[20220531]測試quiz night.txt
[20220531]測試quiz night.txt
--//自己重複測試看看。
1.環境:
SCOTT@book> @ver1
PORT_STRING VERSION BANNER
------------------------------ -------------- --------------------------------------------------------------------------------
x86_64/Linux 2.4.xx 11.2.0.4.0 Oracle Database 11g Enterprise Edition Release 11.2.0.4.0 - 64bit Production
2.測試:
/* Formatted on 2022/5/31 10:41:31 (QP5 v5.269.14213.34769) */
CREATE TABLE interr_skuplannparam
(
atpdur NUMBER (38)
,depdmdopt NUMBER (38)
,externalskusw NUMBER (1)
,firstreplendate DATE
,lastfrzstart DATE
,lastplanstart DATE
,plandur NUMBER (38)
,planleadtime NUMBER (38)
,planleadtimerule NUMBER (38)
,planshipfrzdur NUMBER (38)
,restrictdur NUMBER (38)
,allocbatchsw NUMBER (1)
,cmpfirmdur NUMBER (38)
,custservicelevel FLOAT (126)
,maxchangefactor FLOAT (126)
,mfgleadtime NUMBER (38)
,recschedrcptsdur NUMBER (38)
,cpppriority NUMBER (38)
,cpplocksw NUMBER (1)
,criticalmaterialsw NUMBER (1)
,aggexcesssupplyrule NUMBER (38)
,aggundersupplyrule NUMBER (38)
,bufferleadtime NUMBER (38)
,maxoh FLOAT (126)
,maxcovdur NUMBER (38)
,drpcovdur NUMBER (38)
,drpfrzdur NUMBER (38)
,drprule NUMBER (38)
,drptimefencedate DATE
,drptimefencedur NUMBER (38)
,incdrpqty FLOAT (126)
,mindrpqty FLOAT (126)
,mpscovdur NUMBER (38)
,mfgfrzdur NUMBER (38)
,mpsrule NUMBER (38)
,mpstimefencedate DATE
,mpstimefencedur NUMBER (38)
,incmpsqty FLOAT (126)
,minmpsqty FLOAT (126)
,shrinkagefactor NUMBER (38)
,item VARCHAR2 (50 CHAR)
,loc VARCHAR2 (50 CHAR)
,expdate DATE
,atprule NUMBER (38)
,prodcal VARCHAR2 (50 CHAR)
,prodstartdate DATE
,prodstopdate DATE
,orderingcost FLOAT (126)
,holdingcost FLOAT (126)
,eoq FLOAT (126)
,ff_trigger_control NUMBER (38)
,workingcal VARCHAR2 (50 CHAR)
,lookaheaddur NUMBER
,orderpointrule NUMBER
,orderskudetailsw NUMBER (1)
,supsdmindmdcovdur NUMBER (38)
,orderpointminrule NUMBER (38)
,orderpointminqty FLOAT (126)
,orderpointmindur NUMBER (38)
,orderuptolevelmaxrule NUMBER (38)
,orderuptolevelmaxqty FLOAT (126)
,orderuptolevelmaxdur NUMBER (38)
,aggskurule NUMBER (38)
,fwdbuymaxdur NUMBER (38)
,costuom NUMBER (38)
,cumleadtimedur NUMBER (38)
,cumleadtimeadjdur NUMBER (38)
,cumleadtimerule NUMBER (38)
,roundingfactor FLOAT (126)
,limitplanarrivpublishsw NUMBER (1)
,limitplanarrivpublishdur NUMBER
,maxohrule NUMBER (1)
,integration_stamp DATE
,integration_jobid VARCHAR2 (32 CHAR) NOT NULL
,error_str VARCHAR2 (2000 CHAR)
,error_stamp DATE
);
SCOTT@book> alter table interr_skuplannparam modify integration_jobid default 'INT_JOB';
Table altered.
--//truncate table interr_skuplannparam;
insert into interr_skuplannparam (atpdur)
select 0
from all_objects
where rownum <= 10000
/
commit;
execute dbms_stats.gather_table_stats(user,'interr_skuplannparam')
SCOTT@book> SELECT AVG_ROW_LEN FROM USER_TABLES WHERE TABLE_NAME='INTERR_SKUPLANNPARAM';
AVG_ROW_LEN
-----------
20
--//怎麼是20呢,我在測試前做了估算開始因為長度指示器佔1,數字0編碼0x80,佔1個位元組,72個NULL,integration_jobid預設值是
--//INT_JOB佔7,加上長度指示器1.
--//2+72+1+7 = 82應該是82。後面的null值oracle不儲存。
--//一位讀者的解析:
The result of the last query cannot be determined with the given information – unless interr_skuplannparam is the only
table in the current schema. If it is the only table, the expected result is 1 (row contains not null values) + 1 (size
of Oracle's representation of 0) + (7+1) (length of the not-null column in bytes plus one byte to store the string's
length) + 10*1 (because there are 10 date columns) = 20.
