问题:
我正在创建一个 Trac 报告,显示在我们图书馆的每个章节的开发周期的每个阶段有多少票。 一张工单代表一项工作,通常是一个单独的例程。
例如,在第 X 章的同行评审阶段,即将发布(里程碑)的门票有多少。
共有 10 个发展阶段和 47 个章节。
给定的 MySQL 查询适用于所有 10 个开发阶段,但仅适用于一章且长 25 行,因此所有章节的整个查询超过 1200 行。
Trac 给出的错误是 KeyError: 'numrows' 查询变得很大。
当直接在 MySQL 中输入查询时,给出的错误是 Out of resources when opening file (Errcode: 24) (23)
问题:
重构 - 这可以做得“更好”吗?sql 大师们,有一些聪明的技巧/高级技术吗?
方法 - 我是否需要完全不同的方法?
配置 - MySQL 和/或 Trac 是否可以配置为接受非常大的查询
注释:
表中的数据很小,在明显的大小限制下执行查询不会花费很长时间。
查询从 Trac 系统传递到 MySQL,这对可以执行的操作施加了一些限制,例如,只能从 trac 发送单个查询以生成报告。
可以查看 Trac 报告的示例 here.
查询中的 %c%* 只是我在通过脚本生成查询时用于替换实际章节的唯一字符串。
SELECT '%c%' as Chapter,
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status IN ('new','assigned') ) AS 'New',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='document_interface' ) AS 'Document\
Interface',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='interface_development' ) AS 'Inter\
face Development',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='interface_check' ) AS 'Interface C\
heck',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='document_routine' ) AS 'Document R\
outine',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='full_development' ) AS 'Full Devel\
opment',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='peer_review_1' ) AS 'Peer Review O\
ne',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%'AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='peer_review_2' ) AS 'Peer Review Tw\
o',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='qa' ) AS 'QA',
(SELECT count(ticket.id) AS Matches FROM engine.ticket INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%'AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine' AND ticket.status='closed' ) AS 'Closed',
count(id) AS Total,
ticket.id AS _id
FROM engine.ticket
INNER JOIN engine.ticket_custom ON ticket.id = ticket_custom.ticket
WHERE ticket_custom.name='chapter' AND ticket_custom.value LIKE '%c%' AND type='New material' AND milestone='1.1.12' AND component NOT LIKE 'internal_engine'
最佳答案
不是对每个计数都进行子查询,而是使用 case 从已为查询获取的数据中计数:
select '%c%' as Chapter,
sum(case when ticket.status IN ('new','assigned') then 1 else 0 end) as 'New',
sum(case when ticket.status='document_interface' then 1 else 0 end) as 'DocumentInterface',
sum(case when ticket.status='interface_development' then 1 else 0 end) as 'Interface Development',
sum(case when ticket.status='interface_check' then 1 else 0 end) as 'Interface Check',
sum(case when ticket.status='document_routine' then 1 else 0 end) as 'Document Routine',
sum(case when ticket.status='full_development' then 1 else 0 end) as 'Full Development',
sum(case when ticket.status='peer_review_1' then 1 else 0 end) as 'Peer Review One',
sum(case when ticket.status='peer_review_2' then 1 else 0 end) as 'Peer Review Two',
sum(case when ticket.status='qa' then 1 else 0 end) as 'QA',
sum(case when ticket.status='closed' then 1 else 0 end) as 'Closed',
count(id) as Total,
ticket.id as _id
from
engine.ticket
inner join engine.ticket_custom on ticket.id = ticket_custom.ticket
where
ticket_custom.name='chapter' and
ticket_custom.value LIKE '%c%' and
type='New material' and
milestone='1.1.12' and
component NOT LIKE 'internal_engine'
关于mysql - 非常大的 SQL 查询的问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12607667/
我正在用Ruby编写一个简单的程序来检查域列表是否被占用。基本上它循环遍历列表,并使用以下函数进行检查。require'rubygems'require'whois'defcheck_domain(domain)c=Whois::Client.newc.query("google.com").available?end程序不断出错(即使我在google.com中进行硬编码),并打印以下消息。鉴于该程序非常简单,我已经没有什么想法了-有什么建议吗?/Library/Ruby/Gems/1.8/gems/whois-2.0.2/lib/whois/server/adapters/base.
