Linux 如果文件中有匹配的字符串，则获取部分行的文本

Question

在 Linux 中，我有一个文件 test.log 与此类似：

2021/11/18 17:19:18,034 INFO {"queueName":"queue/RequestQueue",{"threadName":"WorkManager(2)-702","correlationID":"b67601e81bfd","requestData":"TYPE_1, REQUEST 1"}
2021/11/18 17:19:18,036 INFO {"queueName":"queue/ResponseQueue","correlationID":"TMYHxIyeYo","responseData":"TYPE_1, RESPONSE 1"} 
2021/11/18 17:19:18,038 INFO {"queueName":"queue/RequestQueue",{"threadName":"WorkManager(2)-885","correlationID":"j9BNzbbv3E","requestData":"TYPE_2, REQUEST 2"}
2021/11/18 17:19:19,172 INFO {"queueName":"queue/RequestQueue",{"threadName":"WorkManager(2)-183","correlationID":"d29d2d7cf5a4","requestData":"TYPE_1, REQUEST 3"}
2021/11/18 17:19:20,784 INFO {"queueName":"queue/esbRequestQueue",{"threadName":"WorkManager(2)-762","correlationID":"4235eba2765a","requestData":"TYPE_1, REQUEST 4"}

我想搜索此文件的所有行并获取包含字符串 ["requestData":"TYPE_1] 的行，并将 requestData 的所有内容写入另一个名为 [=20= 的文件]。test.log 文件中的每个匹配行将写入 type_1.txt 文件中的一行。下面是我在 type_1.txt 文件中的预期结果：

TYPE_1, REQUEST 1
TYPE_1, REQUEST 3
TYPE_1, REQUEST 4

我的问题是有没有可以做到这一点的命令？我是 Linux 的新手，所以请帮助我。

Answer 1

试试这个命令：

grep -Po '(?<="requestData":")TYPE_1[^"]*' test.log > type_1.txt

Linux 如果文件中有匹配的字符串，则获取部分行的文本

Linux get part of line's text if matched string in files

linux

string

grep

centos

file