Acct1 = 'Y' AND Acct2 = 'N' AND Acct3 = 'N' AND Acct4 = 'N' AND Acct5 = 'N' AND ((Acct6 = 'N' OR Acct7 = 'N' AND Acct1 = 'Y') AND Formatted= 'N' AND Acct9 = 'N' AND (Acct10 = 'N' AND Acct11 = 'N') AND EditableField= 'N' )@H_403_3@我输入此子句的数据将来自Csv文件,如下所示.
Country,Type,Usage,Acct1,Acct2,Acct3,Acct4,Acct5,Acct6,Acct7,Formatted,Acct9,Acct10,Acct11,EditableField USA,Premium,Corporate,Y,N,Mexico,USA,@H_403_3@我将不得不根据条款中定义的条件过滤掉文件中的记录.这是一个简单子句的示例,但是会有比这更多的内部条件,并且只要用户需要就可以更改子句,并且记录必须按顺序传递10个这样的子句.
所以我正在寻找一种动态解释该子句并将其应用于传入记录的方法.请提供您有关如何设计/任何示例(如果有)的建议.
解决方法
首先,让我们编写一个tokenizer,将输入字符串拆分为标记.这是令牌类型:
private static enum TokenType { WHITESPACE,AND,OR,EQUALS,LEFT_PAREN,RIGHT_PAREN,IDENTIFIER,LITERAL,EOF }@H_403_3@令牌类本身:
private static class Token { final TokenType type; final int start; // start position in input (for error reporting) final String data; // payload public Token(TokenType type,int start,String data) { this.type = type; this.start = start; this.data = data; } @Override public String toString() { return type + "[" + data + "]"; } }@H_403_3@为了简化标记化,我们创建一个regexp,它从输入字符串中读取下一个标记:
private static final Pattern TOKENS = Pattern.compile("(\\s+)|(AND)|(OR)|(=)|(\\()|(\\))|(\\w+)|\'([^\']+)\'");@H_403_3@请注意,它有许多组,每个TokenType有一个组,顺序相同(首先是WHITESPACE,然后是AND,依此类推).最后是tokenizer方法:
private static TokenStream tokenize(String input) throws ParseException { Matcher matcher = TOKENS.matcher(input); List<Token> tokens = new ArrayList<>(); int offset = 0; TokenType[] types = TokenType.values(); while (offset != input.length()) { if (!matcher.find() || matcher.start() != offset) { throw new ParseException("Unexpected token at " + offset,offset); } for (int i = 0; i < types.length; i++) { if (matcher.group(i + 1) != null) { if (types[i] != TokenType.WHITESPACE) tokens.add(new Token(types[i],offset,matcher.group(i + 1))); break; } } offset = matcher.end(); } tokens.add(new Token(TokenType.EOF,input.length(),"")); return new TokenStream(tokens); }@H_403_3@我正在使用java.text.ParseException.这里我们应用正则表达式匹配器直到输入结束.如果它在当前位置不匹配,我们抛出异常.否则,我们会查找找到的匹配组并从中创建一个令牌,忽略WHITESPACE令牌.最后,我们添加一个EOF标记,表示输入的结束.结果作为特殊的TokenStream对象返回.这是TokenStream类,它将帮助我们进行解析:
private static class TokenStream { final List<Token> tokens; int offset = 0; public TokenStream(List<Token> tokens) { this.tokens = tokens; } // consume next token of given type (throw exception if type differs) public Token consume(TokenType type) throws ParseException { Token token = tokens.get(offset++); if (token.type != type) { throw new ParseException("Unexpected token at " + token.start + ": " + token + " (was looking for " + type + ")",token.start); } return token; } // consume token of given type (return null and don't advance if type differs) public Token consumeIf(TokenType type) { Token token = tokens.get(offset); if (token.type == type) { offset++; return token; } return null; } @Override public String toString() { return tokens.toString(); } }@H_403_3@所以我们有一个标记器,hoorah.您现在可以使用System.out.println进行测试(tokenize(“Acct1 =’Y’AND(Acct2 =’N’或Acct3 =’N’)”));
现在让我们编写解析器,它将创建表达式的树状表示.首先是所有树节点的接口Expr:
public interface Expr { public boolean evaluate(Map<String,String> data); }@H_403_3@它唯一的方法用于评估给定数据集的表达式,如果数据集匹配则返回true.
