我有一个看起来像这样的文件
0: jdbc:hive2://10.241.1.8:10000/> select number,exported_at from audit group by number,exported_at order by number limit 10;
789116704,20150907T130500Z
789190212,20150907T130500Z
789255093,20150907T130500Z
789282142,20150907T130500Z
789359510,20150907T130500Z
789363793,20150907T130500Z
789363843,20150907T130500Z
789369808,20150907T130500Z
789437014,20150907T130500Z
789437071,20150907T130500Z
0: jdbc:hive2://10.241.1.8:10000/>
我只想保留数字后跟逗号的那一行。即我的输出应该是这样的
789116704,20150907T130500Z
789190212,20150907T130500Z
789255093,20150907T130500Z
789282142,20150907T130500Z
789359510,20150907T130500Z
789363793,20150907T130500Z
789363843,20150907T130500Z
789369808,20150907T130500Z
789437014,20150907T130500Z
789437071,20150907T130500Z
我正在寻找可以完成这项工作的 sed 命令。非常感谢任何帮助。谢谢
最佳答案
grep -E '[0-9]+,' data.txt
其中 data.txt 是包含您的输入数据的文件
关于linux - sed 命令仅过滤后跟逗号的一组数字,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36768013/