返回单个匹配而不是一个长匹配正则表达式

Question

很确定在 SO 上必须有一个答案，但我的 google fu 让我失望了。

我有一个 js 文件，其中包含一个 javascript 字典数组，开头为：

var a = t.locales = [{
        countryCode: "AF",
        countryName: "Afghanistan"
    }, {
        countryCode: "AL",
        countryName: "Albania"
    },

在返回中没有空格（相对于上面显示的布局），因此带有国家名称的脚本部分将是以下内容的长版本：

[{countryCode:"AF",countryName:"Afghanistan"},{countryCode:"AL",countryName:"Albania"},{countryCode:"DZ",countryName:"Algeria"},{countryCode:"AS",countryName:"American Samoa"},{countryCode:"AD",countryName:"Andorra"},{countryCode:"AO",countryName:"Angola"},{countryCode:"AI",countryName:"Anguilla"},{countryCode:"AG",countryName:"Antigua & Barbuda"},{countryCode:"AR",countryName:"Argentina"},{countryCode:"AM",countryName:"Armenia"},{countryCode:"AW",countryName:"Aruba"},{countryCode:"AU",countryName:"Australia"},{countryCode:"AT",countryName:"Austria"},{countryCode:"AZ",countryName:"Azerbaijan"},{countryCode:"BS",countryName:"Bahamas"},{countryCode:"BH",countryName:"Bahrain"},{countryCode:"BD",countryName:"Bangladesh"},{countryCode:"BB",countryName:"Barbados"},{countryCode:"BY",countryName:"Belarus"},{countryCode:"BE",countryName:"Belgium"},{countryCode:"BZ",countryName:"Belize"},{countryCode:"BJ",countryName:"Benin"},{countryCode:"BM",countryName:"Bermuda"},{countryCode:"BT",countryName:"Bhutan"},{countryCode:"BO",countryName:"Bolivia"},{countryCode:"BQ",countryName:"Bonaire"},{countryCode:"BA",countryName:"Bosnia & Herzegovina"},{countryCode:"BW",countryName:"Botswana"}]

我想用正则表达式输出国家名称，例如'Afghanistan','Albania'......我似乎无法编写正则表达式模式，它会返回一个匹配列表而不是一个大的长匹配。

例如，

countryName:"(.*)"

这会返回一个贪婪的单一匹配，它不是单个国家的列表。

我很欣赏这可能是一件非常简单的事情，但是我尝试过的所有不同的正则表达式都失败了，例如p = re.compile(r'(?<=countryCode:")(.*)[^"]') 。任何人都可以提供适当的正则表达式和解释吗？

注意这是一个特定的我该如何处理正则表达式问题，而不是它是否是适合这项工作的工具。

基本上，我认为我需要在每次国家名称之后的“之前”定义一个模式（例如，而不是在最后一个国家名称之后的“，或者在某些情况下更远）

预期结果是来自该对象的国家/地区列表，例如

['Afghanistan','Albania',.....]

Python：

import re, requests

r = requests.get('https://www.nexmo.com/static/bundle.js')
p = re.compile(r'(?<=countryCode:")(.*)[^"]')     
countries = p.findall(r.text)
print(countries)

Answer 1

使用模式r'countryName:\"(.*?)\"'

例如：

import re
data = '[{countryCode:"AF",countryName:"Afghanistan"},{countryCode:"AL",countryName:"Albania"},{countryCode:"DZ",countryName:"Algeria"},{countryCode:"AS",countryName:"American Samoa"},{countryCode:"AD",countryName:"Andorra"},{countryCode:"AO",countryName:"Angola"},{countryCode:"AI",countryName:"Anguilla"},{countryCode:"AG",countryName:"Antigua & Barbuda"},{countryCode:"AR",countryName:"Argentina"},{countryCode:"AM",countryName:"Armenia"},{countryCode:"AW",countryName:"Aruba"},{countryCode:"AU",countryName:"Australia"},{countryCode:"AT",countryName:"Austria"},{countryCode:"AZ",countryName:"Azerbaijan"},{countryCode:"BS",countryName:"Bahamas"},{countryCode:"BH",countryName:"Bahrain"},{countryCode:"BD",countryName:"Bangladesh"},{countryCode:"BB",countryName:"Barbados"},{countryCode:"BY",countryName:"Belarus"},{countryCode:"BE",countryName:"Belgium"},{countryCode:"BZ",countryName:"Belize"},{countryCode:"BJ",countryName:"Benin"},{countryCode:"BM",countryName:"Bermuda"},{countryCode:"BT",countryName:"Bhutan"},{countryCode:"BO",countryName:"Bolivia"},{countryCode:"BQ",countryName:"Bonaire"},{countryCode:"BA",countryName:"Bosnia & Herzegovina"},{countryCode:"BW",countryName:"Botswana"}]'
countries = re.findall(r'countryName:\"(.*?)\"', data)
print(countries)

输出：

['Afghanistan',
 'Albania',
 'Algeria',
 'American Samoa',
 'Andorra',
 'Angola',
 'Anguilla',
 'Antigua & Barbuda',
 'Argentina',
 'Armenia',
 'Aruba',
 'Australia',
 'Austria',
 'Azerbaijan',
 'Bahamas',
 'Bahrain',
 'Bangladesh',
 'Barbados',
 'Belarus',
 'Belgium',
 'Belize',
 'Benin',
 'Bermuda',
 'Bhutan',
 'Bolivia',
 'Bonaire',
 'Bosnia & Herzegovina',
 'Botswana']

Answer 2

您需要更改您的正则表达式以使用(?<=countryName: ")[^"]+ 而不是您当前的。由于您当前使用的是.*，它贪婪地匹配所有内容，因此将匹配所有可能的内容，这就是您的情况。

试试这些 Python 代码，

import re

s = '''[{
        countryCode: "AF",
        countryName: "Afghanistan"
    }, {
        countryCode: "AL",
        countryName: "Albania"
    },'''

print(re.findall(r'(?<=countryName: ")[^"]+', s))

打印，

['Afghanistan', 'Albania']

Answer 3

使用您的第一个变体的非贪婪版本：

p = re.compile(r'countryName:"(.*?)"')     
countries = p.findall(text)

使用像"(.*)" 这样的贪婪匹配的问题是它会匹配到last " 的末尾。

{countryCode:"AF",countryName:"Afghanistan"},{countryCode:"AL",countryName:"Albania"}
                  ^match  ^ capture start ^ still matches .*      final match of " ^

但是，您希望它以最小匹配结束 - 由非贪婪匹配表示

{countryCode:"AF",countryName:"Afghanistan"},{countryCode:"AL",countryName:"Albania"}
                  ^match  ^ capture start ^ first match of "