最基本的表达式是EqualsExpr,它类似于Acct1 =’Y’或’Y’= Acct1:
private static class EqualsExpr implements Expr { private final String identifier,literal; public EqualsExpr(TokenStream stream) throws ParseException { Token token = stream.consumeIf(TokenType.IDENTIFIER); if(token != null) { this.identifier = token.data; stream.consume(TokenType.EQUALS); this.literal = stream.consume(TokenType.LITERAL).data; } else { this.literal = stream.consume(TokenType.LITERAL).data; stream.consume(TokenType.EQUALS); this.identifier = stream.consume(TokenType.IDENTIFIER).data; } } @Override public String toString() { return identifier+"='"+literal+"'"; } @Override public boolean evaluate(Map<String,String> data) { return literal.equals(data.get(identifier)); } }@H_403_3@接下来我们将定义SubExpr类,它是EqualsExpr或括号中更复杂的东西(如果我们看到括号):
private static class SubExpr implements Expr { private final Expr child; public SubExpr(TokenStream stream) throws ParseException { if(stream.consumeIf(TokenType.LEFT_PAREN) != null) { child = new OrExpr(stream); stream.consume(TokenType.RIGHT_PAREN); } else { child = new EqualsExpr(stream); } } @Override public String toString() { return "("+child+")"; } @Override public boolean evaluate(Map<String,String> data) { return child.evaluate(data); } }@H_403_3@接下来是AndExpr,它是由AND运算符连接的一组SubExpr表达式:
private static class AndExpr implements Expr { private final List<Expr> children = new ArrayList<>(); public AndExpr(TokenStream stream) throws ParseException { do { children.add(new SubExpr(stream)); } while(stream.consumeIf(TokenType.AND) != null); } @Override public String toString() { return children.stream().map(Object::toString).collect(Collectors.joining(" AND ")); } @Override public boolean evaluate(Map<String,String> data) { for(Expr child : children) { if(!child.evaluate(data)) return false; } return true; } }@H_403_3@为简洁起见,我在toString中使用Java-8 Stream API.如果您不能使用Java-8,可以使用for循环重写它或完全删除toString.
最后我们定义OrExpr,它是由OR连接的一组AndExpr(通常OR的优先级低于AND).它与AndExpr非常相似:
private static class OrExpr implements Expr { private final List<Expr> children = new ArrayList<>(); public OrExpr(TokenStream stream) throws ParseException { do { children.add(new AndExpr(stream)); } while(stream.consumeIf(TokenType.OR) != null); } @Override public String toString() { return children.stream().map(Object::toString).collect(Collectors.joining(" OR ")); } @Override public boolean evaluate(Map<String,String> data) { for(Expr child : children) { if(child.evaluate(data)) return true; } return false; } }@H_403_3@最后的解析方法:
public static Expr parse(TokenStream stream) throws ParseException { OrExpr expr = new OrExpr(stream); stream.consume(TokenType.EOF); // ensure that we parsed the whole input return expr; }@H_403_3@因此,您可以解析表达式以获取Expr对象,然后根据CSV文件的行对其进行评估.我假设您能够将CSV行解析为Map< String,String>.这是用法示例:
Map<String,String> data = new HashMap<>(); data.put("Acct1","Y"); data.put("Acct2","N"); data.put("Acct3","Y"); data.put("Acct4","N"); Expr expr = parse(tokenize("Acct1 = 'Y' AND (Acct2 = 'Y' OR Acct3 = 'Y')")); System.out.println(expr.evaluate(data)); // true expr = parse(tokenize("Acct1 = 'N' OR 'Y' = Acct2 AND Acct3 = 'Y'")); System.out.println(expr.evaluate(data)); // false@H_403_3@