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如何优化HTML而不是SQL查询的呈现?
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Stack Overflow用户
提问于 2013-05-17 23:07:45
回答 1查看 735关注 0票数 1

所以我用的是微型剖析器,它给了我一些很好的统计数据。

然而,我注意到的一件事是,我得到了很多SQL调用的最小值,现在最大的事情是呈现各种部分和HTML。

例如,以下是我所面临的两个不同问题的例子:

Mini-Profiler

代码语言:javascript
复制
GET http://localhost:3000/   14.0   +0.0
  Executing action: index    9.5    +9.0
   Rendering: home/index     7.8    +16.0
    Rendering: home/_row     7.9    +22.0
   Rendering: layouts/application    1038.7 +32.0
    Rendering: layouts/_navigation   6.0    +586.0

Development.log

代码语言:javascript
复制
Started GET "/" for 127.0.0.1 at 2013-05-17 18:00:26 -0500
Processing by HomeController#index as HTML
  Item Load (0.6ms)  SELECT "items".* FROM "items" WHERE "items"."is_approved" = 't'
  ActsAsTaggableOn::Tag Load (0.6ms)  SELECT "tags".* FROM "tags" INNER JOIN "taggings" ON "tags"."id" = "taggings"."tag_id" WHERE "taggings"."taggable_id" = 13 AND "taggings"."taggable_type" = 'Item' AND (taggings.context = 'tags' AND taggings.tagger_id IS NULL)
  Rendered home/_row.html.erb (8.0ms)
  Rendered home/index.html.erb within layouts/application (15.7ms)
  Rendered layouts/_social_media.html.erb (0.5ms)
   (0.4ms)  SELECT items.id FROM "items" 
  ActsAsTaggableOn::Tag Load (0.7ms)  SELECT tags.*, taggings.tags_count AS count FROM "tags" JOIN (SELECT taggings.tag_id, COUNT(taggings.tag_id) AS tags_count FROM "taggings" INNER JOIN items ON items.id = taggings.taggable_id WHERE (taggings.taggable_type = 'Item' AND taggings.context = 'tags') AND (taggings.taggable_id IN(13)) GROUP BY taggings.tag_id HAVING COUNT(taggings.tag_id) > 0) AS taggings ON taggings.tag_id = tags.id ORDER BY count LIMIT 5
  Rendered layouts/_navigation.html.erb (6.1ms)
  Rendered layouts/_site_nav.html.erb (0.6ms)
  Rendered layouts/_messages.html.erb (0.2ms)
  Rendered layouts/_footer.html.erb (0.1ms)
Completed 200 OK in 1069ms (Views: 1066.0ms | ActiveRecord: 2.3ms)

从顶部的Mini输出,显示主application.html.erb显著增加了加载时间。

这里是另一个应用程序的另一个例子,其中视图是花费最多时间呈现的:

代码语言:javascript
复制
Started GET "/" for 127.0.0.1 at 2013-05-17 17:55:01 -0500
  User Load (0.2ms)  SELECT "users".* FROM "users" WHERE "users"."id" = 1 LIMIT 1
Processing by HomeController#index as HTML
  Category Load (0.2ms)  SELECT "categories".* FROM "categories" LIMIT 6
  Banner Load (0.2ms)  SELECT "banners".* FROM "banners" INNER JOIN "banner_types" ON "banner_types"."id" = "banners"."banner_type_id" WHERE (banner_types.name = 'Featured')
  Banner Load (0.2ms)  SELECT "banners".* FROM "banners" INNER JOIN "banner_types" ON "banner_types"."id" = "banners"."banner_type_id" WHERE (banner_types.name = 'Side')
  Product Load (0.3ms)  SELECT "products".* FROM "products" 
  Vendor Load (0.3ms)  SELECT "vendors".* FROM "vendors" WHERE "vendors"."id" IN (12, 11, 10)
  Vendor Load (0.2ms)  SELECT "vendors".* FROM "vendors" 
  User Load (0.2ms)  SELECT "users".* FROM "users" 
  Rendered home/_popular_products.html.erb (16.1ms)
  Rendered home/_popular_stores.html.erb (2.4ms)
  Rendered home/index.html.erb within layouts/application (26.4ms)
  Piggybak::Sellable Load (0.2ms)  SELECT "sellables".* FROM "sellables" WHERE "sellables"."id" = 1 LIMIT 1
  Rendered layouts/_login_nav.html.erb (8.2ms)
  Rendered layouts/_navigation.html.erb (0.8ms)
  CACHE (0.0ms)  SELECT "vendors".* FROM "vendors" 
  Rendered layouts/_store_dropdown.html.erb (2.2ms)
  Rendered layouts/_header.html.erb (18.1ms)
  Rendered layouts/_messages.html.erb (0.3ms)
  Rendered layouts/_footer.html.erb (0.9ms)
Completed 200 OK in 242ms (Views: 209.8ms | ActiveRecord: 1.9ms)

当然,这个特定的时间只是209.8ms,但是在不同的加载时间,它都和5,000ms一样高。

如何优化这些视图和部分的呈现?或者,我能用什么工具至少找出是什么导致了长时间的负载,这样我才能慢慢地处理掉它呢?

EN

回答 1

Stack Overflow用户

发布于 2013-05-17 23:17:13

在第二个示例中,我看到了对以下方面的查询:

  • 所有“特色”类型的横幅
  • 所有“侧”类型的横幅
  • 产品
  • 所有供应商
  • 所有用户

如果您的部分必须迭代大量的对象集合,它将大大降低呈现时间。

注意调用集合(如Vendor.all.each do |v| )的迭代器。不应在视图中调用模型类方法。在所需对象的控制器中构建一个集合,然后发送到实例变量中的视图。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/16619095

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