使用 Bash 删除文本文件中的列？
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Removing columns in text file with Bash?
提问by Homeschooldev
I have a to write a script that remove the Idle column from the output of finger.
我必须编写一个脚本，从finger 的输出中删除空闲列。
>finger
Login Name TTY Idle Login Time Office Phone
Billy Billy Howard *con 6:55 Fri 19:03
Billy Billy Howard s00 5 Fri 19:11
Billy Billy Howard s00 Sat 00:27
I tried remove the extra spaces with tr and then using cut with a delimiter of a space to remove the column, but since Idle can have no value I sometimes get the wrong value since tr delimited the spaces were the idle time should be... Does anyone know how I can remove the Idle column?
我尝试用 tr 删除多余的空格，然后使用带有空格分隔符的 cut 来删除列，但是由于 Idle 可能没有值，我有时会得到错误的值，因为 tr 分隔了空格，空闲时间应该是...有谁知道如何删除空闲列？
采纳答案by Hai Vu
This solution is not perfect: the column position and width might change. If they are constant, the following command will do the trick by removing text columns 34 to 39 inclusively:
此解决方案并不完美：列位置和宽度可能会发生变化。如果它们是常量，则以下命令将通过删除文本列 34 到 39 来实现这一点：
finger | colrm 34 39
回答by potong
This might work for you:
这可能对你有用：
finger | sed 's/\(.\{35\}\).....//'
or this:
或这个：
finger | cut --complement -c36-40
回答by Taras
If "\t"is used as column delimiter then you can get rid of 4th column using awkand delete doubled "\t"using sed. For example:
如果"\t"用作列分隔符，那么您可以使用awk删除第 4 列并删除加倍"\t"使用sed。例如：
finger | awk -F"\t" -v 'OFS=\t' '{ =""; print ##代码##}' | sed 's/\t\{2,\}/\t/'
