PostgreSQL代码分析,查询优化部分,process_duplicate_ors
PostgreSQL代码分析,查询优化部分。
/* * process_duplicate_ors * Given a list of exprs which are ORed together, try to apply * the inverse OR distributive law. * * Returns the resulting expression (could be an AND clause, an OR * clause, or maybe even a single subexpression). */ /* * 假设我们有四个表,分别是TEST_A,TEST_B,TEST_C,TEST_D,每个表有一列分别是A,B,C,D, * 也就是TEST_A有一个A列,TEST_B有一个B列,以此类推。 * * 这个函数处理这种情况,对于一个选择,SELECT * FROM TEST_A,TEST_B,TEST_C,TEST_D * WHERE (A=1 AND B=1) OR (A=1 AND C=1) OR (A=1 AND D=1); * * 语句中的WHERE条件: * (A=1 AND B=1) OR (A=1 AND C=1) OR (A=1 AND D=1) * 可以改写为: * (A=1)AND (B=1 OR C=1 OR D=1) * 这就是这个函数的主要功能。 * * 这个函数的参数是一个list, 对于上述的WHERE条件,orlist的结构如下: * orlist中有一个元素,是OR_EXPR类型的BoolExpr,BoolExpr中的结构如下: typedef struct BoolExpr { Expr xpr; = 略 BoolExprType boolop; = OR_EXPR List *args; = OR中的3个条件,即(A=1 AND B=1) OR (A=1 AND C=1) OR (A=1 AND D=1) bool plusFlag; = 略 } BoolExpr; * * 下面分析函数的具体实现,大致的步骤为: * 1)分析每个OR中的公共项, 2)提取公共项, 3)合并剩余项为AND。 */ static Expr * process_duplicate_ors(List *orlist) { List *reference = NIL; int num_subclauses = 0; List *winners; List *neworlist; ListCell *temp; if (orlist == NIL) return NULL; /* probably can't happen */ /* 如果只有一个。。。。,那就算了吧 */ if (list_length(orlist) == 1) /* single-expression OR (can this * happen?) */ return linitial(orlist); /* * Choose the shortest AND clause as the reference list --- obviously, any * subclause not in this clause isn't in all the clauses. If we find a * clause that's not an AND, we can treat it as a one-element AND clause, * which necessarily wins as shortest. */ /* * “找最短”。 * 在WHERE语句中: * (A=1 AND B=1) OR (A=1 AND C=1) OR (A=1 AND D=1) * OR操作串联了3个子语句,找到其中最短的一个,因为如果有公共项,那么最短的那个也一定 * 包含公共项,那么通过找到最短的那个,在后面的操作里能减少 比较 的次数。 * 在上面的WHERE语句中,3个子语句的长度相同,按照如下执行过程,找到的应该是(A=1 AND B=1), * 即第一个。 */ foreach(temp, orlist) { Expr *clause = (Expr *) lfirst(temp); if (and_clause((Node *) clause)) { List *subclauses = ((BoolExpr *) clause)->args; int nclauses = list_length(subclauses); /* * 这里判断子语句里的长度,比如对于(A=1 AND B=1)子语句, * 他实际上是一个AND连接起来的两个 子子语句, 那么他的长度就是2。 * * 通过nclauses记录最短的子语句,如果有更短的(nclauses < num_subclauses), * 那么就替换成最短的。 */ if (reference == NIL || nclauses < num_subclauses) { reference = subclauses; num_subclauses = nclauses; } } else { /* * 还有一种情况, 就是可能子句不是一个AND语句,这样看上去不大符合规则, * 那么把他看做一个整体,那这个就是最短元素。 * ****************************** * 如果代码执行到这里,那么只有两种情况: * 一种是 ... WHERE (A=1 AND B=1) OR (A=1 AND C=1) OR (A=1)。 * 一种是 ... WHERE ((A=1 OR C=1) AND B=1) OR (A=1 OR C=1). * 如果是这两种情况,都可以做如下简化: * 第一种情况简化为 A=1 * 第二种情况化简为 (A=1 OR C=1) * * 第三种情况待补充... */ reference = list_make1(clause); break; } } /* * Just in case, eliminate any duplicates in the reference list. */ /* 找到最短的, 存到List */ reference = list_union(NIL, reference); /* * Check each element of the reference list to see if it's in all the OR * clauses. Build a new list of winning clauses. */ /* * “找公共项”。 * * NOTE:这时候就能体现“找最短”带来的优势,外层循环次数会少一些。 * * 如果WHERE语句是: * (A=1 AND B=1) OR (A=1 AND C=1) OR (A=1 AND D=1) * “找最短”中找到的一定是(A=1 AND B=1)。 * 则外层会有两次循环...(foreach(temp, reference)),两次循环的变量分别为 * A=1 和 B=1。 * 内层有三次循环...