我有两个等效查询，用于提取特定区域（ace）和城市（pro_com）中建筑物（表a）与最近的高速公路（表v中的高速公路）之间的平均距离。
这是CTE版本
WITH subq AS (
SELECT a.n, a.geom as g1, unnest(ARRAY(SELECT v.geom as g2
FROM atlas_sezioni2 as v
where v.code = '12230' and a.pro_com = v.pro_com and a.code <> v.code
ORDER BY a.geom <-> v.geom LIMIT 15)) as g2
FROM atlas_sezioni2 a
where a.pro_com = 15146 and a.ace = 1 and a.code IN('11100', '11210', '11220', '11230', '11240', '11300', '12100', '14200')
)
select avg(dist) from (
select distinct on(n) n, dist
from (
SELECT n, ST_Distance_Sphere(g1, g2) as dist FROM subq
) disttable
order by n, dist asc
) final;
在CTE中，我提取最近的15条高速公路并计算距离，以便使用GIST索引（http://workshops.boundlessgeo.com/postgis-intro/knn.html）。 而CTE的解释是：
Aggregate (cost=37342.10..37342.11 rows=1 width=8)
CTE subq
-> Index Scan using atlas_sezioni2_code_ace_pro_com_n_idx on atlas_sezioni2 a (cost=0.29..29987.90 rows=20900 width=236211)
Index Cond: (((code)::text = ANY ('{11100,11210,11220,11230,11240,11300,12100,14200}'::text[])) AND (ace = 1) AND (pro_com = 15146::numeric))
SubPlan 1
-> Limit (cost=141.04..141.08 rows=15 width=236190)
-> Sort (cost=141.04..141.21 rows=69 width=236190)
Sort Key: ((a.geom <-> v.geom))
-> Index Scan using atlas_sezioni2_code_ace_pro_com_n_idx on atlas_sezioni2 v (cost=0.28..139.35 rows=69 width=236190)
Index Cond: (((code)::text = '12230'::text) AND (a.pro_com = pro_com))
Filter: ((a.code)::text <> (code)::text)
-> Unique (cost=7247.20..7351.70 rows=200 width=72)
-> Sort (cost=7247.20..7299.45 rows=20900 width=72)
Sort Key: subq.n, (_st_distance(geography(subq.g1), geography(subq.g2), 0::double precision, false))
-> CTE Scan on subq (cost=0.00..5747.50 rows=20900 width=72)
(15 rows)
这与子查询等效：
select avg(dist) from (
select distinct on(n) n, dist
from (
SELECT n, ST_Distance_Sphere(g1, g2) as dist FROM (
SELECT a.n, a.geom as g1, unnest(ARRAY(SELECT v.geom as g2
FROM atlas_sezioni2 as v
where v.code = '12230' and a.pro_com = v.pro_com and a.code <> v.code
ORDER BY a.geom <-> v.geom LIMIT 15)) as g2
FROM atlas_sezioni2 a
where a.pro_com = 15146 and a.ace = 1 and a.code IN('11100', '11210', '11220', '11230', '11240', '11300', '12100', '14200')
) subq
) disttable
order by n, dist asc
) final
及其解释
Aggregate (cost=6366298.35..6366298.36 rows=1 width=8)
-> Unique (cost=6365932.60..6366037.10 rows=20900 width=236230)
-> Sort (cost=6365932.60..6365984.85 rows=20900 width=236230)
Sort Key: subq.n, (_st_distance(geography(subq.g1), geography(subq.g2), 0::double precision, false))
-> Subquery Scan on subq (cost=0.29..35526.40 rows=20900 width=236230)
-> Index Scan using atlas_sezioni2_code_ace_pro_com_n_idx on atlas_sezioni2 a (cost=0.29..29987.90 rows=20900 width=236211)
Index Cond: (((code)::text = ANY ('{11100,11210,11220,11230,11240,11300,12100,14200}'::text[])) AND (ace = 1) AND (pro_com = 15146::numeric))
SubPlan 1
-> Limit (cost=141.04..141.08 rows=15 width=236190)
-> Sort (cost=141.04..141.21 rows=69 width=236190)
Sort Key: ((a.geom <-> v.geom))
-> Index Scan using atlas_sezioni2_code_ace_pro_com_n_idx on atlas_sezioni2 v (cost=0.28..139.35 rows=69 width=236190)
Index Cond: (((code)::text = '12230'::text) AND (a.pro_com = pro_com))
Filter: ((a.code)::text <> (code)::text)
(14 rows)
我知道CTE是优化的边界围栏（Postgres不会在CTE和它们之外的查询之间进行优化），但这很奇怪。为什么表现会以这种方式吹灭？
答案 0 :(得分：2)
正如@CraigRinger所说，我应该检查一下分析。事实上，来自＆＃34;解释分析＆＃34;我们看到第一个是：
WebView webView;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
webView = (WebView) findViewById(R.id.webview);
String url="http://www.google.com";
webView.setWebViewClient(new WebViewClient() {
@Override
public boolean shouldOverrideUrlLoading(WebView view, String url) {
view.loadUrl(url);
return true;
}
});
webView.loadUrl(url);
}
而子查询是：
Aggregate (cost=58406.66..58406.67 rows=1 width=8) (actual time=138191.294..138191.295 rows=1 loops=1)
CTE subq
-> Bitmap Heap Scan on atlas_sezioni2 a (cost=9.93..51052.46 rows=20900 width=236211) (actual time=2.814..308.667 rows=3705 loops=1)
Recheck Cond: (ace = 1)
Filter: ((pro_com = 15146::numeric) AND ((code)::text = ANY ('{11100,11210,11220,11230,11240,11300,12100,14200}'::text[])))
Rows Removed by Filter: 4
Heap Blocks: exact=42
-> Bitmap Index Scan on atlas_sezioni2_ace_idx (cost=0.00..9.88 rows=251 width=0) (actual time=0.110..0.110 rows=251 loops=1)
Index Cond: (ace = 1)
SubPlan 1
-> Limit (cost=240.70..240.74 rows=15 width=236190) (actual time=0.630..0.636 rows=15 loops=247)
-> Sort (cost=240.70..240.87 rows=69 width=236190) (actual time=0.627..0.630 rows=15 loops=247)
Sort Key: ((a.geom <-> v.geom))
Sort Method: top-N heapsort Memory: 26kB
-> Bitmap Heap Scan on atlas_sezioni2 v (cost=4.56..239.01 rows=69 width=236190) (actual time=0.045..0.518 rows=73 loops=247)
Recheck Cond: ((code)::text = '12230'::text)
Filter: (((a.code)::text <> (code)::text) AND (a.pro_com = pro_com))
Heap Blocks: exact=6916
-> Bitmap Index Scan on atlas_sezioni2_code_idx (cost=0.00..4.55 rows=73 width=0) (actual time=0.030..0.030 rows=73 loops=247)
Index Cond: ((code)::text = '12230'::text)
-> Unique (cost=7247.20..7351.70 rows=200 width=72) (actual time=138190.527..138191.243 rows=247 loops=1)
-> Sort (cost=7247.20..7299.45 rows=20900 width=72) (actual time=138190.526..138190.800 rows=3705 loops=1)
Sort Key: subq.n, (_st_distance(geography(subq.g1), geography(subq.g2), 0::double precision, false))
Sort Method: quicksort Memory: 270kB
-> CTE Scan on subq (cost=0.00..5747.50 rows=20900 width=72) (actual time=159.739..138182.891 rows=3705 loops=1)
Planning time: 2.623 ms
Execution time: 138217.574 ms
(27 rows)
所以它只在解释:)中表现更好。真正的表现并没有改变。