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PostgreSQL 源码解读(60)- 查询语句#45(make_one_rel函数#10-...

原创 PostgreSQL 作者:husthxd 时间:2018-09-30 17:06:48 0 删除 编辑

这一小节继续介绍查询物理优化中的create_index_paths->choose_bitmap_and,该函数执行Bitmap AND操作后创建位图索引扫描访问路径(BitmapAndPath)节点。
关于Bitmap Scan的相关知识,请参照PostgreSQL DBA(6) - SeqScan vs IndexScan vs BitmapHeapScan这篇文章.

下面是BitmapAnd访问路径的样例:

testdb=# explain verbose select t1.* 
testdb-# from t_dwxx t1 
testdb-# where (dwbh > '10000' and dwbh < '15000') AND (dwdz between 'DWDZ10000' and 'DWDZ15000');
QUERY PLAN                                
                                                    
----------------------------------------------------------------------------------------------
 Bitmap Heap Scan on public.t_dwxx t1  (cost=32.33..88.38 rows=33 width=20)
   Output: dwmc, dwbh, dwdz
   Recheck Cond: (((t1.dwbh)::text > '10000'::text) AND ((t1.dwbh)::text < '15000'::text) AND ((t1.dwdz)::text >= 'DWDZ10000'
::text) AND ((t1.dwdz)::text <= 'DWDZ15000'::text))
   ->  BitmapAnd  (cost=32.33..32.33 rows=33 width=0)  -->BitmapAnd
         ->  Bitmap Index Scan on t_dwxx_pkey  (cost=0.00..13.86 rows=557 width=0)
               Index Cond: (((t1.dwbh)::text > '10000'::text) AND ((t1.dwbh)::text < '15000'::text))
         ->  Bitmap Index Scan on idx_dwxx_dwdz  (cost=0.00..18.21 rows=592 width=0)
               Index Cond: (((t1.dwdz)::text >= 'DWDZ10000'::text) AND ((t1.dwdz)::text <= 'DWDZ15000'::text))
(8 rows)

一、数据结构

Cost相关
注意:实际使用的参数值通过系统配置文件定义,而不是这里的常量定义!

 typedef double Cost; /* execution cost (in page-access units) */

 /* defaults for costsize.c's Cost parameters */
 /* NB: cost-estimation code should use the variables, not these constants! */
 /* 注意:实际值通过系统配置文件定义,而不是这里的常量定义! */
 /* If you change these, update backend/utils/misc/postgresql.sample.conf */
 #define DEFAULT_SEQ_PAGE_COST  1.0       //顺序扫描page的成本
 #define DEFAULT_RANDOM_PAGE_COST  4.0      //随机扫描page的成本
 #define DEFAULT_CPU_TUPLE_COST  0.01     //处理一个元组的CPU成本
 #define DEFAULT_CPU_INDEX_TUPLE_COST 0.005   //处理一个索引元组的CPU成本
 #define DEFAULT_CPU_OPERATOR_COST  0.0025    //执行一次操作或函数的CPU成本
 #define DEFAULT_PARALLEL_TUPLE_COST 0.1    //并行执行,从一个worker传输一个元组到另一个worker的成本
 #define DEFAULT_PARALLEL_SETUP_COST  1000.0  //构建并行执行环境的成本
 
 #define DEFAULT_EFFECTIVE_CACHE_SIZE  524288    /*先前已有介绍, measured in pages */

 double      seq_page_cost = DEFAULT_SEQ_PAGE_COST;
 double      random_page_cost = DEFAULT_RANDOM_PAGE_COST;
 double      cpu_tuple_cost = DEFAULT_CPU_TUPLE_COST;
 double      cpu_index_tuple_cost = DEFAULT_CPU_INDEX_TUPLE_COST;
 double      cpu_operator_cost = DEFAULT_CPU_OPERATOR_COST;
 double      parallel_tuple_cost = DEFAULT_PARALLEL_TUPLE_COST;
 double      parallel_setup_cost = DEFAULT_PARALLEL_SETUP_COST;
 
 int         effective_cache_size = DEFAULT_EFFECTIVE_CACHE_SIZE;
 Cost        disable_cost = 1.0e10;//1后面10个0,通过设置一个巨大的成本,让优化器自动放弃此路径
 
 int         max_parallel_workers_per_gather = 2;//每次gather使用的worker数

PathClauseUsage

 /* Per-path data used within choose_bitmap_and() */
 typedef struct
 {
     Path       *path;           /* 访问路径链表,IndexPath, BitmapAndPath, or BitmapOrPath */
     List       *quals;          /* 限制条件子句链表,the WHERE clauses it uses */
     List       *preds;          /* 部分索引谓词链表,predicates of its partial index(es) */
     Bitmapset  *clauseids;      /* 位图集合,quals+preds represented as a bitmapset */
 } PathClauseUsage;

