如果使用awk在文件中找到匹配项，则在匹配后打印第"n"行的第"m"列 [英] print "m"th column of "n"th line after a match if found in a file using awk

问题描述

在文件中找到特定的图案后，我需要打印第六行.在第6行中，我只想打印第5列.我可以使用以下命令来解决此问题的第一部分，

I need to print 6th line after a particular pattern is found in a file. In that 6th line, I want to print only the 5th column. I can do the first part of this problem using the following command,

awk 'c&&!--c;/pattern/{c=6}' file

但是我找不到一种方法来修改它以仅打印此第6行的第5列.任何帮助将不胜感激.

but I can't find a way to modify it to print just the 5th column of this 6th line instead. Any help will be greatly appreciated.

推荐答案

您可以使用内置的

You can use the built-in NR variable for this

awk '/pattern/ { nrs[NR + 6] = 1; } NR in nrs { print $5; delete nrs[NR] }'

这将测试pattern，并在其行号加六个(NR + 6)的数组中进行输入.然后，我们对该数组进行简单查找，以查看当前行号是否是我们要打印的行号(nrs[NR] == 1)，然后打印第5列(print $5)，然后清理该数组.

This will test for pattern and make an entry in an array of the it's line number plus six (NR + 6). We then do a simple lookup on that array to see if our current line-number is one we want to print (nrs[NR] == 1) and then print the 5th column (print $5) and then clean up the array.

此解决方案考虑了以下事实:在给定的6行范围内，模式可能会多次出现.

This solution accounts for the fact that a pattern might occur multiple times within any given 6 line range.

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