无法弄清楚如何用AppleScript实现REGEX [英] Can't figure out how to implement REGEX with AppleScript

问题描述

我写了一个regex命令来查找并输出字符串中数字行的第一个实例:

I wrote a regex command to find and output the first instance of a line of digits in a string:

find:    ^[^\d]*(\d+).*
replace: $1

问题是，为了在AppleScript中实际使用它，我知道的唯一方法是调用Shell脚本并使用sed.我不知道如何以这种方式实际使用我的正则表达式.我已经尝试了好几个小时，没有任何运气.这与我所能接近的差不多，但是它返回字符串中的所有数字，而不是第一组数字:

The problem is that in order to actually utilize this in AppleScript, the only way I know of doing this is with calling a shell script and using sed. I can't figure out how to actually use my regex in this way. I've tried for hours without any luck. This is as close as I can get, but it returns ALL the numbers in a string, rather than the first group of numbers:

set num to do shell script "sed 's/[^0-9]*//g' <<< " & quoted form of input

我真正想要的是一种使用AppleScript与正则表达式一起工作并找到匹配替换项($ 1，$ 2等)的方法.

What I would really like is a way to use AppleScript to just WORK with regex and found match replacement ($1, $2, etc).

推荐答案

请注意， sed 不支持PCRE速记字符类，例如 \ d ，也不支持正则表达式在括号表达式内转义.

Note that sed does not support PCRE shorthand character classes like \d, nor does it support regex escapes inside bracket expressions.

此外，由于使用的是 sed 的POSIX BRE风格(不使用 -r 或 -E 选项)，因此可以定义捕获组中，您需要 \(... \)，而不是(...).

Also, since you use POSIX BRE flavor of sed (no -r or -E option is used), to define a capturing group, you need \(...\), not (...).

此外， + 与POSIX BRE模式中的文字 + 符号匹配，您需要对其进行转义，但为了安全起见，只需展开 a + 到 aa * .

Also, a + is matching a literal + symbol in POSIX BRE pattern, you need to escape it, but to play it safe, you can just expand a+ to aa*.

sed 中的替换反向引用语法为 \ +数字.

使用此POSIX BRE解决方案:

Use this POSIX BRE solution:

sed 's/^[^0-9]*\([0-9][0-9]*\).*/\1/'

，或者，如果使用 -E 或 -r 选项，则为POSIX ERE解决方案:

or, if you use -E or -r option, a POSIX ERE solution:

sed -E 's/^[^0-9]*([0-9]+).*/\1/'

详细信息

• ^ -字符串的开头
• [^ 0-9] * -0+个除数字以外的字符(也可以使用 [[:digit:]] * )
• \(-捕获组#1的开始(在替换模式中用 \ 1 占位符指代)(在ERE中，(将启动一个捕获组)
• [0-9] [0-9] * = [0-9] \ + (BRE)= [0-9] + (ERE)-1个以上数字
• \)-捕获组的末尾(在POSIX ERE中，))
• .* -该行的其余部分.

• ^ - start of string
• [^0-9]* - 0+ chars other than digits (also, you may use [[:digit:]]*)
• \( - start of a capturing group #1 (referred to with the \1 placeholder from the replacement pattern) (in ERE, ( will start a capturing group)
• [0-9][0-9]* = [0-9]\+ (BRE) = [0-9]+ (ERE) - 1+ digits
• \) - end of the capturing group (in POSIX ERE, ))
• .* - the rest of the line.

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