spring-retry PatternMatcher 源码

spring-retry PatternMatcher 代码

文件路径：/src/main/java/org/springframework/classify/PatternMatcher.java

/*
 * Copyright 2006-2022 the original author or authors.
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *      https://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
package org.springframework.classify;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

import org.springframework.util.Assert;

/**
 * @author Dave Syer
 * @author Dan Garrette
 * @param <S> the type of the thing to match a pattern on
 */
public class PatternMatcher<S> {

	private Map<String, S> map = new HashMap<>();

	private List<String> sorted = new ArrayList<>();

	/**
	 * Initialize a new {@link PatternMatcher} with a map of patterns to values
	 * @param map a map from String patterns to values
	 */
	public PatternMatcher(Map<String, S> map) {
		super();
		this.map = map;
		// Sort keys to start with the most specific
		this.sorted = new ArrayList<>(map.keySet());
		this.sorted.sort((o1, o2) -> {
			String s1 = o1; // .replace('?', '{');
			String s2 = o2; // .replace('*', '}');
			return s2.compareTo(s1);
		});
	}

	/**
	 * Lifted from AntPathMatcher in Spring Core. Tests whether or not a string matches
	 * against a pattern. The pattern may contain two special characters:<br>
	 * '*' means zero or more characters<br>
	 * '?' means one and only one character
	 * @param pattern pattern to match against. Must not be <code>null</code>.
	 * @param str string which must be matched against the pattern. Must not be
	 * <code>null</code>.
	 * @return <code>true</code> if the string matches against the pattern, or
	 * <code>false</code> otherwise.
	 */
	public static boolean match(String pattern, String str) {
		char[] patArr = pattern.toCharArray();
		char[] strArr = str.toCharArray();
		int patIdxStart = 0;
		int patIdxEnd = patArr.length - 1;
		int strIdxStart = 0;
		int strIdxEnd = strArr.length - 1;
		char ch;

		boolean containsStar = pattern.contains("*");

		if (!containsStar) {
			// No '*'s, so we make a shortcut
			if (patIdxEnd != strIdxEnd) {
				return false; // Pattern and string do not have the same size
			}
			for (int i = 0; i <= patIdxEnd; i++) {
				ch = patArr[i];
				if (ch != '?') {
					if (ch != strArr[i]) {
						return false;// Character mismatch
					}
				}
			}
			return true; // String matches against pattern
		}

		if (patIdxEnd == 0) {
			return true; // Pattern contains only '*', which matches anything
		}

		// Process characters before first star
		while ((ch = patArr[patIdxStart]) != '*' && strIdxStart <= strIdxEnd) {
			if (ch != '?') {
				if (ch != strArr[strIdxStart]) {
					return false;// Character mismatch
				}
			}
			patIdxStart++;
			strIdxStart++;
		}
		if (strIdxStart > strIdxEnd) {
			// All characters in the string are used. Check if only '*'s are
			// left in the pattern. If so, we succeeded. Otherwise failure.
			for (int i = patIdxStart; i <= patIdxEnd; i++) {
				if (patArr[i] != '*') {
					return false;
				}
			}
			return true;
		}

		// Process characters after last star
		while ((ch = patArr[patIdxEnd]) != '*' && strIdxStart <= strIdxEnd) {
			if (ch != '?') {
				if (ch != strArr[strIdxEnd]) {
					return false;// Character mismatch
				}
			}
			patIdxEnd--;
			strIdxEnd--;
		}
		if (strIdxStart > strIdxEnd) {
			// All characters in the string are used. Check if only '*'s are
			// left in the pattern. If so, we succeeded. Otherwise failure.
			for (int i = patIdxStart; i <= patIdxEnd; i++) {
				if (patArr[i] != '*') {
					return false;
				}
			}
			return true;
		}

		// process pattern between stars. padIdxStart and patIdxEnd point
		// always to a '*'.
		while (patIdxStart != patIdxEnd && strIdxStart <= strIdxEnd) {
			int patIdxTmp = -1;
			for (int i = patIdxStart + 1; i <= patIdxEnd; i++) {
				if (patArr[i] == '*') {
					patIdxTmp = i;
					break;
				}
			}
			if (patIdxTmp == patIdxStart + 1) {
				// Two stars next to each other, skip the first one.
				patIdxStart++;
				continue;
			}
			// Find the pattern between padIdxStart & padIdxTmp in str between
			// strIdxStart & strIdxEnd
			int patLength = (patIdxTmp - patIdxStart - 1);
			int strLength = (strIdxEnd - strIdxStart + 1);
			int foundIdx = -1;
			strLoop: for (int i = 0; i <= strLength - patLength; i++) {
				for (int j = 0; j < patLength; j++) {
					ch = patArr[patIdxStart + j + 1];
					if (ch != '?') {
						if (ch != strArr[strIdxStart + i + j]) {
							continue strLoop;
						}
					}
				}

				foundIdx = strIdxStart + i;
				break;
			}

			if (foundIdx == -1) {
				return false;
			}

			patIdxStart = patIdxTmp;
			strIdxStart = foundIdx + patLength;
		}

		// All characters in the string are used. Check if only '*'s are left
		// in the pattern. If so, we succeeded. Otherwise failure.
		for (int i = patIdxStart; i <= patIdxEnd; i++) {
			if (patArr[i] != '*') {
				return false;
			}
		}

		return true;
	}

	/**
	 * <p>
	 * This method takes a String key and a map from Strings to values of any type. During
	 * processing, the method will identify the most specific key in the map that matches
	 * the line. Once the correct is identified, its value is returned. Note that if the
	 * map contains the wildcard string "*" as a key, then it will serve as the "default"
	 * case, matching every line that does not match anything else.
	 *
	 * <p>
	 * If no matching prefix is found, a {@link IllegalStateException} will be thrown.
	 *
	 * <p>
	 * Null keys are not allowed in the map.
	 * @param line An input string
	 * @return the value whose prefix matches the given line
	 */
	public S match(String line) {

		S value = null;
		Assert.notNull(line, "A non-null key must be provided to match against.");

		for (String key : this.sorted) {
			if (PatternMatcher.match(key, line)) {
				value = this.map.get(key);
				break;
			}
		}

		if (value == null) {
			throw new IllegalStateException("Could not find a matching pattern for key=[" + line + "]");
		}
		return value;

	}

}

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