【发布时间】:2017-10-11 15:09:57
【问题描述】:
这是我的语法,我想输入为
alter table ;
一切正常,但当我给予时
altasder table; alter table ;
它按预期在第一个字符串上给了我一个错误,但我想要解析第二个命令,忽略第一个'altasder table;'
grammar Hello;
start : compilation;
compilation : sql*;
sql : altercommand;
altercommand : ALTER TABLE SEMICOLON;
ALTER: 'alter';
TABLE: 'table';
SEMICOLON : ';';
我怎样才能实现它???
我已经使用了 DefualtError 策略,但它仍然没有工作
import org.antlr.v4.runtime.DefaultErrorStrategy;
import org.antlr.v4.runtime.Parser;
import org.antlr.v4.runtime.RecognitionException;
import org.antlr.v4.runtime.TokenStream;
import org.antlr.v4.runtime.misc.IntervalSet;
public class CustomeErrorHandler extends DefaultErrorStrategy {
@Override
public void recover(Parser recognizer, RecognitionException e) {
// TODO Auto-generated method stub
super.recover(recognizer, e);
TokenStream tokenStream = (TokenStream)recognizer.getInputStream();
if (tokenStream.LA(1) == HelloParser.SEMICOLON )
{
IntervalSet intervalSet = getErrorRecoverySet(recognizer);
tokenStream.consume();
consumeUntil(recognizer, intervalSet);
}
}
}
主类: 公共类主 {
public static void main(String[] args) throws IOException {
ANTLRInputStream ip = new ANTLRInputStream("altasdere table ; alter table ;");
HelloLexer lex = new HelloLexer(ip);
CommonTokenStream token = new CommonTokenStream(lex);
HelloParser parser = new HelloParser(token);
parser.setErrorHandler(new CustomeErrorHandler());
System.out.println(parser.start().toStringTree(parser));
}
}
我的输出:
line 1:0 token recognition error at: 'alta'
line 1:4 token recognition error at: 's'
line 1:5 token recognition error at: 'd'
line 1:6 token recognition error at: 'e'
line 1:7 token recognition error at: 'r'
line 1:8 token recognition error at: 'e'
line 1:9 token recognition error at: ' '
(start compilation)
为什么它不移动到第二个命令?
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标签: antlr antlr4 antlr3 antlr2