Question

我试图找出一个基本的Node.js服务器。我希望它在客户端尝试连接到“根” URL（正确的术语是btw？）时发送index.html和styles.css文件。这是代码。

var http = require ('http');
var fs = require ('fs');

function send404response(response){
    response.writeHead(404, {"Content-Type":"text/plain"});
    response.write("page not found");
    response.end();
}
function onRequest (request, response) {
    if(request.method == "GET" && request.url == '/'){
        response.writeHead(200, {"Content-Type":"text/html"});
        fs.createReadStream("./index.html").pipe(response);

        response.writeHead(200, {"Content-Type":"text/css"}); 
        fs.createReadStream("./styles.css").pipe(response);


    }else{
        send404response(response);
    }
}
http.createServer(onRequest).listen(8888)

console.log('the server is running...');

styles.css无法加载。客户端错误：GET http://localhost:8888/styles.css net::ERR_ABORTED 404 (Not Found)

Answer 1

我想我知道问题出在哪里。因此，您正在将每个请求发送到函数onRequest，即使链接标记请求也到达了那里。由于IF条件的值为false，因此它会发送您指定的404错误。尝试使用快递服务器并创建路由，或添加其他条件以检查该文件是否存在于路径中（如果存在），然后返回该文件作为响应。

Answer 2

在我的特定情况下有效的实现，我使用了swich构造来处理不同的请求：

var http = require ('http');
var fs = require ('fs');

function send404response(response){
    response.writeHead(404, {"Content-Type":"text/plain"});
    response.write("page not found");
    response.end();
}
function onRequest (request, response) {
    if(request.method == "GET"){

        console.log("request URL is: " + request.url);
        switch (request.url){
            case "/":
            response.writeHead(200, {"Content-Type":"text/html"});
            fs.createReadStream("./index.html").pipe(response);
            break;
            case "/styles.css":
            response.writeHead(200, {"Content-Type":"text/css"});
            fs.createReadStream("./styles.css").pipe(response);
            break;
        }
    }else{
        send404response(response);
    }
}
http.createServer(onRequest).listen(8888)

console.log('the server is running...');

Answer 3

int base = 1;
for (int line = 0; line < n; line++)
{
    int sum = base + line;
    for (int col = 0; col < n - line; col++)
    {
        sum = sum + col;
        printf("%d ", sum);
        sum = sum + line;
    }
    base = base + line + 1;
    printf("\n");
}

在上面的代码片段中，function onRequest (request, response) { if(request.method == "GET" && request.url == '/'){ response.writeHead(200, {"Content-Type":"text/html"}); fs.createReadStream("./index.html").pipe(response); response.writeHead(200, {"Content-Type":"text/css"}); fs.createReadStream("./styles.css").pipe(response); }else{ send404response(response); } }条件存在问题。您只检查if网址。当您的html文件尝试加载CSS时，它将向服务器发出GET请求，并且请求url为/。但是在上面的代码中，您根本不允许路由。根据您的逻辑，您仅提供/cssFileName.css网址，所有其他路由都给/错误。因此，您还需要考虑404网址来提供您的CSS文件。因此理想情况下，您的代码应该像这样...

/cssFileName.css

但理想情况下，对于所有CSS和素材资源（图片，视频），都应从静态文件夹中提供。因此，请改用function onRequest (request, response) { if(request.method == "GET" && request.url == '/'){ response.writeHead(200, {"Content-Type":"text/html"}); fs.createReadStream("./index.html").pipe(response); } else if (request.method == "GET" && request.url == '/cssFileName.css') response.writeHead(200, {"Content-Type":"text/css"}); fs.createReadStream("./styles.css").pipe(response); }else{ send404response(response); } }。供参考https://expressjs.com/en/starter/static-files.html

如何将CSS文件和HTML文件一起发送？

3 个答案: