sql-execution-plan

I want to know the principle of "Bitmap heap scan", I know this often happens when I execute a query with OR in the condition. Who can explain the principle behind a "Bitmap heap scan"? The best explanation comes from Tom Lane , which is the algorithm's author unless I'm mistaking. See also the wikipedia article . In short, it's a bit like a seq scan. The difference is that, rather than visiting every disk page, a bitmap index scan ANDs and ORs applicable indexes together, and only visits the disk pages that it needs to. This is different from an index scan, where the index is visited row by

I'm a relative novice when it comes to databases. We are using MySQL and I'm currently trying to speed up a SQL statement that seems to take a while to run. I looked around on SO for a similar question but didn't find one. The goal is to remove all the rows in table A that have a matching id in table B. I'm currently doing the following: DELETE FROM a WHERE EXISTS (SELECT b.id FROM b WHERE b.id = a.id); There are approximately 100K rows in table a and about 22K rows in table b. The column 'id' is the PK for both tables. This statement takes about 3 minutes to run on my test box - Pentium D, XP

I'm trying to optimize a query but don't quite understand some of the information returned from Explain Plan . Can anyone tell me the significance of the OPTIONS and COST columns? In the OPTIONS column, I only see the word FULL. In the COST column, I can deduce that a lower cost means a faster query. But what exactly does the cost value represent and what is an acceptable threshold? Jeffrey Kemp The output of EXPLAIN PLAN is a debug output from Oracle's query optimiser. The COST is the final output of the Cost-based optimiser (CBO), the purpose of which is to select which of the many different

Is there any difference in the performance of the following three SQL statements? SELECT * FROM tableA WHERE EXISTS (SELECT * FROM tableB WHERE tableA.x = tableB.y) SELECT * FROM tableA WHERE EXISTS (SELECT y FROM tableB WHERE tableA.x = tableB.y) SELECT * FROM tableA WHERE EXISTS (SELECT 1 FROM tableB WHERE tableA.x = tableB.y) They all should work and return the same result set. But does it matter if the inner SELECT selects all fields of tableB, one field, or just a constant? Is there any best practice when all statements behave equal? The truth about the EXISTS clause is that the SELECT

In a SQL Server Execution plan what is the difference between an Index Scan and an Index Seek I'm on SQL Server 2005. Justin An index scan is where SQL server reads the whole of the index looking for matches - the time this takes is proportional to the size of the index. An index seek is where SQL server uses the b-tree structure of the index to seek directly to matching records (see http://mattfleming.com/node/192 for an idea on how this works) - time taken is only proportional to the number of matching records. In general an index seek is preferable to an index scan (when the number of

When attempting to understand how a SQL statement is executing, it is sometimes recommended to look at the explain plan. What is the process one should go through in interpreting (making sense) of an explain plan? What should stand out as, "Oh, this is working splendidly?" versus "Oh no, that's not right." David Aldridge I shudder whenever I see comments that full tablescans are bad and index access is good. Full table scans, index range scans, fast full index scans, nested loops, merge join, hash joins etc. are simply access mechanisms that must be understood by the analyst and combined with

I'll preface this by saying that I hate XML, horrible stuff to work with, but necessary sometimes. My current issue is that I'm trying to take the XML from an execution plan (supplied by a user, so could be any size) and shred this into a table for further manipulation. I'm down to two options at the moment; I could work out the maximum amount of nodes available for an execution plan (I suspect this would be a lot) and create the whole script that could be used for any XML input. This would be a one time thing so not an issue. The alternative would be to work out the number of nodes