二、源码解读

choose_bitmap_and函数
create_index_paths->choose_bitmap_and函数,该函数给定非空的位图访问路径链表,执行AND操作后合并到一条路径中,最终得到位图索引扫描访问路径节点.

 /*
  * choose_bitmap_and
  *      Given a nonempty list of bitmap paths, AND them into one path.
  *      给定非空的位图访问路径链表,执行AND操作后合并到一条路径中
  *
  * This is a nontrivial decision since we can legally use any subset of the
  * given path set.  We want to choose a good tradeoff between selectivity
  * and cost of computing the bitmap.
  * 这是一个非常重要的策略,因为这样可以合法地使用给定路径集的任何子集。
  * 
  * The result is either a single one of the inputs, or a BitmapAndPath
  * combining multiple inputs.
  * 输出结果要么是输出的其中之一,要么是融合多个输入之后的BitmapAndPath
  */
 static Path *
 choose_bitmap_and(PlannerInfo *root, RelOptInfo *rel, List *paths)
 {
     int         npaths = list_length(paths);
     PathClauseUsage **pathinfoarray;
     PathClauseUsage *pathinfo;
     List       *clauselist;
     List       *bestpaths = NIL;
     Cost        bestcost = 0;
     int         i,
                 j;
     ListCell   *l;
 
     Assert(npaths > 0);         /* else caller error */
     if (npaths == 1)
         return (Path *) linitial(paths);    /* easy case */
 
     /*
      * In theory we should consider every nonempty subset of the given paths.
      * In practice that seems like overkill, given the crude nature of the
      * estimates, not to mention the possible effects of higher-level AND and
      * OR clauses.  Moreover, it's completely impractical if there are a large
      * number of paths, since the work would grow as O(2^N).
      * 理论上,我们应该考虑给定路径的所有非空子集。在实践中,
      * 考虑到估算的不确定性和成本,以及更高级别的AND和OR约束可能产生的影响,这样的做法并不合适.
      * 此外,它并不切合实际,如果有大量的路径,这项工作的复杂度会是指数级的O(2 ^ N)。
      *
      * As a heuristic, we first check for paths using exactly the same sets of
      * WHERE clauses + index predicate conditions, and reject all but the
      * cheapest-to-scan in any such group.  This primarily gets rid of indexes
      * that include the interesting columns but also irrelevant columns.  (In
      * situations where the DBA has gone overboard on creating variant
      * indexes, this can make for a very large reduction in the number of
      * paths considered further.)
      * 作为一种启发式方法,首先使用完全相同的WHERE子句+索引谓词条件集检查路径,
      * 并去掉这类条件组中除成本最低之外的所有路径。
      * 这主要是去掉了包含interesting列和不相关列的索引。
      * (在DBA过度创建索引的情况下,这会大大减少进一步考虑的路径数量。)
      * 
      * We then sort the surviving paths with the cheapest-to-scan first, and
      * for each path, consider using that path alone as the basis for a bitmap
      * scan.  Then we consider bitmap AND scans formed from that path plus
      * each subsequent (higher-cost) path, adding on a subsequent path if it
      * results in a reduction in the estimated total scan cost. This means we
      * consider about O(N^2) rather than O(2^N) path combinations, which is
      * quite tolerable, especially given than N is usually reasonably small
      * because of the prefiltering step.  The cheapest of these is returned.
      * 然后,我们首先使用成本最低的扫描路径对现存的路径进行排序,
      * 对于每个路径,考虑单独使用该路径作为位图扫描的基础。
      * 然后我们考虑位图和从该路径形成的扫描加上每个后续的(更高成本的)路径,
      * 如果后续路径导致估算的总扫描成本减少,那么就添加一个后续路径。
      * 这意味着我们只需要处理O(N ^ 2),而不是O(2 ^ N)个路径组合,
      * 这样的成本完全可以接受,特别是N通常相当小时。函数返回成本最低的路径。
      * 
      * We will only consider AND combinations in which no two indexes use the
      * same WHERE clause.  This is a bit of a kluge: it's needed because
      * costsize.c and clausesel.c aren't very smart about redundant clauses.
      * They will usually double-count the redundant clauses, producing a
      * too-small selectivity that makes a redundant AND step look like it
      * reduces the total cost.  Perhaps someday that code will be smarter and
      * we can remove this limitation.  (But note that this also defends
      * against flat-out duplicate input paths, which can happen because
      * match_join_clauses_to_index will find the same OR join clauses that
      * extract_restriction_or_clauses has pulled OR restriction clauses out
      * of.)
      * 我们将只考虑没有两个索引同时使用相同的WHERE子句的AND组合。
      * 这是一个有点蹩脚的做法:之所以这样是因为cost.c和clausesel.c未能足够聪明的处理多余的子句。
      * 它们通常会重复计算冗余子句,从而产生很小的选择性,使冗余子句看起来像是减少了总成本。
      * 也许有一天,代码会变得更聪明,我们可以消除这个限制。
      * (但是要注意,这也可以防止完全重复的输入路径,
      * 因为match_join_clauses_to_index会找到相同的OR连接子句,而这些子句
      * 已通过extract_restriction_or_clauses函数提升到外面去了.)
      *
      * For the same reason, we reject AND combinations in which an index
      * predicate clause duplicates another clause.  Here we find it necessary
      * to be even stricter: we'll reject a partial index if any of its
      * predicate clauses are implied by the set of WHERE clauses and predicate
      * clauses used so far.  This covers cases such as a condition "x = 42"
      * used with a plain index, followed by a clauseless scan of a partial
      * index "WHERE x >= 40 AND x < 50".  The partial index has been accepted
      * only because "x = 42" was present, and so allowing it would partially
      * double-count selectivity.  (We could use predicate_implied_by on
      * regular qual clauses too, to have a more intelligent, but much more
      * expensive, check for redundancy --- but in most cases simple equality
      * seems to suffice.)
      * 出于同样的原因,我们不会组合索引谓词子句与另一个重复的子句。
      * 在这里,有必要更加严格 : 如果部分索引的任何谓词子句
      * 隐含在WHERE子句中,则不能使用此索引。
      * 这里包括了形如使用普通索引的“x = 42”和使用部分索引“x >= 40和x < 50”的情况。
      * 部分索引被接受,是因为存在“x = 42”,因此允许它部分重复计数选择性。
      * (我们也可以在普通的qual子句上使用predicate_implied_by函数,
      * 这样就可以更智能但更昂贵地检查冗余——但在大多数情况下,简单的等式似乎就足够了。)
      */
 
     /*
      * Extract clause usage info and detect any paths that use exactly the
      * same set of clauses; keep only the cheapest-to-scan of any such groups.
      * The surviving paths are put into an array for qsort'ing.
      * 提取子句使用信息并检测使用完全相同子句集的所有路径;
      * 只保留这类路径中成本最低的,这些路径被放入一个数组中进行qsort'ing
      */
     pathinfoarray = (PathClauseUsage **)
         palloc(npaths * sizeof(PathClauseUsage *));//数组
     clauselist = NIL;
     npaths = 0;
     foreach(l, paths)//遍历paths
     {
         Path       *ipath = (Path *) lfirst(l);
 
         pathinfo = classify_index_clause_usage(ipath, &clauselist);//归类路径信息
         for (i = 0; i < npaths; i++)
         {
             if (bms_equal(pathinfo->clauseids, pathinfoarray[i]->clauseids))
                 break;//只要发现子句集一样,就继续执行
         }
         if (i < npaths)//发现相同的
         {
             /* duplicate clauseids, keep the cheaper one */
             //相同的约束条件,只保留成本最低的
             Cost        ncost;
             Cost        ocost;
             Selectivity nselec;
             Selectivity oselec;
 
             cost_bitmap_tree_node(pathinfo->path, &ncost, &nselec);//计算成本
             cost_bitmap_tree_node(pathinfoarray[i]->path, &ocost, &oselec);
             if (ncost < ocost)
                 pathinfoarray[i] = pathinfo;
         }
         else//没有发现条件一样的,添加到数组中
         {
             /* not duplicate clauseids, add to array */
             pathinfoarray[npaths++] = pathinfo;
         }
     }
 
     /* If only one surviving path, we're done */
     if (npaths == 1)//结果只有一条,则返回之
         return pathinfoarray[0]->path;
 
