java - 如何在java中打开文件并解析每一行

Question

我有一个包含如下数据的文本文件:

sender id: 1/sequence number: 0/depth: 600/sending time: 13/05/2018 11:17:17 ص/data: Hello I am AUV: 1 this is aggregation # 0 from my sideEnd-End Delay:0.357072263241232 sender id: 0/sequence number: 0/depth: 700/sending time: 13/05/2018 11:17:17 ص/data: Hello I am AUV: 0 this is aggregation # 0 from my sideEnd-End Delay:0.293693984840975 sender id: 2/sequence number: 0/depth: 500/sending time: 13/05/2018 11:17:17 ص/data: Hello I am AUV: 2 this is aggregation # 0 from my sideEnd-End Delay:0.421466419584594 sender id: 3/sequence number: 0/depth: 400/sending time: 13/05/2018 11:17:17 ص/data: Hello I am AUV: 3 this is aggregation # 0 from my sideEnd-End Delay:0.486473186901286 sender id: 3/sequence number: 1/depth: 400/sending time: 13/05/2018 11:17:18 ص/data: Hello I am AUV: 3 this is aggregation # 1 from my sideEnd-End Delay:0.486473186901286 sender id: 1/sequence number: 0/depth: 600/sending time: 13/05/2018 11:17:21 ص/data: Hello I am AUV: 1 this is aggregation # 0 from my sideEnd-End Delay:0.32221205728025 sender id: 0/sequence number: 0/depth: 700/sending time: 13/05/2018 11:17:21 ص/data: Hello I am AUV: 0 this is aggregation # 0 from my sideEnd-End Delay:0.267329486404354

每行以结束延迟结束，例如 line#1:

sender id: 1/sequence number: 0/depth: 600/sending time: 13/05/2018 11:17:17 ص/data: Hello I am AUV: 1 this is aggregation # 0 from my sideEnd-End Delay:0.357072263241232

我需要将此文件分成其他文件，其中每个文件对应于特定的发件人 ID。

每个文件将仅包含每个发件人 ID 的端到端延迟值，例如 0.357072263241232。

我有 4 个发件人，因此将生成 4 个文件...

有什么帮助或提示吗？

我没有正则表达式的经验!

Answer 1

尝试正则表达式:sender\sid:\s(\d+).*?End-End\sDelay:(\d+(?:\.\d+)?)

Demo

组 1 将具有发件人 ID

第 2 组将具有端到端延迟值