Explain in detail the steps/processes that occur from the moment you type a URL in a browser and hit enter
In an extremely rough and simplified sketch, assuming the simplest possible
HTTP request, no proxies and IPv4 (this would work similarly for IPv6-only
client, but I have yet to see such workstation):
browser checks cache; if requested object is in cache and is fresh, skip to
#9browser asks OS for server's IP address
OS makes a DNS lookup and replies the IP address to the browser
browser opens a TCP connection to server (this step is much more
complex with HTTPS)
browser sends the HTTP request through TCP connection
browser receives HTTP response and may close the TCP connection, or
reuse it for another request
browser checks if the response is a redirect (3xx result status codes),
authorization request (401), error (4xx and 5xx), etc.; these are handled
differently from normal responses (2xx)
if cacheable, response is stored in cache
browser decodes response (e.g. if it's gzipped)
browser determines what to do with response (e.g. is it a HTML page, is it
an image, is it a sound clip?)
browser renders response, or offers a download dialog for unrecognized
types
Again, discussion of each of these points have filled countless pages; take this
as a starting point. Also, there are many other things happening in parallel to
this (processing typed-in address, adding page to browser history, displaying
progress to user, notifying plugins and extensions, rendering the page while
it's downloading, pipelining, connection tracking for keep-alive, etc.).
Explain in detail the steps/processes that occur from the moment you type a URL in a browser and hit enter的更多相关文章
- Cracking the coding interview--问题与解答
http://www.hawstein.com/posts/ctci-solutions-contents.html 作者:Hawstein出处:http://hawstein.com/posts/c ...
- fcagte.exe应用程序错误
原文:What is Fcagte.exe and How To Fix It? Overview of Fcagte.exe What Is Fcagte.exe? Fcagte.exe is a ...
- Vue2 实现树形菜单(多级菜单)功能模块
结构示意图 ├── index.html ├── main.js ├── router │ └── index.js # 路由配置文件 ├── components # 组件目录 │ ├── App. ...
- React后台管理系统-后台接口封装
1新建文件夹 service ,里边建4个文件,分别是statistic-service.jsx 首页数据统计接口, user-service.jsx用户接口, product-service.jsx ...
- Install the high performance Nginx web server on Ubuntu
Look out Apache, there's a web server – Nginx (pronounced Engine X) – that means to dismantle you as ...
- Building Applications with Force.com and VisualForce(Dev401)(十):Designing Applications for Multiple Users: Building Business Processes that You Want
Dev401-011: Building Business Processes that You Want Course Objectives1.Describe the capabilities o ...
- MYSQL EXPLAIN执行计划命令详解(支持更新中)
本文来自我的github pages博客http://galengao.github.io/ 即www.gaohuirong.cn 摘要: 本篇是根据官网中的每个一点来翻译.举例.验证的:英语不好,所 ...
- SQL EXPLAIN解析
本文转载自MySQL性能优化最佳实践 - 08 SQL EXPLAIN解析 什么是归并排序? 如果需要排序的数据超过了sort_buffer_size的大小,说明无法在内存中完成排序,就需要写到临时文 ...
- mysql explain的使用(优化查询)
explain显示了mysql如何使用索引来处理select语句以及连接表.可以帮助选择更好的索引和写出更优化的查询语句. 1.创建数据库 创建的sql语句如下: /* Navicat MySQL D ...
随机推荐
- Two Sum & Add Two Numbers
Two Sum 题目:https://leetcode.com/problems/two-sum/ class Solution(object): def twoSum(self, nums, tar ...
- Odoo下拉动作列表
- acm入门编成题
http://wenku.baidu.com/view/c8f2f64acf84b9d528ea7aee.html
- windows2003 iis+dedecms 4.0701版本,登录后台显示空白
又可能是,i_user对 system32没有权限导致
- .net web获取自己的ip地址
using System;using System.Text;using System.Web;using System.Text.RegularExpressions; namespace MxWe ...
- 【洛谷P1351】联合权值
我们枚举中间点,当连的点数不小于2时进行处理 最大值好搞 求和:设中间点 i 所连所有点权之和为sum 则对于每个中间点i的联合权值之和为: w[j]*(sum-w[j])之和 #include< ...
- Find Minimum in Rotated Sorted Array leetcode
原题链接 直接贴代码,这道题是 search in rotated sorted array leetcode 的前面部分! class Solution { public: int findMin( ...
- [Leetcode][JAVA] Populating Next Right Pointers in Each Node II
Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tre ...
- delphi 获取颜色值的RGB
前言:http://www.cnblogs.com/studypanp/p/5002953.html 获取的颜色值 前面获取到一个像素点的颜色值后(十六进制),比如说(黄色):FFD1C04C(共八位 ...
- hdu 5101 n集合选2个不同集合数使和大于k
http://acm.hdu.edu.cn/showproblem.php?pid=5101 给n个集合,选择两个来自不同集合的数,加和大于k,问有多少种选择方案. 答案=从所有数中选择的两个加和大于 ...