首页 \ 问答 \ Node.js是否真的异步执行后台I / O任务？(Does Node.js really do the background I/O tasks asynchronously?)

Node.js是否真的异步执行后台I / O任务？(Does Node.js really do the background I/O tasks asynchronously?)

 编辑：我们可以关闭。 是不是真的异步，非阻塞javascript不可能？  
 
var PATH = require ("path");
var URL = require ("url");

var sleep = function (ms){
    var start = new Date ().getTime ();
    while ((new Date ().getTime () - start) < ms);
}

require ("http").createServer (function (req, res){
    if (URL.parse (req.url).pathname === "/1"){
        console.log ("tab 1: I'm in");
        PATH.exists ("test", function (exists){

            sleep (5000);

            res.writeHead (200, {"Content-Type": "text/plain"});
            res.end ("1");
            console.log ("tab 1: I'm done");
        });
    }else{
        console.log ("tab 2: I'm in");
        res.writeHead (200, {"Content-Type": "text/plain"});
        res.end ("2");
        console.log ("tab 2: I'm done");
    }
}).listen (80);
 
 
  将内容复制到文件中。  
  执行文件。  
  在浏览器中打开一个新选项卡。 将url设置为localhost/1 。 不要走了。  
  在浏览器中打开一个新选项卡。 将url设置为localhost/2 。 不要走了。  
  返回第一个标签。 按Enter键并在更改到第二个选项卡后立即按Enter键。  
 
 结果：  
 
  控制台日志： 
 表1：我在 
 表1：我已经完成了 
 表2：我在 
 表2：我已经完成了  
  标签1等待5秒钟以接收结果“1”。  
  标签2也必须等待5秒钟，因为标签1正在睡眠5秒钟。  
 
 文档说除了代码之外所有都是异步的。 只有一个线程。 一次只能有一个请求。 请求已入队。  
 I / O调用应该是异步的，对吧？ 那么，如果回调来自异步I / O进程，为什么选项卡2必须等待选项卡1？  
 谢谢。 

Edit: We can close. Isn't truly asynchronous, non-blocking javascript impossible? 
 
var PATH = require ("path");
var URL = require ("url");

var sleep = function (ms){
    var start = new Date ().getTime ();
    while ((new Date ().getTime () - start) < ms);
}

require ("http").createServer (function (req, res){
    if (URL.parse (req.url).pathname === "/1"){
        console.log ("tab 1: I'm in");
        PATH.exists ("test", function (exists){

            sleep (5000);

            res.writeHead (200, {"Content-Type": "text/plain"});
            res.end ("1");
            console.log ("tab 1: I'm done");
        });
    }else{
        console.log ("tab 2: I'm in");
        res.writeHead (200, {"Content-Type": "text/plain"});
        res.end ("2");
        console.log ("tab 2: I'm done");
    }
}).listen (80);
 
 
 Copy the content into a file. 
 Execute the file. 
 Open a new tab in browser. Set the url to localhost/1. Don't go yet. 
 Open a new tab in browser. Set the url to localhost/2. Don't go yet. 
 Go back to the first tab. Press enter and immediately after change to the second tab and press enter. 
 
Result: 
 
 console log:
 tab 1: I'm in
 tab 1: I'm done
 tab 2: I'm in
 tab 2: I'm done  
 Tab 1 waits 5 seconds to receive the result "1". 
 Tab 2 also has to wait 5 seconds because tab 1 is sleeping for 5 seconds. 
 
The docs says that all is asynchronous except the code. Only one thread. Only one request at a time. Requests are enqueued. 
I/O calls are supposed to be asynchronous, right? Then why tab 2 has to wait to tab 1 if the callback comes from an asynchronous I/O process? 
Thanks.

原文：https://stackoverflow.com/questions/9896980

更新时间：2023-08-23 07:08

最满意答案

 通常可以将复杂的update转换为merge ：  
merge into proc_try k
using
   ( select jobs
     from   ( with a(namehost) as
                   ( select b.namehost
                     from   definition   a
                            join schema.nodes b
                                 on  b.namehost = a.idnode
                                 or  (b.nodeid = a.idnode and nodetype <> 'R')
                     group  by b.namehost
                     having sum(1 + lengthb(namejob)) - 1 > 4000 )
              select namehost
                   , 'TOOLONG' as jobs
              from   a
              union all
              select p.namehost
                   , listagg(p.namejob, ',') within group(order by p.namejob) as jobs
              from   ( select distinct
                              b.namejob, a.namehost
                       from   schema.nodes a
                              left join definition b
                                   on  b.idnode in (a.nodeid, a.namehost)
                                   and b.application not like '@NOTINCLUDE'
                       where  a.namehost not in (select * from a)
                       and    nodetype not like 'R'
                     ) p
              group  by p.namehost
            ) random
   ) new_jobs
on (k.namehost = new_jobs.namehost)
when matched then update set k.detail = new_jobs.jobs;
 
