<?xml version="1.0" encoding="UTF-8"?>
<rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:wfw="http://wellformedweb.org/CommentAPI/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
	xmlns:slash="http://purl.org/rss/1.0/modules/slash/"
	>

<channel>
	<title>Rise of the PHX &#187; boring code</title>
	<atom:link href="http://riseofthephx.com/category/boring-code/feed/" rel="self" type="application/rss+xml" />
	<link>http://riseofthephx.com</link>
	<description>Web musings in the Valley of the Sun.</description>
	<lastBuildDate>Mon, 19 Jul 2010 18:29:49 +0000</lastBuildDate>
	<generator>http://wordpress.org/?v=2.9.2</generator>
	<language>en</language>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
			<item>
		<title>Find pesky duplicates in mysql</title>
		<link>http://riseofthephx.com/2009/12/14/find-pesky-duplicates-in-mysql/</link>
		<comments>http://riseofthephx.com/2009/12/14/find-pesky-duplicates-in-mysql/#comments</comments>
		<pubDate>Mon, 14 Dec 2009 17:00:23 +0000</pubDate>
		<dc:creator>gturnbul</dc:creator>
				<category><![CDATA[boring code]]></category>

		<guid isPermaLink="false">http://riseofthephx.com/?p=129</guid>
		<description><![CDATA[Let&#8217;s say you had an app that had a ton of users and were suspicious that there were pesky duplicates.  Here is a quick mysql select to help put your mind at ease.

select username,count(*) as n
 from users
group by username
having  n > 1;

or search by first and last name:

select concat(firstname,' ',lastname) as name, [...]]]></description>
			<content:encoded><![CDATA[<p>Let&#8217;s say you had an app that had a ton of users and were suspicious that there were pesky duplicates.  Here is a quick mysql select to help put your mind at ease.</p>
<p><code><br />
select username,count(*) as n<br />
 from users<br />
group by username<br />
having  n > 1;<br />
</code></p>
<p>or search by first and last name:</p>
<p><code><br />
select concat(firstname,' ',lastname) as name, count(*) as n<br />
 from users<br />
group by name<br />
having  n > 1;<br />
</code></p>
]]></content:encoded>
			<wfw:commentRss>http://riseofthephx.com/2009/12/14/find-pesky-duplicates-in-mysql/feed/</wfw:commentRss>
		<slash:comments>0</slash:comments>
		</item>
	</channel>
</rss>