     /* Sort the surviving paths by index access cost */
     qsort(pathinfoarray, npaths, sizeof(PathClauseUsage *),
           path_usage_comparator);//以索引访问成本排序现存路径
 
     /*
      * For each surviving index, consider it as an "AND group leader", and see
      * whether adding on any of the later indexes results in an AND path with
      * cheaper total cost than before.  Then take the cheapest AND group.
      * 对于现存的索引,把它视为"AND group leader",
      * 并查看是否添加了以后的索引后,会得到一个总成本比以前更低的AND路径。
      * 选择成本最低的AND组.
      * 
      */
     for (i = 0; i < npaths; i++)//遍历这些路径
     {
         Cost        costsofar;
         List       *qualsofar;
         Bitmapset  *clauseidsofar;
         ListCell   *lastcell;
 
         pathinfo = pathinfoarray[i];//PathClauseUsage结构体
         paths = list_make1(pathinfo->path);//路径链表
         costsofar = bitmap_scan_cost_est(root, rel, pathinfo->path);//当前的成本
         qualsofar = list_concat(list_copy(pathinfo->quals),
                                 list_copy(pathinfo->preds));
         clauseidsofar = bms_copy(pathinfo->clauseids);
         lastcell = list_head(paths);    /* 用于快速删除,for quick deletions */
 
         for (j = i + 1; j < npaths; j++)//扫描后续的路径
         {
             Cost        newcost;
 
             pathinfo = pathinfoarray[j];
             /* Check for redundancy */
             if (bms_overlap(pathinfo->clauseids, clauseidsofar))
                 continue;       /* 多余的路径,consider it redundant */
             if (pathinfo->preds)//部分索引?
             {
                 bool        redundant = false;
 
                 /* we check each predicate clause separately */
                 //单独检查每一个谓词
                 foreach(l, pathinfo->preds)
                 {
                     Node       *np = (Node *) lfirst(l);
 
                     if (predicate_implied_by(list_make1(np), qualsofar, false))
                     {
                         redundant = true;
                         break;  /* out of inner foreach loop */
                     }
                 }
                 if (redundant)
                     continue;
             }
             /* tentatively add new path to paths, so we can estimate cost */
             //尝试在路径中添加新路径,这样我们就可以估算成本
             paths = lappend(paths, pathinfo->path);
             newcost = bitmap_and_cost_est(root, rel, paths);//估算成本
             if (newcost < costsofar)//新成本更低
             {
                 /* keep new path in paths, update subsidiary variables */
                 costsofar = newcost;
                 qualsofar = list_concat(qualsofar,
                                         list_copy(pathinfo->quals));//添加此条件
                 qualsofar = list_concat(qualsofar,
                                         list_copy(pathinfo->preds));//添加此谓词
                 clauseidsofar = bms_add_members(clauseidsofar,
                                                 pathinfo->clauseids);//添加此子句ID
                 lastcell = lnext(lastcell);
             }
             else
             {
                 /* reject new path, remove it from paths list */
                 paths = list_delete_cell(paths, lnext(lastcell), lastcell);//去掉新路径
             }
             Assert(lnext(lastcell) == NULL);
         }
 
         /* Keep the cheapest AND-group (or singleton) */
         if (i == 0 || costsofar < bestcost)//单条路径或者取得最小的成本
         {
             bestpaths = paths;
             bestcost = costsofar;
         }
 
         /* some easy cleanup (we don't try real hard though) */
         list_free(qualsofar);
     }
 
     if (list_length(bestpaths) == 1)
         return (Path *) linitial(bestpaths);    /* 无需AND路径,no need for AND */
     return (Path *) create_bitmap_and_path(root, rel, bestpaths);//生成BitmapAndPath
 }
 
//-------------------------------------------------------------------------- bitmap_scan_cost_est
 /*
  * Estimate the cost of actually executing a bitmap scan with a single
  * index path (no BitmapAnd, at least not at this level; but it could be
  * a BitmapOr).
  */
 static Cost
 bitmap_scan_cost_est(PlannerInfo *root, RelOptInfo *rel, Path *ipath)
 {
     BitmapHeapPath bpath;
     Relids      required_outer;
 
     /* Identify required outer rels, in case it's a parameterized scan */
     required_outer = get_bitmap_tree_required_outer(ipath);
 
