update x set y = null 需要很长时间
在工作中，我有一个大表(大约 300 万行，例如 40-50 列).我有时需要清空一些列并用新数据填充它们.没想到的是
At work, I have a large table (some 3 million rows, like 40-50 columns). I sometimes need to empty some of the columns and fill them with new data. What I did not expect is that
UPDATE table1 SET y = null
比用生成的数据填充列花费的时间要多得多，例如，在 sql 查询中来自同一表的其他列或从子查询中的其他表查询.我是否一次遍历所有表行(如上面的更新查询中)或我是否使用游标逐行遍历表(使用 pk)都没有关系.无论我在工作中使用大表还是创建一个小测试表并用数十万个测试行填充它都没有关系.将列设置为 null 总是比用一些动态数据更新列(每行不同)花费的时间更长(在整个测试中，我遇到了 2 到 10 的因素).
takes much more time than filling the column with data which is generated, for example, in the sql query from other columns of the same table or queried from other tables in a subquery. It does not matter if I go through all table rows at once (like in the update query above) or if I use a cursor to go through the table row by row (using the pk). It does not matter if I use the large table at work or if I create a small test table and fill it with some hundredthousands of test-rows. Setting the column to null always takes way longer (Throughout the tests, I encountered factors of 2 to 10) than updating the column with some dynamic data (which is different for each row).
这是什么原因?将列设置为空时，Oracle 会做什么?或者 - 我的推理错误是什么?
Whats the reason for this? What does Oracle do when setting a column to null? Or - what's is my error in reasoning?
感谢您的帮助！
P.S.:我使用的是 oracle 11g2，并且使用 plsql developer 和 oracle sql developer 都发现了这些结果.
P.S.: I am using oracle 11g2, and found these results using both plsql developer and oracle sql developer.
推荐答案
总结
我认为更新为空值比较慢，因为 Oracle(错误地)试图利用它存储空值的方式，导致它频繁地重新组织块中的行(堆块压缩")，从而产生大量额外的撤消和重做.
I think updating to null is slower because Oracle (incorrectly) tries to take advantage of the way it stores nulls, causing it to frequently re-organize the rows in the block ("heap block compress"), creating a lot of extra UNDO and REDO.
null 有什么特别之处?
来自 Oracle 数据库概念:
如果空值位于具有数据值的列之间，则它们将存储在数据库中.在这些情况下，它们需要 1 个字节来存储列的长度(零).
"Nulls are stored in the database if they fall between columns with data values. In these cases they require 1 byte to store the length of the column (zero).
一行中的尾随空值不需要存储，因为新的行标题表示前一行中的其余列为空值.例如，如果表的最后三列为空，则不会存储这些列的信息.在多列的表中，更可能包含空值的列应该最后定义以节省磁盘空间."
Trailing nulls in a row require no storage because a new row header signals that the remaining columns in the previous row are null. For example, if the last three columns of a table are null, no information is stored for those columns. In tables with many columns, the columns more likely to contain nulls should be defined last to conserve disk space."
测试
基准更新非常困难，因为不能仅从更新语句中衡量更新的真实成本.例如，日志开关将不是每次更新都会发生，延迟块清理会在以后发生.要准确测试更新，应该进行多次运行，每次运行都应该重新创建对象，并且应该丢弃高低值.
Benchmarking updates is very difficult because the true cost of an update cannot be measured just from the update statement. For example, log switches will not happen with every update, and delayed block cleanout will happen later. To accurately test an update, there should be multiple runs, objects should be recreated for each run, and the high and low values should be discarded.
为简单起见，下面的脚本不会抛出高低结果，仅测试具有单列的表.但是不管列数多少，数据多少，更新哪一列，问题依然存在.
For simplicity the script below does not throw out high and low results, and only tests a table with a single column. But the problem still occurs regardless of the number of columns, their data, and which column is updated.
我使用了来自 http://www.oracle-developer.net/utilities 的 RunStats 实用程序.php比较update-to-a-value和update-to-a-null的资源消耗.
I used the RunStats utility from http://www.oracle-developer.net/utilities.php to compare the resource consumption of updating-to-a-value with updating-to-a-null.
create table test1(col1 number);
BEGIN
dbms_output.enable(1000000);
runstats_pkg.rs_start;
for i in 1 .. 10 loop
execute immediate 'drop table test1 purge';
execute immediate 'create table test1 (col1 number)';
execute immediate 'insert /*+ append */ into test1 select 1 col1
from dual connect by level <= 100000';
commit;
execute immediate 'update test1 set col1 = 1';
commit;
end loop;
runstats_pkg.rs_pause;
runstats_pkg.rs_resume;
for i in 1 .. 10 loop
execute immediate 'drop table test1 purge';
execute immediate 'create table test1 (col1 number)';
execute immediate 'insert /*+ append */ into test1 select 1 col1
from dual connect by level <= 100000';
commit;
execute immediate 'update test1 set col1 = null';
commit;
end loop;
runstats_pkg.rs_stop();
END;
/
结果
有很多不同之处，我认为这些是最相关的四个:
There are dozens of differences, these are the four I think are most relevant:
Type Name Run1 Run2 Diff
----- ---------------------------- ------------ ------------ ------------
TIMER elapsed time (hsecs) 1,269 4,738 3,469
STAT heap block compress 1 2,028 2,027
STAT undo change vector size 55,855,008 181,387,456 125,532,448
STAT redo size 133,260,596 581,641,084 448,380,488
解决方案?
我能想到的唯一可能的解决方案是启用表压缩.压缩表不会发生尾随空存储技巧.因此，即使 Run2 的堆块压缩"数字变得更高，从 2028 年到 23208 年，我想它实际上并没有做任何事情.在启用表压缩的情况下，两次运行之间的重做、撤消和经过时间几乎相同.
The only possible solution I can think of is to enable table compression. The trailing-null storage trick doesn't happen for compressed tables. So even though the "heap block compress" number gets even higher for Run2, from 2028 to 23208, I guess it doesn't actually do anything. The redo, undo, and elapsed time between the two runs is almost identical with table compression enabled.
然而，表压缩有很多潜在的缺点.更新为 null 会运行得更快，但其他所有更新的运行速度至少会稍微慢一些.
However, there are lots of potential downsides to table compression. Updating to a null will run much faster, but every other update will run at least slightly slower.
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