Question

如何在angularjs和mysql以及node.js上传图像？

<html>

   <head>
      <script src = "http://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js"></script>
   </head>

   <body ng-app = "myApp">

      <div ng-controller = "myCtrl">
         <input type = "file" file-model = "myFile"/>
         <button ng-click = "uploadFile()">upload me</button>
      </div>

      <script>
         var myApp = angular.module('myApp', []);

         myApp.directive('fileModel', ['$parse', function ($parse) {
            return {
               restrict: 'A',
               link: function(scope, element, attrs) {
                  var model = $parse(attrs.fileModel);
                  var modelSetter = model.assign;

                  element.bind('change', function(){
                     scope.$apply(function(){
                        modelSetter(scope, element[0].files[0]);
                     });
                  });
               }
            };
         }]);

         myApp.service('fileUpload', ['$http', function ($http) {
            this.uploadFileToUrl = function(file, uploadUrl){
               var fd = new FormData();
               fd.append('file', file);

               $http.post(uploadUrl, fd, {
                  transformRequest: angular.identity,
                  headers: {'Content-Type': undefined}
               })

               .success(function(){
               })

               .error(function(){
               });
            }
         }]);

         myApp.controller('myCtrl', ['$scope', 'fileUpload', function($scope, fileUpload){
            $scope.uploadFile = function(){
               var file = $scope.myFile;

               console.log('file is ' );
               console.dir(file);

               var uploadUrl = "/fileUpload";
               fileUpload.uploadFileToUrl(file, uploadUrl);
            };
         }]);

      </script>

   </body>
</html>

Answer 1

AngularJs是一种客户端语言。所以它无法将数据保存到MySql。

如果任何数据存储在数据库或服务器中，则必须使用任何服务器端语言。

您正在创建一个Web服务，可以从客户端（AngularJS）添加响应并发送正确的响应。

使用NodeJs在Mysql中上传文件

fs.open(temp_path, 'r', function (status, fd) {
    if (status) {
        console.log(status.message);
        return;
    }
    var buffer = new Buffer(getFilesizeInBytes(temp_path));
    fs.read(fd, buffer, 0, 100, 0, function (err, num) {
    var query = "INSERT INTO `files` SET ?",
    values = {
               file_type: 'img',
               file_size: buffer.length,
               file: buffer
     };
    mySQLconnection.query(query, values, function (er, da) {
      if (er)throw er;
    });
  });
});

function getFilesizeInBytes(filename) {
 var stats = fs.statSync(filename)
 var fileSizeInBytes = stats["size"]
 return fileSizeInBytes
}

您好如何使用angularjs + mysql + node.js上传图像

1 个答案: