我正在尝试学习nodejs并偶然发现此错误
TypeError:回调不是函数
当我尝试使用此命令调用服务器时。
curl http://localhost:8090/albums.json
这是我的server.js
的代码
var http = require('http'),
fs = require('fs');
function load_album(album_name, callback) {
fs.readdir("albums/", +album_name, (err, files) => {
if (err) {
if (err.code == "ENOENT") {
callback(make_error("no_such_album", "That album doesn't exist"));
} else {
callback(make_error("can't load album", "The server is broken"));
}
} else {
//callback(null, files);
var only_files = [];
var path = 'albums/${album_name}/';
var iterator = (index) => {
if (index == files.length) {
var obj = {
short_name: album_name,
photos: only_files
};
callback(null, obj);
return;
}
fs.stat(path + files[index], (err, stats) => {
if (!err && stats.isFile()) {
only_files.push(files[index]);
}
iterator(index + 1);
});
};
iterator(0);
}
});
}
function handle_incoming_request(req, res) {
console.log("incoming request: " + req.method + " " + req.url);
if (req.url == '/albums.json') {
load_album((err, albums) => {
if (err) {
res.writeHead(500, {
"Content-Type": "application/json "
});
res.end(JSON.stringify({
code: "cant_load_albums",
message: err.message
}));
} else {
var output = {
error: null,
data: {
albums: albums
}
};
res.writeHead(200, {
"Content-Type": "application/json"
});
res.end(JSON.stringify(output) + "\n");
}
});
} else if (req.url.substr(0, 7) == '/albums' && req.url.substr(req.url.length - 5) == '.json') {
//user is requesting contents of album
load_album(req.url.substr(7, req.url.length - 12), (err, photos) => {
if (err) {
res.writeHead(500, {
"Content-type": "application/json"
});
res.end(JSON.stringify(err));
} else {
var output = {
error: null,
data: {
photos: photos
}
};
res.writeHead(200, {
"Content-Type": application / json
});
res.end(JSON.stringify(output) + "\n");
}
});
} else {
res.writeHead(404, {
"Content-type": "application/json"
});
res.end(JSON.stringify({
code: "no_such_page",
message: "No such page"
}));
}
}
var s = http.createServer(handle_incoming_request);
s.listen(8090);

你能告诉我我的代码出了什么问题,我收到错误告诉我回调不是一个功能吗?
谢谢
对于更多格式化代码,您可以访问https://jsfiddle.net/02dbx6m9/
答案 0 :(得分:5)
var http = require('http'),
fs = require('fs');
function load_album(album_name, callback) {
fs.readdir("albums/", +album_name, (err, files) => {
if (err) {
if (err.code == "ENOENT") {
callback(make_error("no_such_album", "That album doesn't exist"));
} else {
callback(make_error("can't load album", "The server is broken"));
}
} else {
//callback(null, files);
var only_files = [];
var path = 'albums/${album_name}/';
var iterator = (index) => {
if (index == files.length) {
var obj = {
short_name: album_name,
photos: only_files
};
callback(null, obj);
return;
}
fs.stat(path + files[index], (err, stats) => {
if (!err && stats.isFile()) {
only_files.push(files[index]);
}
iterator(index + 1);
});
};
iterator(0);
}
});
}
function handle_incoming_request(req, res) {
console.log("incoming request: " + req.method + " " + req.url);
if (req.url == '/albums.json') {
load_album("ALBUM NAME", (err, albums) => {
if (err) {
res.writeHead(500, {
"Content-Type": "application/json "
});
res.end(JSON.stringify({
code: "cant_load_albums",
message: err.message
}));
} else {
var output = {
error: null,
data: {
albums: albums
}
};
res.writeHead(200, {
"Content-Type": "application/json"
});
res.end(JSON.stringify(output) + "\n");
}
});
} else if (req.url.substr(0, 7) == '/albums' && req.url.substr(req.url.length - 5) == '.json') {
//user is requesting contents of album
load_album("Album Name", req.url.substr(7, req.url.length - 12), (err, photos) => {
if (err) {
res.writeHead(500, {
"Content-type": "application/json"
});
res.end(JSON.stringify(err));
} else {
var output = {
error: null,
data: {
photos: photos
}
};
res.writeHead(200, {
"Content-Type": application / json
});
res.end(JSON.stringify(output) + "\n");
}
});
} else {
res.writeHead(404, {
"Content-type": "application/json"
});
res.end(JSON.stringify({
code: "no_such_page",
message: "No such page"
}));
}
}
var s = http.createServer(handle_incoming_request);
s.listen(8090);
您忘记在load_album方法中传递相册名称参数。这就是为album_name参数分配实际回调的原因,而回调参数仍未定义。
答案 1 :(得分:2)
以下是问题的根本原因:
load_album((err, albums) => {
// ...
});
该功能的签名需要两个参数,但您只能传递第一个参数:
function load_album(album_name, callback) {}
其中,一旦被调用,callback
将被取消定义,但您仍然试图将其视为可调用对象。这是一个更简洁的例子,说明如何重现错误:
function foo(bar, baz) {
baz()
}
foo(() => {})