bash 在bash中给定文本中的两个给定不同分隔符之间提取文本
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Extract text between two given different delimiters in a given text in bash
提问by dineshdileep
I have a line of text which looks like hh^ay-pau+h@ow, I want to extract the text between -and +which in this case is pau. This should be done in bash. Any help would be appreciated.
EDIT: I want to extract the text between the first occurence of the tokens
PS: My google search didn't take me anywhere. I apologize if this question is already asked.
我有一行文字看起来像hh^ay-pau+h@ow，我想提取文本之间-，并+在这种情况下pau。这应该在 bash 中完成。任何帮助，将不胜感激。编辑：我想提取第一次出现的标记之间的文本 PS：我的谷歌搜索没有带我去任何地方。如果已经问过这个问题，我深表歉意。
回答by Bernhard
The way to do this in pure bash, is by using parameter expansions in bash
在 pure 中执行此操作的方法bash是在 bash 中使用参数扩展
$ a=hh^ay-pau+h@ow
$ b=${a%%+*}
$ c=${b#*-}
$ echo $c
pau
b: remove everything including and behind the first +occurence
c: remove everything excluding and before the first -ocurrence
b：删除第一次出现之后的+所有内容 c：删除第一次-出现之前和之前的所有内容
More info about substring removing in bash parameter expansion
有关在 bash 参数扩展中删除子字符串的更多信息
回答by Nicolae Namolovan
If you have only one occurence of -and +you can use cut:
如果您只出现一次-并且+可以使用 cut：
$ echo "hh^ay-pau+h@ow" | cut -d "-" -f 2 | cut -d "+" -f 1
pau
回答by iruvar
Assuming one occurence of +and -, you can stick to bash
假设+和 出现一次-，您可以坚持bash
IFS=+- read -r _ x _ <<<'hh^ay-pau+h@ow'
echo $x
pau
回答by Vytenis Bivainis
Try
尝试
grep -Po "(?<=\-).*?(?=\+)"
For example,
例如，
echo "hh^ay-pau+h@ow" | grep -Po "(?<=\-).*?(?=\+)"
回答by eduffy
If you're guarenteed to only have one -and one +.
如果你保证只有一-和一+。
% echo "hh^ay-pau+h@ow" | sed -e 's/.*-//' -e 's/+.*//'
pau
回答by Emilio Galarraga
echo "hh^ay-pau+h@ow" | awk -F'-' '{print }' |awk -F'+' '{print }'
