))
fi
done
if [ nbdir -gt 0 ]
then
echo "Subdirs"
else
echo "No-Subdirs"
fi
回答by ?ukasz Starowicz
in my case it doesn't work as @AIG wrote - for empty dir I got subdircount=1 (find returns only dir itself).
在我的情况下，它不像@AIG 写的那样工作 - 对于空目录，我得到 subdircount=1 （find 只返回目录本身）。
what works for me:
什么对我有用：
#!/usr/bin/bash
subdircount=`find /d/temp/ -maxdepth 1 -type d | wc -l`
if [ $subdircount -ge 2 ]
then
echo "not empty"
else
echo "empty"
fi
回答by ennuikiller
Try this as the condition you test against:
试试这个作为你测试的条件：
subdirs=$(ls -d $DIR/.*/ | grep -v "/./\|/../")
subdirs will be empty if there are no subdirectories
如果没有子目录，则 subdirs 将为空
回答by jman
How about:
怎么样：
num_child=`ls -al $DIR | grep -c -v ^d`
If $num_child > 2, then you have child directories. If you do not want hidden directories, replace ls -al with ls -l.
如果 $num_child > 2，则您有子目录。如果您不想隐藏目录，请将 ls -al 替换为 ls -l。
if [ $num_child -gt 2 ]
then
echo "$num_child child directories!"
fi
回答by Anthony Rutledge
Here is the best answer you will ever see on this issue.
这是您将在此问题上看到的最佳答案。
function hasDirs ()
{
declare $targetDir="" # Where targetDir ends in a forward slash /
ls -ld ${targetDir}*/
}
If you are already in the target directory:
如果您已经在目标目录中：
if hasDirs ./
then
fi
If you want to know about another directory:
如果您想了解另一个目录：
if hasDirs /var/local/ # Notice the ending slash
then
fi
