Question

我从我的函数返回变量uploadFile，当我试图在另一个变量中访问它时，它给了我未定义的

function upload(req, res, callback) {
    var dir = 'uploads/';

    if (!fs.existsSync(dir)) {
        fs.mkdirSync(dir);
    }

    console.log(req.files.file1);
    console.log(req.files.file2);

    var uploadFiles = {
        ext1: path.extname(req.files.file1.originalname),
        path1: req.files.file1.path,
        ext2: path.extname(req.files.file2.originalname),
        path2: req.files.file2.path
    }
    return callback(uploadFiles);
}

这是我调用upload function的功能我猜我做错了，我得Callback is not a function为错误...请指导我

function sendMail(req, res) {
    var data = req.body;
     upload(req,res);
// checking the condition if the file has been uploaded
    if (uploadFiles) {
        data_to_send.attachments = [{
            filename: 'file1' + uploadFiles.file1ext,
            filePath: uploadFiles.file1Path 
        }, {
            filename: 'file2' + uploadFiles.file2ext,
            filePath: uploadFiles.file2Path
        }]
    }
    console.log(data_to_send.attachments)
    smtpTransport.sendMail({
            from: data_to_send.from,
            to: data_to_send.to,
            subject: data_to_send.subject,
            atachments: data_to_send.attachments,
            text: data_to_send.text,
            html: data_to_send.html
        },
//.........

Answer 1

问题在于，在您的上传功能中，您没有检查回调是否实际传递（并且未定义）。此外，您没有返回您的值，您实际上正在返回任何回调的返回值。

以下是一些可能对您有所帮助的代码：

// inside your upload function
var uploadFiles = {
        ext1: path.extname(req.files.file1.originalname),
        path1: req.files.file1.path,
        ext2: path.extname(req.files.file2.originalname),
        path2: req.files.file2.path
    }
if (callback) {
    callback(uploadFiles);
}

//inside your sendMail (notice the 3rd parameter passed to upload)
upload(req, res, function (uploadFiles) {
    if (uploadFiles) {
        data_to_send.attachments = [{
            filename: 'file1' + uploadFiles.file1ext,
            filePath: uploadFiles.file1Path 
        }, {
            filename: 'file2' + uploadFiles.file2ext,
            filePath: uploadFiles.file2Path
        }]
    }
    // rest of the code goes here, inside the callback.
});

现在，您将实际收到回调中的文件，如您所愿。

Answer 2

这是一个范围问题。您无法在另一个函数中调用uploadFiles，因为您是在上传中定义它。您可以尝试在上传之外定义它，或者您可以尝试控制台（上传（您传入的参数））。方案三，我完全误解了你的要求。

以下是Javascript中作用域的一个很好的参考： What is the scope of variables in Javascript?

无法在node.js中返回变量

2 个答案: