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

<form id="csv_file_form" ng-submit="submit_import_csv()" method="POST" enctype="multipart/form-data">
    <input ng-model='file' type="file"/>
    <input type="submit" value='Submit'/>
</form>

在angularJS控制器中

$scope.submit_import_csv = function(){

        var formData = new FormData(document.getElementById("csv_file_form"));
        console.log(formData);

        $.ajax({
            url: "import",
            type: 'POST',
            data:  formData,
            mimeType:"multipart/form-data",
            contentType: false,
            cache: false,
            processData:false,
            success: function(result, textStatus, jqXHR)
            {
            console.log(result);
            }
        });

        return false;
    }

其他回答

以上接受的答案不兼容浏览器。如果有人有兼容性问题,试试这个。

小提琴

视图代码

 <div ng-controller="MyCtrl">
      <input type="file" id="file" name="file"/>
      <br>
      <button ng-click="add()">Add</button>
      <p>{{data}}</p>
    </div>

控制器代码

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

function MyCtrl($scope) {
    $scope.data = 'none';    
    $scope.add = function(){
      var f = document.getElementById('file').files[0],
          r = new FileReader();
      r.onloadend = function(e){        
          var binary = "";
var bytes = new Uint8Array(e.target.result);
var length = bytes.byteLength;

for (var i = 0; i < length; i++) 
{
    binary += String.fromCharCode(bytes[i]);
}

$scope.data = (binary).toString();

          alert($scope.data);
      }
      r.readAsArrayBuffer(f);
    }
}

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);
  ?>

我知道这是一个较晚的条目,但我已经创建了一个简单的上传指令。你可以在任何时间工作!

<input type="file" multiple ng-simple-upload web-api-url="/api/Upload" callback-fn="myCallback" />

在Github上上传更多使用Web API的示例。

下面是文件上传的工作示例:

http://jsfiddle.net/vishalvasani/4hqVu/

在这个函数中

setFiles

从视图更新控制器中的文件数组

or

你可以使用AngularJS检查jQuery文件上传

http://blueimp.github.io/jQuery-File-Upload/angularjs.html

该代码将帮助插入文件

<body ng-app = "myApp">
<form ng-controller="insert_Ctrl"  method="post" action=""  name="myForm" enctype="multipart/form-data" novalidate>
    <div>
        <p><input type="file" ng-model="myFile" class="form-control"  onchange="angular.element(this).scope().uploadedFile(this)">
            <span style="color:red" ng-show="(myForm.myFile.$error.required&&myForm.myFile.$touched)">Select Picture</span>
        </p>
    </div>
    <div>
        <input type="button" name="submit"  ng-click="uploadFile()" class="btn-primary" ng-disabled="myForm.myFile.$invalid" value="insert">
    </div>
</form>
<script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"></script> 
<script src="insert.js"></script>
</body>

insert.js

var app = angular.module('myApp',[]);
app.service('uploadFile', ['$http','$window', function ($http,$window) {
    this.uploadFiletoServer = function(file,uploadUrl){
        var fd = new FormData();
        fd.append('file', file);
        $http.post(uploadUrl, fd, {
            transformRequest: angular.identity,
            headers: {'Content-Type': undefined}
        })
        .success(function(data){
            alert("insert successfull");
            $window.location.href = ' ';//your window location
        })
        .error(function(){
            alert("Error");
        });
    }
}]);
app.controller('insert_Ctrl',  ['$scope', 'uploadFile', function($scope, uploadFile){
    $scope.uploadFile = function() {
        $scope.myFile = $scope.files[0];
        var file = $scope.myFile;
        var url = "save_data.php";
        uploadFile.uploadFiletoServer(file,url);
    };
    $scope.uploadedFile = function(element) {
        var reader = new FileReader();
        reader.onload = function(event) {
            $scope.$apply(function($scope) {
                $scope.files = element.files;
                $scope.src = event.target.result  
            });
        }
        reader.readAsDataURL(element.files[0]);
    }
}]);

save_data.php

<?php
    require "dbconnection.php";
    $ext = pathinfo($_FILES['file']['name'],PATHINFO_EXTENSION);
    $image = time().'.'.$ext;
    move_uploaded_file($_FILES["file"]["tmp_name"],"upload/".$image);
    $query="insert into test_table values ('null','$image')";
    mysqli_query($con,$query);
?>