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Easy way to get filenames and directory names using Bash Shell
Easy way to get filenames and directory names using Bash Shell
前言
还是今天再写一个自动化打包脚本，用到了从路径名中获取最后的文件名。这里记录一下实现过程。当然，最后我也会给出官方的做法。（ps：很囧，实现完了才发现原来Bash Shell有现成的函数）
获取文件名
假设给定的路径名为：
/tmp/csdn/zhengyi/test/zhengyi.txt
awk解法
用“/”做分隔符，然后打印出最后的那一部分。实现代码如下
resFile=`echo /tmp/csdn/zhengyi/test/adb.log | awk -F "/" '{print $NF}'`
官方解法(basename)
Bash Shell本身提供了basename命令，可以直接获取路径名最后的文件名，实现代码如下：
resFile=`basename /tmp/csdn/zhengyi/test/adb.log`
获取目录名
官方解法(dirname)
Bash Shell本身提供了dirname命令，特别方便，可以直接获取路径对应的目录名，实现代码如下：
dirPath=`dirname /tmp/csdn/zhengyi/test/adb.log`
awk解法
可以灵活的使用分隔符，混合正则表达式：
dirPath=`echo /tmp/csdn/zhengyi/test/adb.log | awk -F '/[^/]*$' '{print $1}'`awk+for循环的方法：
echo /tmp/csdn/zhengyi/test/adb.log | awk 'BEGIN{res=""; FS="/";}{ for(i=2;i<=NF-1;i++) res=(res"/"$i);} END{print res}'
更多使用Bash Shell获取文件名和目录名的简单方法相关文章请关注PHP中文网！
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