     /* Set up a dummy BitmapHeapPath */
     bpath.path.type = T_BitmapHeapPath;
     bpath.path.pathtype = T_BitmapHeapScan;
     bpath.path.parent = rel;
     bpath.path.pathtarget = rel->reltarget;
     bpath.path.param_info = get_baserel_parampathinfo(root, rel,
                                                       required_outer);
     bpath.path.pathkeys = NIL;
     bpath.bitmapqual = ipath;
 
     /*
      * Check the cost of temporary path without considering parallelism.
      * Parallel bitmap heap path will be considered at later stage.
      */
     bpath.path.parallel_workers = 0;
     cost_bitmap_heap_scan(&bpath.path, root, rel,
                           bpath.path.param_info,
                           ipath,
                           get_loop_count(root, rel->relid, required_outer));//BitmapHeapPath计算成本
 
     return bpath.path.total_cost;
 }
 
//-------------------------------------------------------------------------- bitmap_and_cost_est

 /*
  * Estimate the cost of actually executing a BitmapAnd scan with the given
  * inputs.
  * 给定输入,估算实际执行BitmapAnd扫描的实际成本
  */
 static Cost
 bitmap_and_cost_est(PlannerInfo *root, RelOptInfo *rel, List *paths)
 {
     BitmapAndPath apath;
     BitmapHeapPath bpath;
     Relids      required_outer;
 
     /* Set up a dummy BitmapAndPath */
     apath.path.type = T_BitmapAndPath;
     apath.path.pathtype = T_BitmapAnd;
     apath.path.parent = rel;
     apath.path.pathtarget = rel->reltarget;
     apath.path.param_info = NULL;   /* not used in bitmap trees */
     apath.path.pathkeys = NIL;
     apath.bitmapquals = paths;
     cost_bitmap_and_node(&apath, root);
 
     /* Identify required outer rels, in case it's a parameterized scan */
     required_outer = get_bitmap_tree_required_outer((Path *) &apath);
 
     /* Set up a dummy BitmapHeapPath */
     bpath.path.type = T_BitmapHeapPath;
     bpath.path.pathtype = T_BitmapHeapScan;
     bpath.path.parent = rel;
     bpath.path.pathtarget = rel->reltarget;
     bpath.path.param_info = get_baserel_parampathinfo(root, rel,
                                                       required_outer);
     bpath.path.pathkeys = NIL;
     bpath.bitmapqual = (Path *) &apath;
 
     /*
      * Check the cost of temporary path without considering parallelism.
      * Parallel bitmap heap path will be considered at later stage.
      */
     bpath.path.parallel_workers = 0;
 
     /* Now we can do cost_bitmap_heap_scan */
     cost_bitmap_heap_scan(&bpath.path, root, rel,
                           bpath.path.param_info,
                           (Path *) &apath,
                           get_loop_count(root, rel->relid, required_outer));//BitmapHeapPath计算成本
 
     return bpath.path.total_cost;
 }
 

//-------------------------------------------------------------------------- create_bitmap_and_path

 /*
  * create_bitmap_and_path
  *    Creates a path node representing a BitmapAnd.
  */
 BitmapAndPath *
 create_bitmap_and_path(PlannerInfo *root,
                        RelOptInfo *rel,
                        List *bitmapquals)
 {
     BitmapAndPath *pathnode = makeNode(BitmapAndPath);
 
     pathnode->path.pathtype = T_BitmapAnd;
     pathnode->path.parent = rel;
     pathnode->path.pathtarget = rel->reltarget;
     pathnode->path.param_info = NULL;   /* not used in bitmap trees */
 
     /*
      * Currently, a BitmapHeapPath, BitmapAndPath, or BitmapOrPath will be
      * parallel-safe if and only if rel->consider_parallel is set.  So, we can
      * set the flag for this path based only on the relation-level flag,
      * without actually iterating over the list of children.
      */
     pathnode->path.parallel_aware = false;
     pathnode->path.parallel_safe = rel->consider_parallel;
     pathnode->path.parallel_workers = 0;
 
     pathnode->path.pathkeys = NIL;  /* always unordered */
 
     pathnode->bitmapquals = bitmapquals;
 