我想为Heroku构建一个Rails3应用程序。他们使用Postgres作为他们的数据库,所以我通过MacPorts安装了postgres9.0。现在我需要一个postgresgem并且共识是出于性能原因你想要pggem。但是我对我得到的错误感到非常困惑当我尝试在rvm下通过geminstall安装pg时。我已经非常明确地指定了所有postgres目录的位置可以找到但仍然无法完成安装:$envARCHFLAGS='-archx86_64'geminstallpg--\--with-pg-config=/opt/local/var/db/postgresql90/defaultdb/po
尝试通过RVM将RubyGems升级到版本1.8.10并出现此错误:$rvmrubygemslatestRemovingoldRubygemsfiles...Installingrubygems-1.8.10forruby-1.9.2-p180...ERROR:Errorrunning'GEM_PATH="/Users/foo/.rvm/gems/ruby-1.9.2-p180:/Users/foo/.rvm/gems/ruby-1.9.2-p180@global:/Users/foo/.rvm/gems/ruby-1.9.2-p180:/Users/foo/.rvm/gems/rub
我的最终目标是安装当前版本的RubyonRails。我在OSXMountainLion上运行。到目前为止,这是我的过程:已安装的RVM$\curl-Lhttps://get.rvm.io|bash-sstable检查已知(我假设已批准)安装$rvmlistknown我看到当前的稳定版本可用[ruby-]2.0.0[-p247]输入命令安装$rvminstall2.0.0-p247注意:我也试过这些安装命令$rvminstallruby-2.0.0-p247$rvminstallruby=2.0.0-p247我很快就无处可去了。结果:$rvminstall2.0.0-p247Search
由于fast-stemmer的问题,我很难安装我想要的任何rubygem。我把我得到的错误放在下面。Buildingnativeextensions.Thiscouldtakeawhile...ERROR:Errorinstallingfast-stemmer:ERROR:Failedtobuildgemnativeextension./System/Library/Frameworks/Ruby.framework/Versions/2.0/usr/bin/rubyextconf.rbcreatingMakefilemake"DESTDIR="cleanmake"DESTDIR=
我知道我可以指定某些字段来使用pluck查询数据库。ids=Item.where('due_at但是我想知道,是否有一种方法可以指定我想避免从数据库查询的某些字段。某种反拔?posts=Post.where(published:true).do_not_lookup(:enormous_field) 最佳答案 Model#attribute_names应该返回列/属性数组。您可以排除其中一些并传递给pluck或select方法。像这样:posts=Post.where(published:true).select(Post.attr
当我尝试安装Ruby时遇到此错误。我试过查看this和this但无济于事➜~brewinstallrubyWarning:YouareusingOSX10.12.Wedonotprovidesupportforthispre-releaseversion.Youmayencounterbuildfailuresorotherbreakages.Pleasecreatepull-requestsinsteadoffilingissues.==>Installingdependenciesforruby:readline,libyaml,makedepend==>Installingrub
我正在尝试使用boilerpipe来自JRuby。我看过guide从JRuby调用Java,并成功地将它与另一个Java包一起使用,但无法弄清楚为什么同样的东西不能用于boilerpipe。我正在尝试基本上从JRuby中执行与此Java等效的操作:URLurl=newURL("http://www.example.com/some-location/index.html");Stringtext=ArticleExtractor.INSTANCE.getText(url);在JRuby中试过这个:require'java'url=java.net.URL.new("http://www
我意识到这可能是一个非常基本的问题,但我现在已经花了几天时间回过头来解决这个问题,但出于某种原因,Google就是没有帮助我。(我认为部分问题在于我是一个初学者,我不知道该问什么......)我也看过O'Reilly的RubyCookbook和RailsAPI,但我仍然停留在这个问题上.我找到了一些关于多态关系的信息,但它似乎不是我需要的(尽管如果我错了请告诉我)。我正在尝试调整MichaelHartl'stutorial创建一个包含用户、文章和评论的博客应用程序(不使用脚手架)。我希望评论既属于用户又属于文章。我的主要问题是:我不知道如何将当前文章的ID放入评论Controller。
首先回顾一下拉格朗日定理的内容:函数f(x)是在闭区间[a,b]上连续、开区间(a,b)上可导的函数,那么至少存在一个,使得:通过这个表达式我们可以知道,f(x)是函数的主体,a和b可以看作是主体函数f(x)中所取的两个值。那么可以有, 也就意味着我们可以用来替换 这种替换可以用在求某些多项式差的极限中。方法: 外层函数f(x)是一致的,并且h(x)和g(x)是等价无穷小。此时,利用拉格朗日定理,将原式替换为 ,再进行求解,往往会省去复合函数求极限的很多麻烦。使用要注意:1.要先找到主体函数f(x),即外层函数必须相同。2.f(x)找到后,复合部分是等价无穷小。3.要满足作差的形式。如果是加