Question

我是Angular JS的新手并尝试上传文件。我的要求是按下按钮提交多部分数据。

我读到ng-model没有在type =＆＃34; file＆＃34;上工作，所以我知道了解决方法并且复制了指令。在完成该指令之后，在发送数据时没有Content-disposition数据集。我的意思是我想在服务器端读取的文件名，内容类型等。这就是为什么我在headerOfFilePart.getFileName()

获得null

我做错了什么。什么是实现我在Angular JS中描述的事情的正确方法。

    <div ng-controller="uploadController">  
        <h2> Add Order </h2>
        Enter your Name: 
            <input type="text" name="name" ng-model="myName"/>
            <input type="file" fileread="vm.uploadme" />
            <input type="button" name="button" ng-click='uploadFile()'/>
    </div>

这是我的JS部分

validationApp.controller('uploadController', function($scope,$http,$location,$window) {
    $scope.uploadFile = function() {

        var fd = new FormData();
        //Take the first selected file
        fd.append("file", $scope.vm.uploadme);
        fd.append("name", $scope.myName);

        uploadUrl = "http://localhost:8080/IPOCCService/rest/UserManager/upload1";
        $http.post(uploadUrl, fd, {
            withCredentials: true,
            headers: {'Content-Type': undefined },
            transformRequest: angular.identity
        }).
        success(function(data, status, headers, config) {
            alert(data);
        }).
        error(function(data, status, headers, config) {
            alert("failure");
        });

    };
});


validationApp.directive("fileread", [function () {
    return {
        scope: {
            fileread: "="
        },
        link: function (scope, element, attributes) {
            element.bind("change", function (changeEvent) {
                var reader = new FileReader();
                reader.onload = function (loadEvent) {
                    scope.$apply(function () {
                        scope.fileread = loadEvent.target.result;
                    });
                };
                reader.readAsDataURL(changeEvent.target.files[0]);
            });
        }
    };
}]);

REST JAVA

    @POST
    @Path("/upload1")
    @Produces({ MediaType.APPLICATION_JSON} )
    @Consumes(MediaType.MULTIPART_FORM_DATA)
    public Response responseMsg3(FormDataMultiPart form) {
        System.out.println("File Uploaded");

        FormDataBodyPart filePart1 = form.getField("name");
        System.out.println(filePart1.getName() + " = " +filePart1.getValue());

        FormDataBodyPart filePart = form.getField("file");

        ContentDisposition headerOfFilePart =  filePart.getContentDisposition();
        InputStream fileInputStream = filePart.getValueAs(InputStream.class);
        String filePath = SERVER_UPLOAD_LOCATION_FOLDER + headerOfFilePart.getFileName();



        // save the file to the server
        saveFile(fileInputStream, filePath);
        String output = "File saved to server location : " + filePath;
        return Response.status(200).entity("true").build();
    }

Answer 1

当您使用FileReader来阅读文件时，仅将文件的内容分配给您的范围：

scope.fileread = loadEvent.target.result;

在您的情况下，只需将文件分配到您的范围：

link: function (scope, element, attributes) {
      element.bind("change", function (changeEvent) {
         scope.$apply(function(){
              scope.fileread = changeEvent.target.files[0];
         // changeEvent.target.files[0] is an HTML5 file object which contains ALL
         // information about the file including fileName, contents,...
         // scope.fileread is now assigned the selected file
         });   
     });
}

Answer 2

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

        element.bind('change', function(){
            function readURL(input) {
                if (input.files && input.files[0]) {
                    var reader = new FileReader();
                    reader.readAsDataURL(input.files[0]);
                }
            }
            readURL(this);
            scope.$apply(function(){
                modelSetter(scope, element[0].files[0]);
            });
        });
    }
};
}]);

<input type="file" file-read="vm.uploadme" />

这条指令对我有用。

Answer 3

你可以使用ng-file-upload模块上传一个关于文件上传的See here

的指令

Angular JS File上传内容处理

3 个答案: