这是我的HTML表单:

<form name="myForm" ng-submit="">
    <input ng-model='file' type="file"/>
    <input type="submit" value='Submit'/>
</form>

我想从本地机器上传一个图像,并想读取上传文件的内容。所有这些我都想用AngularJS来做。

当我试图打印$scope的值时。文件是未定义的。


当前回答

这是

file.html

<html>
   <head>
      <script src = "https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js"></script>
   </head>
   <body ng-app = "app">
      <div ng-controller = "myCtrl">
         <input type = "file" file-model = "myFile"/>
         <button ng-click = "uploadFile()">upload me</button>
      </div>
   </body>
   <script src="controller.js"></script>
</html>

controller.js

     var app = angular.module('app', []);

     app.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(res){
                console.log(res);
           }).error(function(error){
                console.log(error);
           });
        }
     }]);

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

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

           var uploadUrl = "/fileUpload.php";  // upload url stands for api endpoint to handle upload to directory
           fileUpload.uploadFileToUrl(file, uploadUrl);
        };
     }]);

  </script>

fileupload.php

  <?php
    $ext = pathinfo($_FILES['file']['name'],PATHINFO_EXTENSION);
    $image = time().'.'.$ext;
    move_uploaded_file($_FILES["file"]["tmp_name"],__DIR__. ' \\'.$image);
  ?>

其他回答

<input type=file>元素默认不使用ng-model指令。它需要一个自定义指令:

使用ng-model1的select-ng-files指令的工作演示

angular.module("app",[]); angular.module("app").directive("selectNgFiles", function() { return { require: "ngModel", link: function postLink(scope,elem,attrs,ngModel) { elem.on("change", function(e) { var files = elem[0].files; ngModel.$setViewValue(files); }) } } }); <script src="//unpkg.com/angular/angular.js"></script> <body ng-app="app"> <h1>AngularJS Input `type=file` Demo</h1> <input type="file" select-ng-files ng-model="fileList" multiple> <h2>Files</h2> <div ng-repeat="file in fileList"> {{file.name}} </div> </body>


美元http。从FileList中post

$scope.upload = function(url, fileList) {
    var config = { headers: { 'Content-Type': undefined },
                   transformResponse: angular.identity
                 };
    var promises = fileList.map(function(file) {
        return $http.post(url, file, config);
    });
    return $q.all(promises);
};

当发送带有File对象的POST时,重要的是设置'Content-Type': undefined。然后XHR发送方法将检测File对象并自动设置内容类型。

使用简单指令的示例(ng-file-model):

.directive("ngFileModel", [function () {
  return {
      $scope: {
          ngFileModel: "="
      },
      link: function ($scope:any, element, attributes) {
          element.bind("change", function (changeEvent:any) {
              var reader = new FileReader();
              reader.onload = function (loadEvent) {
                  $scope.$apply(function () {
                      $scope.ngFileModel = {
                          lastModified: changeEvent.target.files[0].lastModified,
                          lastModifiedDate: changeEvent.target.files[0].lastModifiedDate,
                          name: changeEvent.target.files[0].name,
                          size: changeEvent.target.files[0].size,
                          type: changeEvent.target.files[0].type,
                          data: changeEvent.target.files[0]
                      };
                  });
              }
              reader.readAsDataURL(changeEvent.target.files[0]);
          });
      }
  }
}])

并使用FormData在函数中上传文件。

var formData = new FormData();
 formData.append("document", $scope.ngFileModel.data)
 formData.append("user_id", $scope.userId)

所有学分都归 https://github.com/mistralworks/ng-file-model

我遇到了一个小问题,你可以在这里查看: https://github.com/mistralworks/ng-file-model/issues/7

最后,这里是一个分叉的回购:https://github.com/okasha93/ng-file-model/blob/patch-1/ng-file-model.js

我们使用了HTML, CSS和AngularJS。下面的例子展示了如何使用AngularJS上传文件。

<html>

   <head>
      <script src = "https://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>

我能够通过使用下面的代码使用AngularJS上传文件:

函数ngUploadFileUpload需要传递的参数的文件是$scope。按你的问题归档。

这里的关键点是使用transformRequest:[]。这将防止$http与文件内容混淆。

       function getFileBuffer(file) {
            var deferred = new $q.defer();
            var reader = new FileReader();
            reader.onloadend = function (e) {
                deferred.resolve(e.target.result);
            }
            reader.onerror = function (e) {
                deferred.reject(e.target.error);
            }

            reader.readAsArrayBuffer(file);
            return deferred.promise;
        }

        function ngUploadFileUpload(endPointUrl, file) {

            var deferred = new $q.defer();
            getFileBuffer(file).then(function (arrayBuffer) {

                $http({
                    method: 'POST',
                    url: endPointUrl,
                    headers: {
                        "accept": "application/json;odata=verbose",
                        'X-RequestDigest': spContext.securityValidation,
                        "content-length": arrayBuffer.byteLength
                    },
                    data: arrayBuffer,
                    transformRequest: []
                }).then(function (data) {
                    deferred.resolve(data);
                }, function (error) {
                    deferred.reject(error);
                    console.error("Error", error)
                });
            }, function (error) {
                console.error("Error", error)
            });

            return deferred.promise;

        }

这里的一些答案建议使用FormData(),但不幸的是,这是一个浏览器对象,在Internet Explorer 9及以下版本中不可用。如果您需要支持这些旧浏览器,您将需要一个备份策略,例如使用<iframe>或Flash。

已经有很多Angular.js模块来执行文件上传。这两种浏览器都明确支持旧浏览器:

https://github.com/leon/angular-upload -使用iframes作为备份 https://github.com/danialfarid/ng-file-upload -使用FileAPI/Flash作为备份

还有一些其他的选择:

https://github.com/nervgh/angular-file-upload/ https://github.com/uor/angular-file https://github.com/twilson63/ngUpload https://github.com/uploadcare/angular-uploadcare

其中一个应该适合你的项目,或者可能会给你一些关于如何自己编写代码的见解。