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127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/.treeinfo HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Fedora HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Server HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Client HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/RedHat HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/CentOS HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/SL HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/directory.yast HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/current/images/MANIFEST HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/images/daily/MANIFEST HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/boot/platform/i86xpv/kernel/unix HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
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127.0.0.1 - - [08/Mar/2011:00:26:35 +0530] "HEAD /sk/.treeinfo HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:35 +0530] "HEAD /sk/Fedora HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:35 +0530] "HEAD /sk/Server HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:35 +0530] "HEAD /sk/Client HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:35 +0530] "HEAD /sk/RedHat HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
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我有一个上面结构的文件，名字是log.txt。 现在我做了
cat log.txt | grep '[*]'
没有输出 然后我做了
cat log.txt | grep '\[* \]'
再次没有输出 但是当我做的时候
cat log.txt | grep '\[.*\]'
然后有输出这是如何工作的？
UPTADE
我基本上是想在上面的日志中查找时间戳。
答案 0 :(得分：6)
cat log.txt | grep '[*]'
此命令行在文件log.txt中搜索其中包含*的行。由于输入中的所有行都没有*，因此您什么也得不到。请注意，正则表达式中的[..]称为字符类，它匹配其中的任何字符。
cat log.txt | grep '\[* \]'
此命令行在文件中搜索具有零个或多个文字[后跟空格并后跟文字]的行。你再也没有这样的台词。在这里，您已经转发了[和]。这使得它们非特殊。所以他们不再形成一个角色类，而是按字面意思对待。
cat log.txt | grep '\[.*\]'
这一行会搜索[后跟任何内容（.*）后跟]的行。由于你有这样的线，你得到输出。
最后
cat file | grep pattern
被称为无用的猫，可以写成：
grep pattern file
答案 1 :(得分：1)
在正则表达式中，“。”意思是“任何角色”而明星“*”的意思是：如果我前面的角色出现0次或更多次，请与我匹配 “[”和“]”用于限制一类字符
所以：
答案 2 :(得分：1)
grep使用正则表达式作为模式。你写的那个（最后一个）意味着
写\[*\]将意味着“'''字符任意次，然后是']'字符”。
由于[*]是一组字符的标记，因此写[]并不意味着什么。
man grep中有正则表达式的简短说明，您可能会发现更多online
答案 3 :(得分：1)