     /* this sets bitmapselectivity as well as the regular cost fields: */
     cost_bitmap_and_node(pathnode, root);//计算成本
 
     return pathnode;
 }

//----------------------------------------------------- cost_bitmap_and_node
 /*
  * cost_bitmap_and_node
  *      Estimate the cost of a BitmapAnd node
  *      估算BitmapAnd节点成本
  *
  * Note that this considers only the costs of index scanning and bitmap
  * creation, not the eventual heap access.  In that sense the object isn't
  * truly a Path, but it has enough path-like properties (costs in particular)
  * to warrant treating it as one.  We don't bother to set the path rows field,
  * however.
  */
 void
 cost_bitmap_and_node(BitmapAndPath *path, PlannerInfo *root)
 {
     Cost        totalCost;
     Selectivity selec;
     ListCell   *l;
 
     /*
      * We estimate AND selectivity on the assumption that the inputs are
      * independent.  This is probably often wrong, but we don't have the info
      * to do better.
      *
      * The runtime cost of the BitmapAnd itself is estimated at 100x
      * cpu_operator_cost for each tbm_intersect needed.  Probably too small,
      * definitely too simplistic?
      */
     totalCost = 0.0;
     selec = 1.0;
     foreach(l, path->bitmapquals)
     {
         Path       *subpath = (Path *) lfirst(l);
         Cost        subCost;
         Selectivity subselec;
 
         cost_bitmap_tree_node(subpath, &subCost, &subselec);
 
         selec *= subselec;
 
         totalCost += subCost;
         if (l != list_head(path->bitmapquals))
             totalCost += 100.0 * cpu_operator_cost;
     }
     path->bitmapselectivity = selec;
     path->path.rows = 0;        /* per above, not used */
     path->path.startup_cost = totalCost;
     path->path.total_cost = totalCost;
 }

三、跟踪分析

测试脚本如下

select t1.* 
from t_dwxx t1 
where (dwbh > '10000' and dwbh < '15000') AND (dwdz between 'DWDZ10000' and 'DWDZ15000');

启动gdb跟踪

(gdb) b choose_bitmap_and
Breakpoint 1 at 0x74e8c2: file indxpath.c, line 1372.
(gdb) c
Continuing.

Breakpoint 1, choose_bitmap_and (root=0x1666638, rel=0x1666a48, paths=0x166fdf0) at indxpath.c:1372
1372    int     npaths = list_length(paths);

输入参数

(gdb) p *paths
$1 = {type = T_List, length = 2, head = 0x166fe20, tail = 0x16706b8}
(gdb) p *(Node *)paths->head->data.ptr_value
$2 = {type = T_IndexPath}
(gdb) p *(Node *)paths->head->next->data.ptr_value
$3 = {type = T_IndexPath}
(gdb) set $p1=(IndexPath *)paths->head->data.ptr_value
(gdb) set $p2=(IndexPath *)paths->head->next->data.ptr_value
(gdb) p *$p1
$4 = {path = {type = T_IndexPath, pathtype = T_IndexScan, parent = 0x1666a48, pathtarget = 0x166d988, param_info = 0x0, 
    parallel_aware = false, parallel_safe = true, parallel_workers = 0, rows = 33, startup_cost = 0.28500000000000003, 
    total_cost = 116.20657683302848, pathkeys = 0x0}, indexinfo = 0x166e420, indexclauses = 0x166f528, 
  indexquals = 0x166f730, indexqualcols = 0x166f780, indexorderbys = 0x0, indexorderbycols = 0x0, 
  indexscandir = ForwardScanDirection, indextotalcost = 18.205000000000002, indexselectivity = 0.059246954595791879}
(gdb) p *$p2
$5 = {path = {type = T_IndexPath, pathtype = T_IndexScan, parent = 0x1666a48, pathtarget = 0x166d988, param_info = 0x0, 
    parallel_aware = false, parallel_safe = true, parallel_workers = 0, rows = 33, startup_cost = 0.28500000000000003, 
    total_cost = 111.33157683302848, pathkeys = 0x0}, indexinfo = 0x1666c58, indexclauses = 0x166fed0, 
  indexquals = 0x166ffc8, indexqualcols = 0x1670018, indexorderbys = 0x0, indexorderbycols = 0x0, 
  indexscandir = ForwardScanDirection, indextotalcost = 13.855, indexselectivity = 0.055688888888888899}

paths中的第1个元素对应(dwbh > '10000' and dwbh < '15000') ,第2个元素对应(dwdz between 'DWDZ10000' and 'DWDZ15000')