不能用給定的資訊來確定最後一個查詢的結果——除非interr_skuplannparam是當前模式中唯一的表。如果是唯一的表,預期結果是1(行
不包含空值)+1(Oracle表示0的大小)+(7+1)(位元組的非空列長度加上一個位元組儲存字串的長度)+10*1(因為有10個日期列)=20。
--//按照讀者的演算法僅僅考慮日期型別的NULL,為什麼呢?
--//透過bbed檢視分析:
SCOTT@book> select rowid from interr_skuplannparam where rownum=1;
ROWID
------------------
AABRWQAAEAAAAkbAAA
SCOTT@book> @ rowid AABRWQAAEAAAAkbAAA
OBJECT FILE BLOCK ROW ROWID_DBA DBA TEXT
---------- ---------- ---------- ---------- -------------------- -------------------- ----------------------------------------
333200 4 2331 0 0x100091B 4,2331 alter system dump datafile 4 block 2331
BBED> set width 160
WIDTH 160
BBED> set dba 4,2331
DBA 0x0100091b (16779547 4,2331)
BBED> x /rnxxxx *kdbr[0]
rowdata[3740] @4873
-------------
flag@4873: 0x2c (KDRHFL, KDRHFF, KDRHFH)
lock@4874: 0x01
cols@4875: 74
col 0[1] @4876: 0
col 1[0] @4878: *NULL*
col 2[0] @4879: *NULL*
col 3[0] @4880: *NULL*
col 4[0] @4881: *NULL*
col 5[0] @4882: *NULL*
col 6[0] @4883: *NULL*
col 7[0] @4884: *NULL*
col 8[0] @4885: *NULL*
col 9[0] @4886: *NULL*
col 10[0] @4887: *NULL*
col 11[0] @4888: *NULL*
col 12[0] @4889: *NULL*
col 13[0] @4890: *NULL*
col 14[0] @4891: *NULL*
col 15[0] @4892: *NULL*
col 16[0] @4893: *NULL*
col 17[0] @4894: *NULL*
col 18[0] @4895: *NULL*
col 19[0] @4896: *NULL*
col 20[0] @4897: *NULL*
col 21[0] @4898: *NULL*
col 22[0] @4899: *NULL*
col 23[0] @4900: *NULL*
col 24[0] @4901: *NULL*
col 25[0] @4902: *NULL*
col 26[0] @4903: *NULL*
col 27[0] @4904: *NULL*
col 28[0] @4905: *NULL*
col 29[0] @4906: *NULL*
col 30[0] @4907: *NULL*
col 31[0] @4908: *NULL*
col 32[0] @4909: *NULL*
col 33[0] @4910: *NULL*
col 34[0] @4911: *NULL*
col 35[0] @4912: *NULL*
col 36[0] @4913: *NULL*
col 37[0] @4914: *NULL*
col 38[0] @4915: *NULL*
col 39[0] @4916: *NULL*
col 40[0] @4917: *NULL*
col 41[0] @4918: *NULL*
col 42[0] @4919: *NULL*
col 43[0] @4920: *NULL*
col 44[0] @4921: *NULL*
col 45[0] @4922: *NULL*
col 46[0] @4923: *NULL*
col 47[0] @4924: *NULL*
col 48[0] @4925: *NULL*
col 49[0] @4926: *NULL*
col 50[0] @4927: *NULL*
col 51[0] @4928: *NULL*
col 52[0] @4929: *NULL*
col 53[0] @4930: *NULL*
col 54[0] @4931: *NULL*
col 55[0] @4932: *NULL*
col 56[0] @4933: *NULL*
col 57[0] @4934: *NULL*
col 58[0] @4935: *NULL*
col 59[0] @4936: *NULL*
col 60[0] @4937: *NULL*
col 61[0] @4938: *NULL*
col 62[0] @4939: *NULL*
col 63[0] @4940: *NULL*
col 64[0] @4941: *NULL*
col 65[0] @4942: *NULL*
col 66[0] @4943: *NULL*
col 67[0] @4944: *NULL*
col 68[0] @4945: *NULL*
col 69[0] @4946: *NULL*
col 70[0] @4947: *NULL*
col 71[0] @4948: *NULL*
col 72[0] @4949: *NULL*
col 73[7] @4950: 0x49 0x4e 0x54 0x5f 0x4a 0x4f 0x42
BBED> p kdbr
sb2 kdbr[0] @118 4773
sb2 kdbr[1] @120 4858
...
--//4858-4773 = 85 , 一行佔85位元組。行頭佔3個位元組
flag@4873: 0x2c (KDRHFL, KDRHFF, KDRHFH)
lock@4874: 0x01
cols@4875: 74
--//而按照前面的計算僅僅計算前面是date型別的NULL值的情況
--//1+1+10+7+1 = 20,為什麼不理解以前也沒有注意這個細節。
--//期待作者的解析。
來自 “ ITPUB部落格 ” ,連結:http://blog.itpub.net/267265/viewspace-2898095/,如需轉載,請註明出處,否則將追究法律責任。
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