[英]regular expression to extract number from string
I want to extract number from string. 我想从字符串中提取数字。 This is the string
这是字符串
#all/30
All I want is 30
. 我只想要
30
。 How can I extract? 如何提取? I try to use :
我尝试使用:
echo "#all/30" | sed 's/.*\/([^0-9])\..*//'
But nothing happen. 但是什么也没发生。 How should I write for the regular expression?
我应该如何写正则表达式? Sorry for bad english.
对不起,英语不好。
You may consider using grep to extract the numbers from a simple string like this. 您可以考虑使用grep从这样的简单字符串中提取数字。
echo "#all/30" | grep -o '[0-9]\+'
-o
option shows only the matching part that matches the pattern. -o
选项仅显示与模式匹配的匹配部分。 You could try the below sed command, 您可以尝试以下sed命令,
$ echo "#all/30" | sed 's/[^0-9]*\([0-9]\+\)[^0-9]*/\1/'
30
[^0-9]*
[^...]
is a negated character class. [^0-9]*
[^...]
是一个否定的字符类。 It matches any character but not the one inside the negated character class. [^0-9]*
matches zero or more non-digit characters. [^0-9]*
匹配零个或多个非数字字符。 \\([0-9]\\+\\)
Captures one or more digit characters. \\([0-9]\\+\\)
捕获一个或多个数字字符。 [^0-9]*
Matches zero or more non-digit characters. [^0-9]*
匹配零个或多个非数字字符。 30
30
echo "all/30" | sed 's/[^0-9]*\/\([0-9][0-9]*\)/\1/'
Avoid writing '.*' as it consumes entire string. 避免写“。*”,因为它会消耗整个字符串。 Default matches are always greedy .
默认匹配项总是贪婪的 。
echo "all/30" | sed 's/[^0-9]*//g'
# OR
echo "all/30" | sed 's#.*/##'
# OR
echo "all/30" | sed 's#.*\([0-9]*\)#\1#'
without more info about possible input string we can only assume that structure is #all/
followed by the number (only) 没有更多关于可能的输入字符串的信息,我们只能假设结构为
#all/
后跟数字(仅)
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