我是lex / yacc的新手,我正在使用lex / yacc编写一个SQL解析器。但是对于特定的JOIN查询(如下所列),我的解析器是故意访问' select_statement'规则而不是'nested_join_statement'规则。
我正在为此查询获取正确的输出:SELECT * FROM sample1 JOIN sample2 ON sample1.C1 = sample2.C4;
(这将转到' join_statement规则',满足其中的simple_join规则)
但是当我尝试这个查询时:SELECT * FROM(SELECT * FROM(SELECT * FROM sample1)T8)AS temp1 JOIN(SELECT * FROM(SELECT * FROM sample2)T9)as temp2 ON temp1.C1 = temp2。 C4;
(理想情况下,这应该转到' nested_join_statement'但它会转到' SELECT选择FROM LPAREN select_statement2 RPAREN VAR'规则在' select语句' ;规则,我收到以下错误消息:错误:语法错误,意外AS,期待VAR)
我将nested_join_statement的优先级放在select_statement上,但仍会出现此错误。我不明白为什么。
lex中的VAR定义为[A-Za-z] [A-Za-z0-9 _# - ] *
任何帮助将不胜感激。我很绝望。
manipulation_statement: NEWLINE
| join_statement SEMICOLON
| nested_join_statement SEMICOLON
| select_statement SEMICOLON { flag=0;q=0;}
| join_statement SEMICOLON NEWLINE
| nested_join_statement SEMICOLON NEWLINE
| select_statement SEMICOLON NEWLINE { flag=0;q=0;}
;
nested_join_statement: two_nest_select_join { for(x=0;x<q;x++) strcpy(sj[x], ""); i=0;j=0;vardot=0;q=0;flag=0;nest=0;}
;
join_statement:
simple_join { for(x=0;x<q;x++) strcpy(sj[x], ""); i=0;j=0;vardot=0;q=0;flag=0;}
| simple_join_nest_select { for(x=0;x<q;x++) strcpy(sj[x], ""); i=0;j=0;vardot=0;q=0;flag=0;}
;
simple_join: SELECT selection FROM VAR JOIN VAR ON VAR DOTS VAR EQUAL VAR DOTS VAR { printf("inside simple join\n");
if(strcmp($4,$8) == 0) join_table($4,$6,sj,q,$10,$14,"","");
else join_table($4,$6,sj,q,$14,$10,"",""); p=q;}
;
simple_join_nest_select: SELECT selection FROM VAR JOIN LPAREN select_statement RPAREN AS VAR ON VAR DOTS VAR EQUAL VAR DOTS VAR {
if (strcmp($4,$12) == 0) join_table($4,"_temp_",sj,q,$14,$18,"","");
else join_table($4,"_temp_",sj,q,$18,$14,"",""); p=q;}
;
two_nest_select_join: SELECT selection FROM LPAREN select_statement RPAREN AS VAR JOIN LPAREN select_statement RPAREN AS VAR ON VAR DOTS VAR EQUAL VAR DOTS VAR {
join_table("temp1","temp2",sj,q,$18,$22,"",""); }
;
select_statement: SELECT selection FROM VAR WHERE where_clause { select_table($4,s,i,whr_var,whr_val);
for(x=0;x<i;x++) strcpy(s[x],"");i=0; strcpy(whr_var,""); strcpy(whr_val,"");j=0;}
| SELECT selection FROM VAR { select_table($4,s,i,whr_var,whr_val);
for(x=0;x<i;x++) strcpy(s[x],"");i=0;j=0;}
| SELECT selection FROM LPAREN select_statement2 RPAREN VAR { printf("inner tab: %s\n", inner_tab); printf("dep:%d\n", dep+1);
for(x=0; x<i;x++) printf("%s\n", col_array[x]);
for(y=0; y<j;y++) printf("%d\n", col_count[y]);
strcpy(inner_tab,""); nest =1; dep=0;
for(x=0; x<i;x++) strcpy(col_array[x], "");
for(y=0; y<j;y++) col_count[x] =0;
i=0;j=0;k=0;m=0;}
;
select_statement2: SELECT selection2 FROM VAR { dep++; inner_tab=malloc(strlen($4)); strcpy(inner_tab,$4); }
| select_list
;
select_list: SELECT selection2 FROM LPAREN select_statement2 RPAREN VAR { dep++; }
;
selection2: ASTERISK { col_array[i] = $1; i++; m++; col_count[j] = m; j++; printf("in level:%d value of k:%d\n",j,m); k=0; m=0;}
| comma_list2 { col_count[j] = m; j++; printf("in level:%d value of k:%d\n",j,m); k=0; m=0; }
;
comma_list2: VAR { col_array[i] = $1; i++; m++;}
| comma_list2 COMMA VAR { col_array[i] = $3; i++; m++;
k=m;}
;
selection: ASTERISK { if(q != 0 || nest == 1) { for(x=0;x<j;x++) col_count[x]=0; i=0; j=0;flag=1;}
s[i] = $1; col_array[i] = $1; i++; col_count[j] = i; j++;
if(q == 0) {sj[q] = $1; q++;} printf("in level t:%d value of k:%d\n",j,i); printf("sj2:%s\n",sj[0]);}
| comma_list { if (flag ==1 ) {col_count[j] = i; j++; printf("in level 2:%d value of k:%d\n",j,i); k=0; m=0; printf("temp:");
for(x=0;x<i;x++) printf("%s ",s[x]); } else {col_count[j] = i; j++;} }
;
comma_list:
VAR { if(q != 0 || nest == 1) { for(x=0;x<j;x++) col_count[x]=0; i=0; j=0;flag=1;}
s[i] = $1; col_array[i] = $1; i++; if(q==0) {sj[q] = $1; q++; }}
| VAR DOTS VAR { strcpy(temp,""); strcat(temp,$1); strcat(temp,$2); strcat(temp,$3);
sj[q] = temp; q++;
printf("temps:"); for(x=0;x<q;x++) printf("%s\n",sj[x]);}
| comma_list COMMA vardot {if (flag == 1) { s[i] = $3; col_array[i] = $3, i++;} else {sj[q] = $3; q++; vardot++;s[i] = $3; col_array[i] = $3, i++;}
}
;
答案 0 :(得分:2)
您可以减少/减少语法中的冲突 - 在选择和选择2之间,以及在comma_list和comma_list2之间。由于'2'版本在你的语法文件中排在第一位,它总是会减少使用这些规则,这意味着只要通过查看所有内容就无法区分select_statement和select_statement2。相关的'FROM'令牌它可能做错了。
在您的示例中,在看到SELECT * FROM (SELECT * FROM
之后,它必须做出此选择(因为parens中的嵌套选择可能是select_for_nest_select_join,也可以是select2用于简单的select_statement),并选择select2,因此将其解析为select_statement,导致您在到达AS
令牌时看到的错误 - 它期待VAR
如果你想解决这个问题,你需要使用更多的前瞻,或者改变语法来摆脱减少/减少冲突。
要使用更多超前预测,您可以使用bison的%glr-parser
选项。由于此处没有歧义,因此您无需添加任何其他消歧代码,但如果您在其他地方存在歧义,则可能会因此而出现运行时错误。
要摆脱减少/减少冲突,你需要摆脱重复的selection2规则 - 如果你把它们全部更改为相应的选择规则,你就摆脱了冲突,但你现在接受了一些结构您以前会拒绝的(例如您可能想要禁止的SELECT * FROM (SELECT * FROM VAR WHERE where_clause) VAR
。