C# 中的 antlr4 解析器 --- 看起来正确但不起作用答案

【问题标题】：an antlr4 parser in C# --- seems right but does not workC# 中的 antlr4 解析器 --- 看起来正确但不起作用
【发布时间】：2014-04-16 14:48:43
【问题描述】：

grammar CSVParser;
//

table   :   line+;

line    :   NAME
',' PEAK ',' 
STARTYEAR ',' 
ENDYEAR ',' LENGTH NEWLINE
;

NEWLINE :   '\r'? '\n'
        ;   

NAME    :   ('"'(~'"')*'"') 
;

PEAK    :   ([0-9]+);

STARTYEAR   :   ([0-9]+);

ENDYEAR :   ([0-9]+);

LENGTH  :   [0-9]+;

如您所见，我想像这样解析 CSV 表：

"ANNUAL REVIEW OF IMMUNOLOGY, VOL 31",0,0,1,1
"",0,0,1,1
"CA-A CANCER JOURNAL FOR CLINICIANS",1,1,2,1
"NATURE CHEMISTRY",1,1,3,2
"NATURE PHOTONICS",1,1,3,2
"ANNUAL REVIEW OF IMMUNOLOGY, VOL 30",1,1,2,1
"PHYSICS TODAY",2,1,3,2
"NATURE BIOTECHNOLOGY",2,2,4,2
"CHEMICAL SOCIETY REVIEWS",2,1,3,2
"NATURE REVIEWS GENETICS",2,2,3,1

但也有例外：

line 1:40 mismatched input '0' expecting STARTYEAR
line 2:5 mismatched input '0' expecting STARTYEAR
line 3:39 mismatched input '1' expecting STARTYEAR
line 4:21 mismatched input '1' expecting STARTYEAR
line 5:21 mismatched input '1' expecting STARTYEAR
line 6:40 mismatched input '1' expecting STARTYEAR
line 7:18 mismatched input '1' expecting STARTYEAR
line 8:25 mismatched input '2' expecting STARTYEAR
line 9:29 mismatched input '1' expecting STARTYEAR
line 10:28 mismatched input '2' expecting STARTYEAR
line 11:31 mismatched input '2' expecting STARTYEAR
line 12:42 mismatched input '2' expecting STARTYEAR
line 13:19 mismatched input '1' expecting STARTYEAR
line 14:40 mismatched input '2' expecting STARTYEAR
line 15:34 mismatched input '2' expecting STARTYEAR
line 16:29 mismatched input '2' expecting STARTYEAR
line 17:40 mismatched input '2' expecting STARTYEAR
line 18:40 mismatched input '2' expecting STARTYEAR
line 19:43 mismatched input '2' expecting STARTYEAR
line 20:40 mismatched input '3' expecting STARTYEAR

怎么了？

哦，Stackoverflow 告诉我添加更多细节。但我认为这里的代码已经足够了，因为这里显示了 csv 文件。

【问题讨论】：

标签： c# regex parsing csv antlr4

【解决方案1】：

您的许多词法分析器规则都匹配相同。当[0-9]+ 匹配时，只会创建PEAK 而不会创建STARTYEAR、ENDYEAR 或LENGTH。词法分析器不会根据解析器“需要”的内容创建标记。

改为这样做：

grammar CSVParser;

table     :   line+ EOF;
line      :   NAME ',' peak ',' startyear ',' endyear ',' length NEWLINE;
peak      :   NUMBER;    
startyear :   NUMBER;
endyear   :   NUMBER;    
length    :   NUMBER;

NEWLINE   :   '\r'? '\n';   
NAME      :   '"' (~'"')* '"';
NUMBER    :   [0-9]+;

【讨论】：