我正在尝试使用 antlr4 解析一个简单的 SQL 子集。
我的语法如下所示:
grammar Query;
query : select;
select : 'select' colname (',' colname)* 'from' tablename;
colname : COLNAME;
tablename : TABLENAME;
COLNAME: [a-z]+ ;
TABLENAME : [a-z]+;
WS : [ \t\n\r]+ -> skip ; // skip spaces, tabs, newlines
我正在使用一个简单的 java 应用程序对此进行测试,如下所示:
import java.io.ByteArrayInputStream;
import java.io.InputStream;
import org.antlr.v4.runtime.*;
import org.antlr.v4.runtime.tree.*;
public class Test {
public static void main(String[] args) throws Exception {
// create a CharStream that reads from standard input
InputStream is = new ByteArrayInputStream("select one,two ,three from table".getBytes());
ANTLRInputStream input = new ANTLRInputStream(is);
// create a lexer that feeds off of input CharStream
QueryLexer lexer = new QueryLexer(input);
// create a buffer of tokens pulled from the lexer
CommonTokenStream tokens = new CommonTokenStream(lexer);
// create a parser that feeds off the tokens buffer
QueryParser parser = new QueryParser(tokens);
ParseTree tree = parser.query(); // begin parsing at init rule
System.out.println(tree.toStringTree(parser)); // print LISP-style tree
}
}
我得到的输出如下:
line 1:27 mismatched input 'table' expecting TABLENAME
(query (select select (colname one) , (colname two) , (colname three) from (tablename table)))
我不明白为什么解析器似乎将“表”作为解析器树中的表名,但我也遇到了错误。我错过了什么?
谢谢
安德鲁