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我正在处理一个日志文件，我想从特定日期打印到它结束。例如，该特定日期是 ($sd=27/Dec/2002)。现在我想搜索这一天并从它打印到日志文件结束！但是如果 27/Dec 不在日志文件中的项目中怎么办？它应该搜索项目 >= $sd (27/Dec) ，我该怎么做？

此代码仅搜索 $Sd 即 27/Dec/2002 ，我想搜索项目 >= $sd

sed -n "$(awk '/'$sd'/ {print NR}' serverlog.log.log | head -1),$ p" serveerlog.log|cut -d: -f1

日志文件示例：

213.64.237.213 - - [23/Dec/2002:03:02:22 +0100]
213.132.36.66 - - [28/Dec/2002:19:33:29 +0100]

并且日志文件已排序！

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1 回答 1

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使用 awk 会很容易。请参见下面的示例：

kent$  cat log.txt
213.64.237.213 - - [20/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [20/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [20/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [20/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [20/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [23/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [23/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [23/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [25/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [25/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [25/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [25/Dec/2002:03:02:22 +0100]
213.132.36.66 - - [28/Dec/2002:19:33:29 +0100]

kent$  sd=21/Dec/2002




kent$  awk -F'[:[]' -v d=$sd '$2>d' log.txt

输出

213.64.237.213 - - [23/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [23/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [23/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [25/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [25/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [25/Dec/2002:03:02:22 +0100]
213.64.237.213 - - [25/Dec/2002:03:02:22 +0100]
213.132.36.66 - - [28/Dec/2002:19:33:29 +0100]

更新

试试这条 awk 行： $sd 是变量。希望它对你有用。

kent$  awk -F'[:[]' -v vd=$sd 'BEGIN{ gsub(/\//," ",vd);"date +%s -d \""vd"\""|getline d} {p=$0;  gsub(/\//," ",$2); "date +%s -d \""$2"\""|getline o;if(o>d) print p}' log.txt
于 2011-09-28T15:31:26.507 回答