不使用sed或awk,只cut,当字段的数量未知或随每一行变化时,我如何得到最后一个字段?


当前回答

这是不可能只使用切割。下面是使用grep的方法:

grep -o '[^,]*$'

用逗号替换其他分隔符。

解释:

-o (--only-matching) only outputs the part of the input that matches the pattern (the default is to print the entire line if it contains a match). [^,] is a character class that matches any character other than a comma. * matches the preceding pattern zero or more time, so [^,]* matches zero or more non‑comma characters. $ matches the end of the string. Putting this together, the pattern matches zero or more non-comma characters at the end of the string. When there are multiple possible matches, grep prefers the one that starts earliest. So the entire last field will be matched.

完整的例子:

如果我们有一个叫data。csv的文件包含

one,two,three
foo,bar

然后输出grep -o '[^,]*$' < data.csv

three
bar

其他回答

选择1

choose支持负索引(语法类似于Python的切片)。

没有awk ? 但是用awk很简单:

echo 'maps.google.com' | awk -F. '{print $NF}'

AWK是一个更强大的工具,可以放在你的口袋里。 -F if用于字段分隔符 NF是字段的数量(也表示最后一个字段的索引)

这是不可能只使用切割。下面是使用grep的方法:

grep -o '[^,]*$'

用逗号替换其他分隔符。

解释:

-o (--only-matching) only outputs the part of the input that matches the pattern (the default is to print the entire line if it contains a match). [^,] is a character class that matches any character other than a comma. * matches the preceding pattern zero or more time, so [^,]* matches zero or more non‑comma characters. $ matches the end of the string. Putting this together, the pattern matches zero or more non-comma characters at the end of the string. When there are multiple possible matches, grep prefers the one that starts earliest. So the entire last field will be matched.

完整的例子:

如果我们有一个叫data。csv的文件包含

one,two,three
foo,bar

然后输出grep -o '[^,]*$' < data.csv

three
bar

使用perl的替代方法是:

perl -pe 's/(.*) (.*)$/$2/' file

在哪里您可以改变\t为任何分隔符的文件是

下面实现一个朋友的建议

#!/bin/bash
rcut(){

  nu="$( echo $1 | cut -d"$DELIM" -f 2-  )"
  if [ "$nu" != "$1" ]
  then
    rcut "$nu"
  else
    echo "$nu"
  fi
}

$ export DELIM=.
$ rcut a.b.c.d
d