bash regexp (sed) 抑制“不匹配”输出
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regexp (sed) suppress "no match" output
提问by yPhil
I'm stuck on that and can't wrap my head around it: How can I tell sed to return the value found, and otherwise shut up?
我被困在这个问题上并且无法解决它：我怎么能告诉 sed 返回找到的值，否则闭嘴？
It's really beyond me: Why would sed return the whole string if he found nothing? Do I have to run anothertest on the returned string to verify it? I tried using "-n" from the (very short) man page but it effectively suppresses all output, including matched strings.
实在是想不通了：sed 一无所获，为什么还要返回整个字符串呢？我是否必须对返回的字符串运行另一个测试来验证它？我尝试使用（非常短的）手册页中的“-n”，但它有效地抑制了所有输出，包括匹配的字符串。
This is what I have now :
这就是我现在所拥有的：
echo plop-02-plop | sed -e 's/^.*\(.\)\([0-9][0-9]\).*$//'
which returns
02(and that is fine and dandy, thank you very much), but:
返回
02（这很好，很花哨，非常感谢），但是：
echo plop-02plop | sed -e 's/^.*\(.\)\([0-9][0-9]\).*$//'
returns
plop-02plop(when it should return this = "" nothing! Dang, you found nothing so be quiet!
For crying out loud !!)
返回
plop-02plop（什么时候应该返回 this = ""什么都没有！Dang，你什么都没发现，所以安静点！大声哭！）
I tried checking for a return value, but this failed too ! Gasp !!
我尝试检查返回值，但这也失败了！喘气！！
$ echo plop-02-plop | sed -e 's/^.*\(.\)\([0-9][0-9]\).*$//' ; echo $?
02
0
$ echo plop-02plop | sed -e 's/^.*\(.\)\([0-9][0-9]\).*$//' ; echo $?
plop-02plop
0
$
This last one I cannot even believe. Is sedreally the tool I should be using? I want to extract a needle from a haystack, and I want a needle or nothing..?
这最后一个我什至无法相信。是sed真的，我应该使用的工具？我想从大海捞针，我想要一根针还是什么都不想要..？
回答by przemoc
sed by default prints all lines.
sed 默认打印所有行。
What you want to do is
你想做的是
/patt/!d;s//repl/
IOW delete lines not matching your pattern, and if they match, extract particular element from it, giving capturing group number for instance. In your case it will be:
IOW 删除与您的模式不匹配的行，如果它们匹配，则从中提取特定元素，例如提供捕获组编号。在您的情况下，它将是：
sed -e '/^.*\(.\)\([0-9][0-9]\).*$/!d;s///'
You can also use -noption to suppress echoing all lines. Then line is printed only when you explicitly state it. In practice scripts using -nare usually longer and more cumbersome to maintain. Here it will be:
您还可以使用-n选项来抑制回显所有行。然后仅当您明确说明时才打印 line。在实践中，使用-n的脚本通常更长，维护起来也更麻烦。这将是：
sed -ne 's/^.*\(.\)\([0-9][0-9]\).*$//p'
There is also grep, but your example shows, why sed is sometimes better.
还有 grep，但您的示例显示了为什么 sed 有时更好。
回答by sehe
Perhaps you can use egrep -o?
也许你可以使用egrep -o？
input.txt:
输入.txt：
blooody
aaaa
bbbb
odor
qqqq
E.g.
例如
sehe@meerkat:/tmp$ egrep -o o+ input.txt
ooo
o
o
sehe@meerkat:/tmp$ egrep -no o+ input.txt
1:ooo
4:o
4:o
Of course egrepwill have slightly different (better?) regex syntax for advanced constructs (back-references, non-greedy operators). I'll let you do the translation, if you like the approach.
当然egrep，高级构造（反向引用、非贪婪运算符）的正则表达式语法会略有不同（更好？）。如果你喜欢这种方法，我会让你做翻译。
