我在数据库中有一个包含4个字符串列的表，每行代表被阻止的用户或组。
referrer | ip | userAgent | email
按组分组我的意思是其中一列（任意）可以有通配符（星号），表示“阻止所有这些”
例如此行
www.google.com | 127.0.0.2 | * | yahoo.com
表示每个用户请求“google.com”作为引荐来源，“127.0.0.2”asIP＆amp; “yahoo.com”作为电子邮件，需要在不考虑UserAgent的情况下被阻止，因为它有通配符
以下代码适用于O（n）复杂度，这对于小尺寸表来说已经足够了，但我的表包含超过百万行
class BlacklistEntry {
private String referrer, ip, userAgent, email;
private static List<BlacklistEntry> cache = new ArrayList<>();
BlacklistEntry(String referrer, String ip, String userAgent, String email) {
this.referrer = referrer;
this.ip = ip;
this.userAgent = userAgent;
this.email = email;
}
private static boolean isBlacklisted(String ref, String ip, String ue, String email) {
final String MATCH_ALL = "*";
return cache.stream()
.filter(e ->
(MATCH_ALL.equals(e.getReferrer()) || e.getReferrer().equals(ref)) &&
(MATCH_ALL.equals(e.getIp()) || e.getIp().equals(ip)) &&
(MATCH_ALL.equals(e.getUserAgent()) || e.getUserAgent().equals(ue)) &&
(MATCH_ALL.equals(e.getEmail()) || e.getEmail().equals(email)))
.findFirst()
.isPresent();
}
public String getReferrer() {
return referrer;
}
public void setReferrer(String referrer) {
this.referrer = referrer;
}
public String getIp() {
return ip;
}
public void setIp(String ip) {
this.ip = ip;
}
public String getUserAgent() {
return userAgent;
}
public void setUserAgent(String userAgent) {
this.userAgent = userAgent;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public static void main(String[] args) {
cache.add(new BlacklistEntry("google.com", "127.0.0.2", "Mozilla", "yahoo.com"));
cache.add(new BlacklistEntry("r1.com", "127.0.0.3", "Mozilla", "*"));
cache.add(new BlacklistEntry("r2.com", "127.0.0.4", "Mozilla", "yahoo.com"));
System.out.println(isBlacklisted("r2.com", "127.0.0.4", "Mozilla", "yahoo.com"));
System.out.println(isBlacklisted("r1.com", "127.0.0.3", "Mozilla", "sould be true"));
System.out.println(isBlacklisted("*", "127.0.0.3", "*", "*"));
}
}
有没有比O（n）更好的东西？我应该考虑使用Lucene吗？
答案 0 :(得分：2)
感谢您的回答
我的第一次尝试是获得所有排列（使用番石榴感谢我的队友使其干净和清晰），使其成为numberParameters ^ 2并检查缓存中的集合是否包含其中任何一个
private static boolean check(Set cache, String ref, String ip, String ue, String mail) {
return Sets.powerSet(ImmutableSet.of(0, 1, 2, 3)).stream().map(set -> {
BlacklistKey key = new BlacklistKey("*", "*", "*", "*");
for (Integer idx : set) {
switch (idx) {
case 0:
key.setReferrer(ref);
break;
case 1:
key.setIp(ip);
break;
case 2:
key.setUserAgent(ue);
break;
case 3:
key.setEmail(mail);
}
}
return key;
}).anyMatch(keys::contains);
}
使用Lucene结束
import org.apache.lucene.analysis.standard.StandardAnalyzer;
import org.apache.lucene.document.Document;
import org.apache.lucene.document.Field;
import org.apache.lucene.document.StringField;
import org.apache.lucene.index.DirectoryReader;
import org.apache.lucene.index.IndexWriter;
import org.apache.lucene.index.IndexWriterConfig;
import org.apache.lucene.index.Term;
import org.apache.lucene.search.BooleanClause;
import org.apache.lucene.search.BooleanQuery;
import org.apache.lucene.search.IndexSearcher;
import org.apache.lucene.search.TermQuery;
import org.apache.lucene.store.RAMDirectory;
import java.io.IOException;
class Blacklist {
private static final String MATCH_ALL = "*";
private static IndexSearcher cache;
private enum Fields {
REFERRER, IP, USER_AGENT, EMAIL
}
private static Document getDocument(String referrer, String ip, String userAgent, String email) {
Document doc = new Document();
doc.add(getStringField(referrer, Fields.REFERRER.name()));
doc.add(getStringField(ip, Fields.IP.name()));
doc.add(getStringField(userAgent, Fields.USER_AGENT.name()));
doc.add(getStringField(email, Fields.EMAIL.name()));
return doc;
}
private static StringField getStringField(String val, String field) {
return new StringField(field, val, Field.Store.NO);
}
private static BooleanQuery createQuery(String referrer, String ip, String userAgent, String email) {
return new BooleanQuery.Builder()
.add(createBooleanQuery(Fields.REFERRER.name(), referrer), BooleanClause.Occur.FILTER)
