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$scope的值时。文件是未定义的。
当前回答
您可以考虑使用IaaS进行文件上传，例如Uploadcare。有一个针对它的Angular包:https://github.com/uploadcare/angular-uploadcare
从技术上讲，它被实现为一个指令，提供了不同的上传选项，以及小部件中上传图像的操作:
<uploadcare-widget
ng-model="object.image.info.uuid"
data-public-key="YOURKEYHERE"
data-locale="en"
data-tabs="file url"
data-images-only="true"
data-path-value="true"
data-preview-step="true"
data-clearable="true"
data-multiple="false"
data-crop="400:200"
on-upload-complete="onUCUploadComplete(info)"
on-widget-ready="onUCWidgetReady(widget)"
value="{{ object.image.info.cdnUrl }}"
/>
更多配置选项:https://uploadcare.com/widget/configure/
其他回答
<input type=file>元素默认不使用ng-model指令。它需要一个自定义指令:
使用ng-model1的select-ng-files指令的工作演示
angular.module("app",[]); angular.module("app").directive("selectNgFiles", function() { return { require: "ngModel", link: function postLink(scope,elem,attrs,ngModel) { elem.on("change", function(e) { var files = elem[0].files; ngModel.$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。
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。
您的文件和json数据同时上传。
// FIRST SOLUTION var _post = function (file, jsonData) { $http({ url: your url, method: "POST", headers: { 'Content-Type': undefined }, transformRequest: function (data) { var formData = new FormData(); formData.append("model", angular.toJson(data.model)); formData.append("file", data.files); return formData; }, data: { model: jsonData, files: file } }).then(function (response) { ; }); } // END OF FIRST SOLUTION // SECOND SOLUTION // If you can add plural file and If above code give an error. // You can try following code var _post = function (file, jsonData) { $http({ url: your url, method: "POST", headers: { 'Content-Type': undefined }, transformRequest: function (data) { var formData = new FormData(); formData.append("model", angular.toJson(data.model)); for (var i = 0; i < data.files.length; i++) { // add each file to // the form data and iteratively name them formData.append("file" + i, data.files[i]); } return formData; }, data: { model: jsonData, files: file } }).then(function (response) { ; }); } // END OF SECOND SOLUTION
该代码将帮助插入文件
<body ng-app = "myApp">
<form ng-controller="insert_Ctrl" method="post" action="" name="myForm" enctype="multipart/form-data" novalidate>
<div>
<p><input type="file" ng-model="myFile" class="form-control" onchange="angular.element(this).scope().uploadedFile(this)">
<span style="color:red" ng-show="(myForm.myFile.$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);
?>
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>
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