 这是未经测试的，因为我没有您的表格或样本数据。  
 编辑：看起来我们可以简化一下，对此：  
merge into proc_try k
using
   ( with overlength (namehost) as
          ( select n.namehost
            from   definition d
                   join schema.nodes n
                        on  n.namehost = d.idnode
                        or  (n.nodeid = d.idnode and nodetype <> 'R')
            group by n.namehost
            having sum(1 + lengthb(n.namejob)) - 1 > 4000 )
     select o.namehost, 'TOOLONG' as jobs
     from   overlength o
     union all
     select sd.namehost
          , listagg(sd.namejob, ',') within group(order by sd.namejob) as jobs
     from   ( select distinct d.namejob, n.namehost
              from   schema.nodes n
                     left join definition d
                          on  d.idnode in (n.nodeid, n.namehost)
                          and d.application not like '@NOTINCLUDE'
              where  n.namehost not in (select o.namehost from overlength o)
              and    n.nodetype not like 'R'
            ) sd
     group  by sd.namehost
   ) new_jobs
on (new_jobs.namehost = k.namehost)
when matched then update set k.detail = new_jobs.jobs;
 
 我仍然看不到什么  
sum(1 + lengthb(namejob)) - 1
 
 尽管如此，是为了做。 看起来这可以简化为  
sum(lengthb(namejob))

You can generally convert a complex update into a merge: 
merge into proc_try k
using
   ( select jobs
     from   ( with a(namehost) as
                   ( select b.namehost
                     from   definition   a
                            join schema.nodes b
                                 on  b.namehost = a.idnode
                                 or  (b.nodeid = a.idnode and nodetype <> 'R')
                     group  by b.namehost
                     having sum(1 + lengthb(namejob)) - 1 > 4000 )
              select namehost
                   , 'TOOLONG' as jobs
              from   a
              union all
              select p.namehost
                   , listagg(p.namejob, ',') within group(order by p.namejob) as jobs
              from   ( select distinct
                              b.namejob, a.namehost
                       from   schema.nodes a
                              left join definition b
                                   on  b.idnode in (a.nodeid, a.namehost)
                                   and b.application not like '@NOTINCLUDE'
                       where  a.namehost not in (select * from a)
                       and    nodetype not like 'R'
                     ) p
              group  by p.namehost
            ) random
   ) new_jobs
on (k.namehost = new_jobs.namehost)
when matched then update set k.detail = new_jobs.jobs;
 
This is untested as I don't have your tables or sample data. 
Edit: Looks like we can simplify it a bit, to this: 
merge into proc_try k
using
   ( with overlength (namehost) as
          ( select n.namehost
            from   definition d
                   join schema.nodes n
                        on  n.namehost = d.idnode
                        or  (n.nodeid = d.idnode and nodetype <> 'R')
            group by n.namehost
            having sum(1 + lengthb(n.namejob)) - 1 > 4000 )
     select o.namehost, 'TOOLONG' as jobs
     from   overlength o
     union all
     select sd.namehost
          , listagg(sd.namejob, ',') within group(order by sd.namejob) as jobs
     from   ( select distinct d.namejob, n.namehost
              from   schema.nodes n
                     left join definition d
                          on  d.idnode in (n.nodeid, n.namehost)
                          and d.application not like '@NOTINCLUDE'
              where  n.namehost not in (select o.namehost from overlength o)
              and    n.nodetype not like 'R'
            ) sd
     group  by sd.namehost
   ) new_jobs
on (new_jobs.namehost = k.namehost)
when matched then update set k.detail = new_jobs.jobs;
 
I still can't see what 
sum(1 + lengthb(namejob)) - 1
 
is meant to do, though. It looks like that could be simplified to 
sum(lengthb(namejob))

Node.js是否真的异步执行后台I / O任务？(Does Node.js really do the background I/O tasks asynchronously?)

最满意答案

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