带有重复的简单正则表达式让我难住了
问候。
这么简单的问题却难倒了我。这里的人非常乐于助人。
我正在尝试匹配包含一些固定文本和随机数字的字符串。
echo blah blah abc123 | grep -o abc
abc
echo blah blah abc123 | grep -o abc[0-9]
abc1
echo blah blah abc123 | grep -o abc[0-9]+
echo blah blah abc123 | grep -o "abc[0-9]+"
echo blah blah abc123 | grep -o "abc[0-9]*"
abc123
echo blah blah abc123 | grep -o abc[0-9]{3}
echo blah blah abc123 | grep -o "abc[0-9]{3}"
* 运算符（匹配零次或多次）是唯一能按我的预期工作的运算符。
为什么 + 运算符（匹配 1 次或多次）不匹配？
为什么特定的重复计数运算符 {3} 不匹配？
如果有影响的话，我正在 Ubuntu 10.10 下的 bash shell 中运行这些示例。
非常感谢。
Greetings.
So simple a problem has me stumped. People here are so helpful.
I am trying to match a string containing some fixed text and random digits.
echo blah blah abc123 | grep -o abc
abc
echo blah blah abc123 | grep -o abc[0-9]
abc1
echo blah blah abc123 | grep -o abc[0-9]+
echo blah blah abc123 | grep -o "abc[0-9]+"
echo blah blah abc123 | grep -o "abc[0-9]*"
abc123
echo blah blah abc123 | grep -o abc[0-9]{3}
echo blah blah abc123 | grep -o "abc[0-9]{3}"
The * operator (matches zero or more times) is the only one that works as I would expect.
Why does the + operator (match 1 or more times) not match?
Why does the specific repetition count operator {3} not match?
I am running these examples in a bash shell under Ubuntu 10.10 if it makes a difference.
Thanks so much.
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当您转义特殊字符时，它们都会起作用：
未转义，正则表达式正在寻找文字
+或{，正如您可能已经推断出的那样。至于为什么必须保持
*不转义，但必须转义+，我不确定。They both work when you escape the special characters:
Unescaped, the regex is looking for a literal
+or{, as you have probably deduced.As to exactly why you have to keep a
*unescaped but you have to escape a+, I'm not sure.