在 NodeJs 中检查文件是否存在的最快方法 [英] Fastest way to check for existence of a file in NodeJs

问题描述

我正在节点和我的 onRequest 侦听器中构建一个超级简单的服务器，我试图根据中的路径确定是否应该提供静态文件(磁盘外)或一些 json(可能从 mongo 中提取)request.url.

I'm building a super simple server in node and in my onRequest listener I'm trying to determine if I should serve a static file (off the disk) or some json (probably pulled from mongo) based on the path in request.url.

目前我正在尝试先统计文件(因为我在其他地方使用了 mtime)，如果没有失败，那么我从磁盘读取内容.像这样:

Currently I'm trying to stat the file first (because I use mtime elsewhere) and if that doesn't fail then I read the contents from disk. Something like this:

fs.stat(request.url.pathname, function(err, stat) {
    if (!err) {
        fs.readFile(request.url.pathname, function( err, contents) {
            //serve file
        });
    }else {
        //either pull data from mongo or serve 404 error
    }
});

除了为 request.url.pathname 缓存 fs.stat 的结果之外，还有什么可以加快这种检查的吗?例如，查看 fs.readFile 是否出错而不是 stat 是否同样快?或者使用 fs.createReadStream 而不是 fs.readFile?或者我可以使用 child_process.spawn 中的内容检查文件吗?基本上，我只是想确保在应该将请求发送到 mongo 以获取数据时，我不会花费任何额外的时间来处理 fileio...

Other than cacheing the result of fs.stat for the request.url.pathname, is there something that could speed this check up? For example, would it be just as fast to see if fs.readFile errors out instead of the stat? Or using fs.createReadStream instead of fs.readFile? Or could I potentially check for the file using something in child_process.spawn? Basically I just want to make sure I'm not spending any extra time messing w/ fileio when the request should be sent to mongo for data...

谢谢！

推荐答案

var fs = require('fs');

fs.exists(file, function(exists) {
  if (exists) {
    // serve file
  } else {
    // mongodb
  }
});

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