使用 sed 从日志文件中 grep 模式

Question

我有以下日志行

2021-10-28 10:26:19,624 INFO [Native-Thread-7f9e6effd700] idol.nifi.connector.GetWeb GetWeb[id=c59c415c-017c-1000-195c-e85818b0a032] Processing: [depth:0] https://qed.qld.gov.au/about-us/rti/disclosure-log/disclosure-log-2015

我想提取日期和单词 Processing: 之后的所有内容，这样看起来像

2021-10-28 10:26:19 [depth:0] https://qed.qld.gov.au/about-us/rti/disclosure-log/disclosure-log-2015

我不确定如何使用 grep 或 sed 实现此目的？

#!/bin/bash cat ./logs/*.log | grep Processing: | grep -E "(\d\d\d\d-\d\d-\d\d \d\d:\d\d:\d\d)"

据我所知。

Answer 1

sed 's/,.*Processing://' file

输出：

2021-10-28 10:26:19 [depth:0] https://qed.qld.gov.au/about-us/rti/disclosure-log/disclosure-log-2015

参见：man sed 和 The Stack Overflow Regular Expressions FAQ

使用 sed 从日志文件中 grep 模式

using sed to grep pattern from log file

grep

sed