(foreach(temp2, orlist)),三次循环的变量分别为 * (A=1 AND B=1) 和 (A=1 AND C=1) 和 (A=1 AND D=1) * * 示例如下: * 假如现在外层循环第一次执行,即查找A=1的公共项,进而假如内层循环也是第一次执行, * 即在(A=1 AND B=1)中查找是否存在A=1这个公共项,发现是存在的(list_member), * 则依次判断内层循环的第二个子句... * * 如上例,具体来说,这些循环分别作的操作是: * 外层第一次: * 判断A=1是否在(A=1 AND B=1),在,判断下一个 * 判断A=1是否在(A=1 AND C=1),在,判断下一个 * 判断A=1是否在(A=1 AND D=1),在,A=1是公共项,记录(winners = lappend...) * 外层第二次: * 判断B=1是否在(A=1 AND B=1),在,判断下一个 * 判断B=1是否在(A=1 AND C=1),不在,跳出循环,下一个不用判断了。 * 判断B=1是否在(A=1 AND D=1),未执行,因为上一个不含公共项,就不可能提取了。 */ winners = NIL; foreach(temp, reference) { Expr *refclause = (Expr *) lfirst(temp); bool win = true; ListCell *temp2; foreach(temp2, orlist) { Expr *clause = (Expr *) lfirst(temp2); if (and_clause((Node *) clause)) { if (!list_member(((BoolExpr *) clause)->args, refclause)) { win = false; break; } } else { if (!equal(refclause, clause)) { win = false; break; } } } if (win) winners = lappend(winners, refclause); } /* * If no winners, we can't transform the OR */ if (winners == NIL) return make_orclause(orlist); /* * Generate new OR list consisting of the remaining sub-clauses. * * If any clause degenerates to empty, then we have a situation like (A * AND B) OR (A), which can be reduced to just A --- that is, the * additional conditions in other arms of the OR are irrelevant. * * Note that because we use list_difference, any multiple occurrences of a * winning clause in an AND sub-clause will be removed automatically. */ /* * “提取公共项”。 * 用list_difference删除公共项,实现细节不在赘述。 */ neworlist = NIL; foreach(temp, orlist) { Expr *clause = (Expr *) lfirst(temp); if (and_clause((Node *) clause)) { List *subclauses = ((BoolExpr *) clause)->args; /* 看这里...看这里..., 消除公共项 */ subclauses = list_difference(subclauses, winners); if (subclauses != NIL) { /* 消除后,剩余的拼接起来,拼接成:(B=1 OR C=1 OR D=1)*/ if (list_length(subclauses) == 1) neworlist = lappend(neworlist, linitial(subclauses)); else neworlist = lappend(neworlist, make_andclause(subclauses)); } else { /* * 这说明子语句中,有一个全部是公共项,也就是如下形式: * ... WHERE (A=1 AND B=1) OR (A=1) * * 这时候公共项是A=1,第一个子句是(A=1 AND B=1),第二个子句是(A=1), * 第二个子句经过list_difference,返回的结果是NULL。 * 对于这种情况,实际上可以化简为:A=1,因为(A=1 AND B=1)一定满足A=1的情况。 */ neworlist = NIL; /* degenerate case, see above */ break; } } else { if (!list_member(winners, clause)) neworlist = lappend(neworlist, clause); else { neworlist = NIL; /* degenerate case, see above */ break; } } } /* * Append reduced OR to the winners list, if it's not degenerate, handling * the special case of one element correctly (can that really happen?). * Also be careful to maintain AND/OR flatness in case we pulled up a * sub-sub-OR-clause. */ if (neworlist != NIL) { if (list_length(neworlist) == 1) winners = lappend(winners, linitial(neworlist)); else /*neworlist里面应该是(B=1 OR C=1 OR D=1),所以用make_orclause */ winners = lappend(winners, make_orclause(pull_ors(neworlist))); } /* * And return the constructed AND clause, again being wary of a single * element and AND/OR flatness. */ if (list_length(winners) == 1) return (Expr *) linitial(winners); else /* 返回的形式是:(A=1)AND (B=1 OR C=1 OR D=1),所以会用make_andclause */ return make_andclause(pull_ands(winners)); }
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