.add(createBooleanQuery(Fields.IP.name(), ip), BooleanClause.Occur.FILTER)
.add(createBooleanQuery(Fields.USER_AGENT.name(), userAgent), BooleanClause.Occur.FILTER)
.add(createBooleanQuery(Fields.EMAIL.name(), email), BooleanClause.Occur.FILTER)
.build();
}
private static BooleanQuery createBooleanQuery(String key, String value) {
return new BooleanQuery.Builder()
.add(new TermQuery(new Term(key, value)), BooleanClause.Occur.SHOULD)
.add(new TermQuery(new Term(key, MATCH_ALL)), BooleanClause.Occur.SHOULD)
.build();
}
private static boolean isBlacklisted(String ref, String ip, String ue, String email) throws IOException {
BooleanQuery query = createQuery(ref, ip, ue, email);
return cache.search(query, 1).totalHits > 0;
}
public static void main(String[] args) throws IOException {
RAMDirectory directory = new RAMDirectory();
IndexWriter writer = new IndexWriter(directory, new IndexWriterConfig(new StandardAnalyzer()));
writer.addDocument(getDocument("ref1", "127.0.0.ip1", "Mozilla UserAgent1", "email.com"));
writer.addDocument(getDocument("ref2", "127.0.0.ip2", "Mozilla UserAgent2", "*"));
writer.close();
DirectoryReader reader = DirectoryReader.open(directory);
cache = new IndexSearcher(reader);
System.out.println(isBlacklisted("ref1", "127.0.0.ip1", "Mozilla UserAgent1", "email.com"));
System.out.println(isBlacklisted("r2.com", "127.0.0.4", "Mozilla", "yahoo.com"));
System.out.println(isBlacklisted("ref2", "127.0.0.ip2", "Mozilla UserAgent2", "this is ignored"));
System.out.println(isBlacklisted("*", "127.0.0.ip2", "Mozilla UserAgent2", "*"));
}
}
答案 1 :(得分：1)
我建议您使用RDB或Lucene。
但是，我认为您可以通过使用二叉树来降低成本。这保证了log（n）时间成本。
<强>程序
BlacklistEntry课程实施Comparator (Java Platform SE 8 ) equals，hashCode并为比较器实施compare TreeSet<BlacklistEntry>并附加您的数据BlacklistEntry#contains(object)，它会将其列入黑名单或不列入黑名单。 此实现为基本提供了有保证的log（n）时间成本 操作（添加，删除和包含）。
<强>示例
BlacklistEntry
package org.stackoverflow.btree;
import java.util.Comparator;
import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
@Data
@AllArgsConstructor
@NoArgsConstructor
public class BlacklistEntry implements Comparator<BlacklistEntry> {
private String referrer, ip, userAgent, email;
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (!(obj instanceof BlacklistEntry))
return false;
BlacklistEntry other = (BlacklistEntry) obj;
// You need to deal with "*" for all elements
// There should be another implementations to solve troublesome
// boiler-plate codes, for example to use reflection or commons.lang.
if (email == null) {
if (other.email != null) {
return false;
}
} else if (!email.equals(other.email) && !(email.equals("*") || other.email.equals("*")) ) {
return false;
}
if (ip == null) {
if (other.ip != null) {
return false;
}
} else if (!ip.equals(other.ip) && !(ip.equals("*") || other.ip.equals("*")) ) {
return false;
}
if (referrer == null) {
if (other.referrer != null) {
return false;
}
} else if (!referrer.equals(other.referrer) && !(referrer.equals("*") || other.referrer.equals("*")) ) {
return false;
}
if (userAgent == null) {
if (other.userAgent != null) {
return false;
}
} else if (!userAgent.equals(other.userAgent) && !(userAgent.equals("*") || other.userAgent.equals("*")) ) {
return false;
}
return true;
}
// I just generated #hashCode with eclipse
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((email == null) ? 0 : email.hashCode());
result = prime * result + ((ip == null) ? 0 : ip.hashCode());
result = prime * result + ((referrer == null) ? 0 : referrer.hashCode());
result = prime * result + ((userAgent == null) ? 0 : userAgent.hashCode());
return result;
}
// if left obj & right obj is equals == true
// --> "*" or just completely matching
// else
// --> detect which is larger than the another one
// (this #compare method is used by binary-tree)
public int compare(BlacklistEntry o1, BlacklistEntry o2) {
if (o1.equals(o2)) {
return 0;
} else {
return (o1.hashCode() < o2.hashCode()) ? -1 : 1;
}
}
}
主要
package org.stackoverflow.btree;
import java.util.TreeSet;
public class Main {
static TreeSet<BlacklistEntry> cache = new TreeSet<BlacklistEntry>(new BlacklistEntry());
