Question

测试字符串是

It is a test data
%Warning: portfast should only be enabled on ports connected to a single
 host. Connecting hubs, concentrators, switches, bridges, etc. to this
 interface when portfast is enabled can cause temporary bridging loops.
 Use with CAUTION
Test is a test.

如果字符串中包含%符号，则该字符串无效。但如果%后面跟着以下字符序列，则有效，“％警告：只应在连接到单个主机的端口上启用portfast。连接集线器，集中器，交换机，网桥等。当启用portfast时，此接口可能会导致临时桥接循环。\ n请小心使用“。即使在这种情况下，'％'符号也只能出现一次。

我的代码是：

public class Abc {

    public static void main(String[] args) {
        String reg="%(Warning: portfast should only be enabled on ports connected to a single\n host. Connecting hubs, concentrators, switches, bridges, etc. to this\n interface when portfast is enabled can cause temporary bridging loops.\n Use with CAUTION)([^%])";
        String str = "%Warning: portfast should only be enabled on ports connected to a single\n host. Connecting hubs, concentrators, switches, bridges, etc. to this\n interface when portfast is enabled can cause temporary bridging loops.\n Use with CAUTIONTest is a testData is invalid%";
        Pattern regEx2 = Pattern
                .compile(reg);
        Matcher matcher = regEx2.matcher(str);
        if (matcher.find()) {
            System.out.println("valid");
        } else {
            System.out.println("invalid");
        }
    }
}

但我没有得到预期的结果。我可以得到合适的正则表达式吗？

Answer 1

也许你可以将你的代码与另一个得到＆＃39;％＆＃39;

的表达式合并

 public static void main(String[] args) {
     String reg = "%Warning: portfast should only be enabled on ports connected to a single\n host. Connecting hubs, concentrators, switches, bridges, etc. to this\n interface when portfast is enabled can cause temporary bridging loops.\n Use with CAUTION";
     String str = "%Warning: portfast should only be enabled on ports connected to a single\n host. Connecting hubs, concentrators, switches, bridges, etc. to this\n interface when portfast is enabled can cause temporary bridging loops.\n Use with CAUTIONTest is a testData is invalid%";
     Pattern regEx2 = Pattern.compile(reg);
     Matcher matcher = regEx2.matcher(str);
     int occurences = str.length() - str.replace("%", "").length();

     if (matcher.find() && occurences <= 1) {
        System.out.println("valid");
     } else {
        System.out.println("invalid");
     }

 }

如果你有StringUtils类，那么你可以使用它的方法 countMatches

Answer 2

好的，您的代码中存在两个问题：

find方法仅尝试在字符串中查找给定模式的出现。在您的情况下，您需要使用matches，它检查完整字符串是否与模式匹配。
([^%]) - 这只匹配一个不是%的字符。要匹配任意数量的字符，请使用([^%]*)。 *表示＆＃34; 0或更多＆＃34;，或者您可能需要+，而这意味着＆＃34; 1或更多＆＃34;。

结果代码如下所示：https://ideone.com/Q2y21b

public static void main(String[] args) {
    String reg="%(Warning: portfast should only be enabled on ports connected to a single\n host. Connecting hubs, concentrators, switches, bridges, etc. to this\n interface when portfast is enabled can cause temporary bridging loops.\n Use with CAUTION)([^%]+)";
    String str = "%Warning: portfast should only be enabled on ports connected to a single\n host. Connecting hubs, concentrators, switches, bridges, etc. to this\n interface when portfast is enabled can cause temporary bridging loops.\n Use with CAUTIONTest is a testData is invalid%";
    Pattern regEx2 = Pattern
            .compile(reg);
    Matcher matcher = regEx2.matcher(str);
    if (matcher.matches()) {
        System.out.println("valid");
    } else {
        System.out.println("invalid");
    }
}

我可以使用以下条件为java中的以下字符串获取适当的正则表达式

2 个答案: