mysql - 类型错误 : Cannot read property 'filename' of undefined--multer

Question

我目前面临着与此类似的问题post 。在引用那里发布的解决方案后，我设法解决了我的最初问题。但是，当我尝试发布小于 1MB 且为 jpg 格式的图像(我在编辑之前设法做到这一点)时，它现在失败并声明 TypeError: Cannot read property 'filename' of undefined .

我的app.js代码:

const upload = multer({
    dest: storage,
    storage: storage,
    limits: {
      fileSize: 1024 * 1024
    },
    fileFilter: function(req, file, callback,error) {
        var ext = path.extname(file.originalname);
        var error_msg = error instanceof multer.MulterError;
        if(ext !== '.jpg') {
             req.fileValidationError = "Not a jpg file!";
             return callback(null, false, req.fileValidationError);
        }
        if(error_msg) {
            req.fileSizeError = "Image more than"
            return callback(null, false, req.fileSizeError)
        }
        callback(null,true)
    }
  });

app.post("/upload", function (req, res, next) {
    upload.single('name')(req, res, function (error) {
        if(req.fileValidationError) {
            res.status(500).send({message:req.fileValidationError});
        }
        else {
            if(error.code === 'LIMIT_FILE_SIZE') {
                req.fileSizeError = "Image more than 1MB!";
                res.status(500).send({message:req.fileSizeError});
            }
            else {
                console.log('File Received!');
                console.log(req.file);
                var sql = "INSERT INTO `file`(name,description,type,size) VALUES('" + req.file.filename + "', '" + (req.file.encoding + "_" + req.file.destination + "_" + req.file.path)+ "', '" + req.file.mimetype + "', '" + req.file.size + "')";
                db.query(sql, (error, results) => {
                    console.log('Inserted Data!');
                });
            const message = "Successfully Uploaded!"
            res.status(200).send({message:message, file_details:req.file})
            }
        }
    })
})

Answer 1

该数组仅包含对象，不包含文件名属性。数组中的第一个对象具有 filename 属性。使用 req.files['image'][0].filename 或 req.files[0].filename 选择其中的第一个也是唯一一个项目，它应该可以工作。