英文:
StackOverflowError when matching longer string using Regex - Optimization
问题
我正在使用正则表达式来匹配传入的文件内容，以检测具有以下模式的 ID：
AXXXXXXXXXX-MID-XX（其中 X = 长度为 10 和 2 的数字值）
这是我的正则表达式：(.|\n|\r)*(A[0-9]{10}-MID-[0-9]{2})(.|\n|\r)*
但是，当内容超过 1500 个字符时，我会收到堆栈溢出错误。
寻求帮助，看看这是否可以进行优化？
这是 Java 代码 -
String pattern1="(.|\n|\r)*(A[0-9]{10}-MID-[0-9]{2})(.|\n|\r)*";if(file_content.matches(pattern1)) {//...做一些操作 <-- 代码永远不会执行到这里。}
英文:
I am using Regex for matching incoming file content to detect an ID which has following pattern
AXXXXXXXXXX-MID-XX (Where X = numeric values with length 10 and 2)
Here's my Regex (.|\n|\r)*(A[0-9]{10}-MID-[0-9]{2})(.|\n|\r)*
But, when the content exceeds like 1500 characters, I get StackOverflow error.
Seeking help here to check if this look like something which can be optimized?
Here's the Java Code -
String pattern1="(.|\n|\r)*(A[0-9]{10}-MID-[0-9]{2})(.|\n|\r)*";if(file_content.matches(pattern1)) {//...Do something <-- The code never reaches here.}
答案1
得分: 0
你不必要地使用了 `(.|\n|\r)*`。以下是更清晰、更高效的方法：import java.util.regex.Matcher;import java.util.regex.Pattern;public class Main {public static void main(String[] args) {Pattern pattern = Pattern.compile("A[0-9]{10}-MID-[0-9]{2}");String fileContent = "A1234567890-MID-12"; // 将其替换为实际内容Matcher matcher = pattern.matcher(fileContent);while (matcher.find()) {String id = matcher.group(); // 如果你想获取IDSystem.out.println(id);// ...做点什么}}}**输出:**A1234567890-MID-12
英文:
You have unnecessarily used (.|\n|\r)*. Given below is a cleaner and more performant way of doing it:
import java.util.regex.Matcher;import java.util.regex.Pattern;public class Main {public static void main(String[] args) {Pattern pattern = Pattern.compile("A[0-9]{10}-MID-[0-9]{2}");String fileContent = "A1234567890-MID-12";// Replace it with actual contentMatcher matcher = pattern.matcher(fileContent);while (matcher.find()) {String id = matcher.group();// If you want to grab the IDSystem.out.println(id);// ...do something}}}
Output:
A1234567890-MID-12
答案2
得分: 0
Pattern pattern = Pattern.compile("A[0-9]{10}-MID-[0-9]{2}");
for(String line : lines) {
Matcher matcher = pattern.matcher(line);
if (matcher.find()) {String id = matcher.group();// ...执行某些操作 < -- 代码从未执行到这里。}
}
英文:
Pattern pattern = Pattern.compile("A[0-9]{10}-MID-[0-9]{2}");for(String line : lines) {Matcher matcher = pattern.matcher(line);if (matcher.find()) {String id = matcher.group();// ...Do something <-- The code never reaches here.}}
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