我有大文件，我需要从中提取特定值。我在找＆＃34; Brontes＆＃34;在其中和服务开始时的字。
2016-09-05 11:23:08,022 ERROR [xxx.xxx.xx] (MSC service thread 1-6) XXXX015875: XXXX AS 7.1.1.Final "Brontes" started (with errors) in 275637ms - Started 1221 of 1307 services (3 services failed or missing dependencies, 71 services are passive or on-demand)
我试图像这样打印：
awk '/Brontes/{print $18}' file
并且结果是正确的：275637ms
但是当字符串看起来略有不同（列数量发生变化）时，结果将不相同。
如何实现相同的结果，但不依赖于列数量？
答案 0 :(得分：1)
示例输入：
echo $ola
2016-09-05 11:23:08,022 ERROR [xxx.xxx.xx] (MSC service thread 1-6) XXXX015875: XXXX AS 7.1.1.Final Brontes started (with errors) in 275637ms - Started 1221 of 1307 services (3 services failed or missing dependencies, 71 services are passive or on-demand)
使用awk的解决方案：扫描每列并打印包含ms的列。
echo $ola |awk '/Brontes/{for(i=1;i<=NF;i++) if( $i ~ /[0-9]+ms/) print $i}'
275637ms
答案 1 :(得分：1)
grep -oP 'Brontes.* \K\d+ms' file
这将打印包含字符ms的行之前以Brontes结尾的所有数字