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有一文本文件GB,其内容如下:
G B
******************P= 0
136.30960 190.38455
******************P= 5
135.76039 208.61451
******************P= 10
133.39609 226.06144
******************P= 15
134.53847 245.16657
******************P= 20
133.88448 262.35811
******************P= 25
133.11973 279.86156
******************P= 30
133.67447 296.10937
******************P= 35
132.70722 313.04688
******************P= 40
130.90885 329.09339
******************P= 45
130.36599 344.65525
******************P= 50
128.04804 359.38771
******************P= 55
126.70398 373.88764
我想提取******************P= 5所在行号,命令行编写如下:
n=$(grep -nH "******************P= "5 /home/wangyl86/work/result/GB | awk -F: '{print $2}')
其中,/home/wangyl86/work/result/GB是GB所在目录路径,执行过程如下:
[wangyl@:~] $ n=$(grep -nH "******************P= "5 /home/wangyl86/work/result/GB | awk -F: '{print $2}')
[wangyl@:~] $ echo $n
4 22 24 26 28
正确结果应该只有4才对,烦请各位老大指教一二, 不胜感激!! |
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