你如何在 bash 中解析文件名？
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How do you parse a filename in bash?
提问by Nick Pierpoint
I have a filename in a format like:
我有一个格式如下的文件名：
system-source-yyyymmdd.dat
system-source-yyyymmdd.dat
I'd like to be able to parse out the different bits of the filename using the "-" as a delimiter.
我希望能够使用“-”作为分隔符来解析文件名的不同位。
回答by Bobby Hyman
You can use the cut commandto get at each of the 3 'fields', e.g.:
您可以使用cut 命令获取 3 个“字段”中的每一个，例如：
$ echo "system-source-yyyymmdd.dat" | cut -d'-' -f2
source
"-d" specifies the delimiter, "-f" specifies the number of the field you require
“-d”指定分隔符，“-f”指定您需要的字段编号
回答by Colas Nahaboo
A nice and elegant (in my mind :-) using only built-ins is to put it into an array
仅使用内置函数的一个漂亮而优雅的（在我看来 :-) 是将它放入一个数组中
var='system-source-yyyymmdd.dat'
parts=(${var//-/ })
Then, you can find the parts in the array...
然后，您可以找到数组中的部分...
echo ${parts[0]} ==> system
echo ${parts[1]} ==> source
echo ${parts[2]} ==> yyyymmdd.dat
Caveat: this will not work if the filename contains "strange" characters such as space, or, heaven forbids, quotes, backquotes...
警告：如果文件名包含“奇怪”字符，如空格，或者，天堂禁止，引号，反引号，这将不起作用。
回答by flight
Depending on your needs, awkis more flexible than cut. A first teaser:
根据您的需要，awk比 cut 更灵活。第一个预告片：
# echo "system-source-yyyymmdd.dat" \
|awk -F- '{printf "System: %s\nSource: %s\nYear: %s\nMonth: %s\nDay: %s\n",
,,substr(,1,4),substr(,5,2),substr(,7,2)}'
System: system
Source: source
Year: yyyy
Month: mm
Day: dd
Problem is that describing awk as 'more flexible' is certainly like calling the iPhone an enhanced cell phone ;-)
问题是，将 awk 描述为“更灵活”肯定就像将 iPhone 称为增强型手机一样；-)
回答by David
回答by Shannon Nelson
Another method is to use the shell's internal parsing tools, which avoids the cost of creating child processes:
另一种方法是使用shell的内部解析工具，这样可以避免创建子进程的成本：
oIFS=$IFS IFS=- file="system-source-yyyymmdd.dat" set $file IFS=$oIFS echo "Source is "
回答by William Pursell
The simplest (and IMO best way) to do this is simply to use read:
最简单的（也是 IMO 最好的方法）就是使用read：
$ IFS=-. read system source date ext << EOF
> foo-bar-yyyymmdd.dat
> EOF
$ echo $system
foo
$ echo $source $date $ext
bar yyyymmdd dat
There are many variations on that theme, many of which are shell dependent:
该主题有许多变体，其中许多与 shell 相关：
bash$ IFS=-. read system source date ext <<< foo-bar-yyyymmdd.dat
bash$ IFS=-. read system source date ext <<< foo-bar-yyyymmdd.dat
echo "$name" | { IFS=-. read system source date ext
echo In all shells, the variables are set here...; }
echo but only in some shells do they retain their value here
