V$DB_CACHE_ADVICE视图到底是个啥意思啊，我看了看9iOCP的文档，越看越晕： [php] Using V$DB_CACHE_ADVICE View The following output shows that if the cache was 212MB, rather than the current size of 304MB, the estimated number of physical reads would be 17 million (17,850,847). Increasing the cache size beyond its current size would not provide a significant benefit. Estd Phys Estd Phys Cache Size (MB) Buffers Read Factor Reads ----------------------- ---------------- ------------- ------------------------------ (10%) 30 3,802 18.70 192,317,943 60 7,604 12.83 131,949,536 91 11,406 7.38 75,865,861 121 15,208 4.97 51,111,658 152 19,010 3.64 37,460,786 182 22,812 2.50 25,668,196 212 26,614 1.74 17,850,847 243 30,416 1.33 13,720,149 273 34,218 1.13 11,583,180 (Current) 304 38,020 1.00 10,282,475 334 41,822 .93 9,515,878 364 45,624 .87 8,909,026 395 49,426 .83 8,495,039 424 53,228 .79 8,116,496 (150%) 456 57,030 .76 7,824,764 … [/php] 这是我翻译的那一段说明： 下列输出显示如果cache是212M，胜于当前的304M，物理读估计数是17,850,847。增加cache大小超过当前值不会有好处。 这里212M时物理读估计数是17,850,847，304M时物理读估计数是10,282,475，不还是304M时好一点吗，怎么212M就胜过了304M呢。