如何使用 sed 删除从第一行开始到遇到模式 '[ERROR] -17-12-2015' 之前的行？

我需要在遇到模式“[ERROR] -17-12-2015”之前删除从第一行到该行的行 目前我正在尝试以下命令，但不幸的是它没有找到模式本身：

  sed '1,/[ERROR] -17-12-2015/d' errLog

这里有什么问题吗？

其次，上面的脚本还将删除包含模式 '[ERROR] -17-12-2015' 的行，是否可以仅删除从第一行到遇到此模式之前的行？

示例输入是：

[ERROR] -09-11-2015 05:22:17 : : XMLrequest failed: You do not have access 
[ERROR] -09-11-2015 05:22:18 : : XMLrequest failed: You do not have access to period 2015/12, XMLrequest received: <?xml version="1.0" encoding="UTF-8"?>
<MatchingRequest version="12.0"><StartBackground>

[ERROR] -17-12-2015 05:22:18 : : XMLrequest failed: You do not have access 
[ERROR] -17-12-2015 05:22:18 : : XMLrequest failed: You do not have access
[ERROR] -17-12-2015 05:22:18 : : XMLrequest failed: You do not have access
[ERROR] -09-11-2015 05:22:18 : : XMLrequest failed: You do not have access , XMLrequest received: <?xml version="1.0" encoding="UTF-8"?>
<MatchingRequest version="12.0">

预期输出：

[ERROR] -17-12-2015 05:22:18 : : XMLrequest failed: You do not have access 
[ERROR] -17-12-2015 05:22:18 : : XMLrequest failed: You do not have access
[ERROR] -17-12-2015 05:22:18 : : XMLrequest failed: You do not have access
[ERROR] -09-11-2015 05:22:18 : : XMLrequest failed: You do not have access , XMLrequest received: <?xml version="1.0" encoding="UTF-8"?>
<MatchingRequest version="12.0">

你可以尝试一下：

$ cat test.txt
2015-12-03 17:20:36
2015-12-03 17:20:39
2015-12-03 17:21:23
[ERROR] -17-12-2015 something something
testing
testing again

$ sed '/\[ERROR\] -17-12-2015/,$!d' test.txt
[ERROR] -17-12-2015 something something
testing
testing again

$ sed '/\[ERROR\] -17-12-2015/,$!d' test.txt > tmpfile && mv tmpfile test.txt

$ cat test.txt
[ERROR] -17-12-2015 something something
testing
testing again

备用：

$ sed -n '/\[ERROR\] -17-12-2015/,$p' test.txt

这意味着仅从与字符串匹配的行开始打印 (p) 到文件末尾 ($)。 -n 表示默认不打印行。