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$scope的值时。文件是未定义的。
这是我的HTML表单:
<form name="myForm" ng-submit="">
<input ng-model='file' type="file"/>
<input type="submit" value='Submit'/>
</form>
我想从本地机器上传一个图像,并想读取上传文件的内容。所有这些我都想用AngularJS来做。
当我试图打印$scope的值时。文件是未定义的。
当前回答
简单地说
在Html中-只添加以下代码
<form name="upload" class="form" data-ng-submit="addFile()">
<input type="file" name="file" multiple
onchange="angular.element(this).scope().uploadedFile(this)" />
<button type="submit">Upload </button>
</form>
当你点击“上传文件按钮”时,这个函数被调用。它将上传文件。你可以安慰它。
$scope.uploadedFile = function(element) {
$scope.$apply(function($scope) {
$scope.files = element.files;
});
}
在控制器中添加更多-以下代码添加到函数中。这个函数在你点击按钮时被调用,这个按钮被用来“命中api (POST)”。它将发送文件(上传)和表单数据到后端。
var url = httpURL + "/reporttojson"
var files=$scope.files;
for ( var i = 0; i < files.length; i++)
{
var fd = new FormData();
angular.forEach(files,function(file){
fd.append('file',file);
});
var data ={
msg : message,
sub : sub,
sendMail: sendMail,
selectUsersAcknowledge:false
};
fd.append("data", JSON.stringify(data));
$http.post(url, fd, {
withCredentials : false,
headers : {
'Content-Type' : undefined
},
transformRequest : angular.identity
}).success(function(data)
{
toastr.success("Notification sent successfully","",{timeOut: 2000});
$scope.removereport()
$timeout(function() {
location.reload();
}, 1000);
}).error(function(data)
{
toastr.success("Error in Sending Notification","",{timeOut: 2000});
$scope.removereport()
});
}
在这种情况下…我添加了下面的代码作为表单数据
var data ={
msg : message,
sub : sub,
sendMail: sendMail,
selectUsersAcknowledge:false
};
其他回答
该代码将帮助插入文件
<body ng-app = "myApp">
<form ng-controller="insert_Ctrl" method="post" action="" name="myForm" enctype="multipart/form-data" novalidate>
<div>
<p><input type="file" ng-model="myFile" class="form-control" onchange="angular.element(this).scope().uploadedFile(this)">
<span style="color:red" ng-show="(myForm.myFile.$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);
?>
我尝试了@ anoz(正确答案)给出的所有替代方案…最好的解决方案是https://github.com/danialfarid/angular-file-upload
一些特点:
进步 多文件 字段 旧浏览器(IE8-9)
这对我来说很好。你只需要注意说明。
在服务器端,我使用NodeJs, Express 4和Multer中间件来管理多部分请求。
你可以使用flow.js实现良好的文件和文件夹上传。
https://github.com/flowjs/ng-flow
点击这里查看演示
http://flowjs.github.io/ng-flow/
它不支持IE7, IE8, IE9,所以你最终必须使用兼容层
https://github.com/flowjs/fusty-flow.js
我已经阅读了所有的线程,HTML5 API解决方案看起来最好。但它会改变我的二进制文件,以一种我没有调查过的方式破坏它们。最适合我的解决方案是:
HTML:
<input type="file" id="msds" ng-model="msds" name="msds"/>
<button ng-click="msds_update()">
Upload
</button>
JS:
msds_update = function() {
var f = document.getElementById('msds').files[0],
r = new FileReader();
r.onloadend = function(e) {
var data = e.target.result;
console.log(data);
var fd = new FormData();
fd.append('file', data);
fd.append('file_name', f.name);
$http.post('server_handler.php', fd, {
transformRequest: angular.identity,
headers: {'Content-Type': undefined}
})
.success(function(){
console.log('success');
})
.error(function(){
console.log('error');
});
};
r.readAsDataURL(f);
}
服务器端(PHP):
$file_content = $_POST['file'];
$file_content = substr($file_content,
strlen('data:text/plain;base64,'));
$file_content = base64_decode($file_content);
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);
?>