bash 对于目录中的文件，只回显文件名（无路径）
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For files in directory, only echo filename (no path)
提问by Mechaflash
How do I go about echoing only the filename of a file if I iterate a directory with a for loop?
如果我用 for 循环迭代一个目录，我如何只回显文件的文件名？
for filename in /home/user/*
do
echo $filename
done;
will pull the full path with the file name. I just want the file name.
将拉出带有文件名的完整路径。我只想要文件名。
回答by SiegeX
If you want a native bashsolution
如果您想要本机bash解决方案
for file in /home/user/*; do
echo "${file##*/}"
done
The above uses Parameter Expansionwhich is native to the shell and does not require a call to an external binary such as basename
以上使用了shell本地的参数扩展，不需要调用外部二进制文件，例如basename
However, might I suggest just using find
但是，我是否建议只使用 find
find /home/user -type f -printf "%f\n"
回答by Sean Bright
Just use basename:
只需使用basename：
echo `basename "$filename"`
The quotes are needed in case $filename contains e.g. spaces.
如果 $filename 包含空格，则需要引号。
回答by Oliver Charlesworth
Use basename:
使用basename：
echo $(basename /foo/bar/stuff)
回答by ryan
Another approach is to use lswhen reading the file list within a directory so as to give you what you want, i.e. "just the file name/s". As opposed to reading the full file path and then extracting the "file name" component in the body of the for loop.
另一种方法是ls在读取目录中的文件列表时使用，以便为您提供所需的内容，即“仅文件名/s”。与读取完整文件路径然后在 for 循环体中提取“文件名”组件相反。
Example below that follows your original:
下面的示例遵循您的原件：
for filename in $(ls /home/user/)
do
echo $filename
done;
If you are running the script in the same directory as the files, then it simply becomes:
如果您在与文件相同的目录中运行脚本，那么它就会变成：
for filename in $(ls)
do
echo $filename
done;
