bash 如何使用 grep 和 regex 匹配特定长度的单词？
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How can I use grep and regex to match a word with specific length?
提问by BubbleMonster
I'm using Linux's terminal and i've got a wordlist which has words like:
我正在使用 Linux 的终端，我有一个单词列表，其中包含以下单词：
filers
filing
filler
filter
finance
funky
fun
finally
futuristic
fantasy
fabulous
fill
fine
And I want to do a grep and a regex to match the find words with the first two letters "fi" and only show the word if it's 6 characters in total.
我想做一个 grep 和一个正则表达式来匹配查找单词与前两个字母“fi”，并且只显示总共 6 个字符的单词。
I've tried:
我试过了：
cat wordlist | grep "^fi"
This shows the words beginning with fi.
这显示了以 fi 开头的单词。
I've then tried:
然后我尝试过：
cat wordlist | grep -e "^fi{6}"
cat wordlist | grep -e "^fi{0..6}"
and plenty more, but it's not bring back any results. Can anyone point me in the right direction?
还有更多，但它不会带回任何结果。任何人都可以指出我正确的方向吗？
回答by choroba
It's fiand four more characters:
它fi和另外四个字符：
grep '^fi....$'
or shorter
或更短
grep '^fi.\{4\}$'
or
或者
grep -E '^fi.{4}$'
$matches at the end of line.
$匹配行尾。
回答by Zoltan Ersek
Solution:
解决方案：
cat wordlist | grep -e "^fi.{4}$"
Your try:
你的尝试：
cat wordlist | grep -e "^fi{6}"
This means f and i six times, the dot added above means any charater, so it's fi and any character 4 times. I've also put an $ to mark the end of the line.
这意味着 f 和 i 六次，上面添加的点表示任何字符，所以它是 fi 和任何字符 4 次。我还放了一个 $ 来标记该行的结尾。
回答by Maroun
Try this:
尝试这个：
grep -P "^fi.{4,}"
Note that since you already have "fi", you only need at least 4 more characters.
请注意，由于您已经有了“fi”，因此您只需要至少 4 个字符。
.denotes any character, and {4,}is to match that character 4 or more times.
.表示任何字符，并且{4,}要匹配该字符 4 次或更多次。
If you write grep -e "^fi{6}"as you did in your example, you're trying to match strings beginning with f, followed by 6 is.
如果您grep -e "^fi{6}"像在示例中那样编写，您将尝试匹配以 开头f，后跟 6i秒的字符串。
