Solved

Help with Inner Join Replace

Posted on 2010-11-16
2
281 Views
Last Modified: 2012-05-10
This query takes 2 seconds to run:

update WideCityList
set Item=1
from widecitylist
inner join (SELECT  Distinct zips.CityName, zips.ProvinceAbbr
  FROM dbo.postalcodes AS [zips]
 INNER JOIN dbo.CalculateBoundary(49.136510, - 122.831860, 45.000000, 'kilometers') AS bounds ON 1 = 1
  WHERE [zips].[Latitude] BETWEEN [bounds].[South] AND [bounds].[North]
   AND [zips].[Longitude] BETWEEN [bounds].[West] AND [bounds].[East]
   AND [zips].[Latitude] <> 0
   AND [zips].[Longitude] <> 0
   AND   (dbo.CalculateDistance(49.136510, - 122.831860, zips.Latitude, zips.Longitude, 'kilometers') <= 21.000000)  
   AND [zips].[CityType] = 'D') c on
c.cityname=widecitylist.citt  and c.provinceabbr = widecitylist.state


and all i do is change the last line to:
 
c.cityname=Replace (widecitylist.city,'-',' ')  and c.provinceabbr = widecitylist.state
       
And now it takes over 20 seconds, which is unacceptable.

What is happening is that the query pulls all fields from dbo.zips in a radius of 21 kilometers of Surrey BC, which is 32,000 (canada has hundreds of thousands of zipcodes, sometimes 10,000+ in a single city).  I then select only the distinct city, which is 24.

From those 24 results, I simply want to update WideCityList item = 1, but I needed the replace function because all cities in WideCityList have a "-" instead of a space for cities with two words.  I dont know why it takes so horribly long to run the query simply by changing the last line...I'm assuming that somewhere in there its doing an inner join using REPLACE on all 32000 results before DISTINCT is processed?  I only want to perform the inner join with Replace against the 24 results of the inside select statement.

hoping this makes sense!   Any ideas?
0
Comment
Question by:arthurh88
2 Comments
 
LVL 58

Accepted Solution

by:
cyberkiwi earned 500 total points
ID: 34151619
Try this form:

update WideCityList
set Item=1
from (SELECT  Distinct zips.CityName, zips.ProvinceAbbr
  FROM dbo.postalcodes AS [zips]
 INNER JOIN dbo.CalculateBoundary(49.136510, - 122.831860, 45.000000, 'kilometers') AS bounds ON 1 = 1
  WHERE [zips].[Latitude] BETWEEN [bounds].[South] AND [bounds].[North]
   AND [zips].[Longitude] BETWEEN [bounds].[West] AND [bounds].[East]
   AND [zips].[Latitude] <> 0
   AND [zips].[Longitude] <> 0
   AND   (dbo.CalculateDistance(49.136510, - 122.831860, zips.Latitude, zips.Longitude, 'kilometers') <= 21.000000)  
   AND [zips].[CityType] = 'D') c
inner join widecitylist on
c.cityname=Replace(widecitylist.city,'-',' ')  and c.provinceabbr = widecitylist.state
option(force order)

If there are only 24 matches, it should be reasonably quick.
If that doesn't work, reverse the condition

inner join widecitylist on
Replace(c.cityname,' ','-')=widecitylist.city  and c.provinceabbr = widecitylist.state

The reversal changes 24 rows, but uses that to do an index lookup in the widecitylist table (assuming one can be used)
0
 

Author Comment

by:arthurh88
ID: 34151631
"inner join widecitylist on
Replace(c.cityname,' ','-')=widecitylist.city  and c.provinceabbr = widecitylist.state"

that worked!!  (the reversal)  thanks so much.  
0

Featured Post

NAS Cloud Backup Strategies

This article explains backup scenarios when using network storage. We review the so-called “3-2-1 strategy” and summarize the methods you can use to send NAS data to the cloud

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

This article explains how to reset the password of the sa account on a Microsoft SQL Server.  The steps in this article work in SQL 2005, 2008, 2008 R2, 2012, 2014 and 2016.
JSON is being used more and more, besides XML, and you surely wanted to parse the data out into SQL instead of doing it in some Javascript. The below function in SQL Server can do the job for you, returning a quick table with the parsed data.
Via a live example, show how to setup several different housekeeping processes for a SQL Server.
Viewers will learn how to use the INSERT statement to insert data into their tables. It will also introduce the NULL statement, to show them what happens when no value is giving for any given column.

770 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question