如何在Node.js中将缓冲区转换为流 [英] How to convert buffer to stream in Nodejs

问题描述

我遇到了在Node.js中将缓冲区转换为流的问题，这是代码：

I confront with a problem about converting buffer into stream in Nodejs.Here is the code:

var fs = require('fs');
var b = Buffer([80,80,80,80]);
var readStream = fs.createReadStream({path:b});

代码引发异常：

TypeError: path must be a string or Buffer

但是文档

fs.createReadStream（path [，options]）

   path< string> | < Buffer> | < URL>

  选项< string> | <对象>

fs.createReadStream(path[, options])
  path <string> | <Buffer> | <URL>
  options <string> | <Object>

任何人都可以回答这个问题吗？非常感谢！

Anybody could answer the question? Thanks very much!

推荐答案

fs 旨在对文件系统。

fs.createReadStream（）期望将路径作为它的第一个参数，该路径可以是文件名，文件的url或文件缓冲区（如此处所述）。您正在传递对象 {path：b} 。

fs is intended to perform operations on the file system.
fs.createReadStream() expect a path as it first argument which could be a file name, an url of a file or a file buffer (as it is nicely explained here). You are passing an object {path:b}.

如果尝试使用 createReadStream（）语法正确，错误变得更加明显：

If you try with the createReadStream() correct syntax, the error become more clear:

var fs = require('fs');
var b = Buffer([80,80,80,80]);
var readStream = fs.createReadStream(b);

console.log(readStream);

// ReadStream {
//   ...
//   path: <Buffer 50 50 50 50>,
//   ...
// }
//
// Error: ENOENT: no such file or directory, open 'PPPP'

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