我想用条件解析输入：
<AD(1 to n)>并<AD> to </AD>下面是我的代码和输入文本。但是，我没有得到预期的输出。我得到的输出如下
[T, > Primary , ddress , /, T, >, > Primary , ddress Line , ~Primary , ddress Line , ~Primary , ddress Line , ~Primary , ddress Line , ~City~State~Zip Code~Country, /, >; , T, > Office , ddress , /, T, >, > Primary , ddress Line , ~Primary , ddress Line , ~Primary , ddress Line , ~Primary , ddress Line , ~City~State~Zip Code~Country, /, >; , T, > Home , ddress , /, T, >, > Primary , ddress Line , ~Primary , ddress Line , ~Primary , ddress Line , ~Primary , ddress Line , ~City~State~Zip Code~Country, /, >]
public class TestProject {
public static void main(String[] args) {
String address ="<AT1> Primary Address </AT1><AD1> Primary Address Line 1 ~Primary Address Line 2 ~Primary Address Line 3~Primary Address Line 4~City~State~Zip Code~Country</AD1>;\n"
+ "<AT2> Office Address </AT2><AD2> Primary Address Line 1 ~Primary Address Line 2 ~Primary Address Line 3~Primary Address Line 4~City~State~Zip Code~Country</AD2>;\n"
+ "<AT3> Home Address </AT3><AD2> Primary Address Line 1 ~Primary Address Line 2 ~Primary Address Line 3~Primary Address Line 4~City~State~Zip Code~Country</AD2>";
System.out.println(captureValues(address));
}
public static Pattern ptn = Pattern.compile("[^<AD0-9]+");
public static List<String> captureValues(String address)
{
Matcher mtch = ptn.matcher(address);
List<String> ips = new ArrayList<String>();
while(mtch.find())
{
ips.add(mtch.group());
}
return ips;
}
}
正则表达式可能不是最好的工具，但是在给定输入的情况下尝试：
Pattern ptn = Pattern.compile("(?<=<AD(\\d)>).+?(?=</AD\\d>)");
意思是：
(?<=<AD(\\d)>)-积极的背后<AD(digit)>，.+? -任何一个或多个字符，但带有延迟量词，因此它将匹配字符的最小数量，以使整个正则表达式匹配，(?=</AD\\d>) -积极向前看 </AD(digit)>第一个数字在捕获组中(\\d)，因此您仍然可以通过mtch.group(1)方法来获取它。如果不需要，可以删除括号。
您正则表达式[^<AD0-9]+是指一个或多个任意字符而不是：数字（0-9）<，A或者D，这就是为什么你会得到无效的输出，它砍你输入的每处地方的正则表达式不匹配。
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