(gdb) set $ri1=(RestrictInfo *)$p1->indexclauses->head->data.ptr_value
(gdb) set $tmp=(RelabelType *)((OpExpr *)$ri1->clause)->args->head->data.ptr_value
(gdb) p *(Var *)$tmp->arg
$17 = {xpr = {type = T_Var}, varno = 1, varattno = 3, vartype = 1043, vartypmod = 104, varcollid = 100, varlevelsup = 0, 
  varnoold = 1, varoattno = 3, location = 76}
(gdb) p *(Node *)((OpExpr *)$ri1->clause)->args->head->next->data.ptr_value
$18 = {type = T_Const}
(gdb) p *(Const *)((OpExpr *)$ri1->clause)->args->head->next->data.ptr_value
$19 = {xpr = {type = T_Const}, consttype = 25, consttypmod = -1, constcollid = 100, constlen = -1, constvalue = 23636608, 
  constisnull = false, constbyval = false, location = 89}

开始遍历paths,提取子句条件并检测是否使用完全相同子句集的所有路径,只保留这些路径中成本最低的,这些路径被放入一个数组中进行qsort.

...
(gdb) 
1444    npaths = 0;
(gdb) 
1445    foreach(l, paths)
(gdb) 

收集信息到PathClauseUsage数组中

...
(gdb) n
1471        pathinfoarray[npaths++] = pathinfo;
(gdb) 
1445    foreach(l, paths)
(gdb) 
1476    if (npaths == 1)
(gdb) p npaths
$26 = 2
(gdb) 

按成本排序

(gdb) n
1480    qsort(pathinfoarray, npaths, sizeof(PathClauseUsage *),

遍历路径,找到成本最低的AND group

1488    for (i = 0; i < npaths; i++)
(gdb) n
1495      pathinfo = pathinfoarray[i];
(gdb) 
1496      paths = list_make1(pathinfo->path);
(gdb) 
1497      costsofar = bitmap_scan_cost_est(root, rel, pathinfo->path);
(gdb) 
1499                  list_copy(pathinfo->preds));

获取当前的成本,设置当前的条件子句

(gdb) p costsofar
$27 = 89.003250000000008
(gdb) n
1498      qualsofar = list_concat(list_copy(pathinfo->quals),

执行AND操作(路径叠加),成本更低,调整当前成本和相关变量

(gdb) n
1531        newcost = bitmap_and_cost_est(root, rel, paths);
(gdb) 
1532        if (newcost < costsofar)
(gdb) p newcost
$30 = 88.375456720095343
(gdb) n
1535          costsofar = newcost;
(gdb) n
1537                      list_copy(pathinfo->quals));
(gdb) 
1536          qualsofar = list_concat(qualsofar,
(gdb) 
1539                      list_copy(pathinfo->preds));

处理下一个AND条件,单个AND条件比上一个条件成本高,保留原来的

1488    for (i = 0; i < npaths; i++)
(gdb) 
1495      pathinfo = pathinfoarray[i];
(gdb) 
1496      paths = list_make1(pathinfo->path);
(gdb) 
1497      costsofar = bitmap_scan_cost_est(root, rel, pathinfo->path);
(gdb) 
1499                  list_copy(pathinfo->preds));
(gdb) p costsofar
$34 = 94.053250000000006
(gdb) n
1498      qualsofar = list_concat(list_copy(pathinfo->quals),
(gdb) 
1500      clauseidsofar = bms_copy(pathinfo->clauseids);
(gdb) 
1501      lastcell = list_head(paths);  /* for quick deletions */
(gdb) 
1503      for (j = i + 1; j < npaths; j++)
(gdb) 
1553      if (i == 0 || costsofar < bestcost)
(gdb) p i
$35 = 1
(gdb) p costsofar
$36 = 94.053250000000006
(gdb) p bestcost
$37 = 88.375456720095343
(gdb) 

构建BitmapAndPath,返回

(gdb) n
1563    if (list_length(bestpaths) == 1)
(gdb) 
1565    return (Path *) create_bitmap_and_path(root, rel, bestpaths);
(gdb) 
1566  }

DONE!

(gdb) n
create_index_paths (root=0x1666638, rel=0x1666a48) at indxpath.c:337
337     bpath = create_bitmap_heap_path(root, rel, bitmapqual,

四、参考资料

allpaths.c
cost.h
costsize.c
PG Document:Query Planning

来自 “ ITPUB博客 ” ,链接:http://blog.itpub.net/6906/viewspace-2374844/,如需转载,请注明出处,否则将追究法律责任。

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