public static boolean isBlacklisted(String ref, String ip, String ue, String email) {
return isBlacklisted(new BlacklistEntry(ref, ip, ue, email));
}
public static boolean isBlacklisted(BlacklistEntry e) {
return cache.contains(e);
}
public static void main(String[] args) {
cache.add(new BlacklistEntry("www.google.com", "127.0.0.2", "*", "yahoo.com") );
System.out.println(isBlacklisted("www.google.com", "127.0.0.1", "IE", "yahoo.com"));
System.out.println(isBlacklisted("www.google.com", "127.0.0.2", "Chrome", "yahoo.com"));
System.out.println(isBlacklisted("www.google.com", "127.0.0.3", "Edge", "yahoo.com"));
System.out.println(isBlacklisted("www.google.com", "127.0.0.4", "Mozilla", "yahoo.com"));
System.out.println(isBlacklisted("www.google.com", "127.0.0.5", "Ruby", "yahoo.com"));
}
}
<强>输出
false
true
false
false
false
答案 2 :(得分：1)
鉴于对于每一行，四列中只有一列可以有一个通配符，一种简单的方法是有五个哈希集：一组匹配所有列，一组用于匹配三个非通配符的每个通配符列。然后，检查黑名单包括查找五组中每一组的条目，这是渐近常数时间（O(1)）。
概念证明代码（当然可以改进）：
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Objects;
import java.util.Set;
public class Blacklist {
private static class BlacklistEntry {
private String[] fields;
private BlacklistEntry(String... fields) {
this.fields = fields;
}
@Override
public int hashCode() {
return Objects.hash(fields);
}
@Override
public boolean equals(final Object obj) {
return obj instanceof BlacklistEntry && Arrays.equals(fields, ((BlacklistEntry)obj).fields);
}
}
private static List<Set<BlacklistEntry>> wildcardSets = new ArrayList<>();
private static final int FIELD_COUNT = 4;
static {
for (int i = 0; i < FIELD_COUNT; i++) {
wildcardSets.add(new HashSet<>());
}
}
private static Set<BlacklistEntry> fullMatchSet = new HashSet<>();
private static boolean isBlacklisted(String... fields) {
if (fullMatchSet.contains(new BlacklistEntry(fields))) {
return true;
}
for (int i = 0; i < FIELD_COUNT; i++) {
if (wildcardSets.get(i).contains(new BlacklistEntry(removeField(fields, i)))) {
return true;
}
}
return false;
}
private static void addEntry(BlacklistEntry entry) {
if (entry.fields.length != FIELD_COUNT) {
throw new RuntimeException("Invalid entry, should contain four fields");
}
for (int i = 0; i < FIELD_COUNT; i++) {
if ("*".equals(entry.fields[i])) {
// Make a copy of the fields list without the wildcard, and add it to the correct set.
String[] fields = removeField(entry.fields, i);
wildcardSets.get(i).add(new BlacklistEntry(fields));
return;
}
}
// No wildcards in field list.
fullMatchSet.add(entry);
}
private static String[] removeField(final String[] fields, final int index) {
String[] newFields = new String[fields.length - 1];
for (int i = 0; i < index; i++) {
newFields[i] = fields[i];
}
for (int i = index + 1; i < FIELD_COUNT; i++) {
newFields[i - 1] = fields[i];
}
return newFields;
}
public static void main(String[] args) {
addEntry(new BlacklistEntry("r1.com", "127.0.0.1", "UA1", "*"));
addEntry(new BlacklistEntry("r2.com", "127.0.0.2", "*", "e1.com"));
addEntry(new BlacklistEntry("r3.com", "*", "UA2", "e2.com"));
addEntry(new BlacklistEntry("*", "127.0.0.3", "UA3", "e3.com"));
addEntry(new BlacklistEntry("r4.com", "127.0.0.4", "UA4", "e4.com"));
// All these should return true
System.out.println(isBlacklisted("r1.com", "127.0.0.1", "UA1", "wildcard"));
System.out.println(isBlacklisted("r2.com", "127.0.0.2", "wildcard", "e1.com"));
System.out.println(isBlacklisted("r3.com", "wildcard", "UA2", "e2.com"));
System.out.println(isBlacklisted("wildcard", "127.0.0.3", "UA3", "e3.com"));
System.out.println(isBlacklisted("r4.com", "127.0.0.4", "UA4", "e4.com"));
// All these should return false
System.out.println(isBlacklisted("r1.com", "127.0.0.1", "no match", "wildcard"));
System.out.println(isBlacklisted("no match", "127.0.0.2", "wildcard", "e1.com"));
System.out.println(isBlacklisted("r3.com", "wildcard", "no match", "e2.com"));
System.out.println(isBlacklisted("wildcard", "127.0.0.3", "UA3", "no match"));
System.out.println(isBlacklisted("r5.com", "127.0.0.5", "UA4", "e5.com"));
}
}