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
超文本标记语言
<input type="file" id="file" name='file' onchange="angular.element(this).scope().profileimage(this)" />
添加'profileimage()'方法到你的控制器
$scope.profileimage = function(selectimage) {
console.log(selectimage.files[0]);
var selectfile=selectimage.files[0];
r = new FileReader();
r.onloadend = function (e) {
debugger;
var data = e.target.result;
}
r.readAsBinaryString(selectfile);
}
其他回答
超文本标记语言
<input type="file" id="file" name='file' onchange="angular.element(this).scope().profileimage(this)" />
添加'profileimage()'方法到你的控制器
$scope.profileimage = function(selectimage) {
console.log(selectimage.files[0]);
var selectfile=selectimage.files[0];
r = new FileReader();
r.onloadend = function (e) {
debugger;
var data = e.target.result;
}
r.readAsBinaryString(selectfile);
}
我能够通过使用下面的代码使用AngularJS上传文件:
函数ngUploadFileUpload需要传递的参数的文件是$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);
?>
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) { $http({ url: your url, method: "POST", headers: { 'Content-Type': undefined }, transformRequest: function (data) { var formData = new FormData(); formData.append("model", angular.toJson(data.model)); for (var i = 0; i < data.files.length; i++) { // add each file to // the form data and iteratively name them formData.append("file" + i, data.files[i]); } return formData; }, data: { model: jsonData, files: file } }).then(function (response) { ; }); } // END OF SECOND SOLUTION
我们使用了HTML, CSS和AngularJS。下面的例子展示了如何使用AngularJS上传文件。
<html>
<head>
<script src = "https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js"></script>
</head>
<body ng-app = "myApp">
<div ng-controller = "myCtrl">
<input type = "file" file-model = "myFile"/>
<button ng-click = "uploadFile()">upload me</button>
</div>
<script>
var myApp = angular.module('myApp', []);
myApp.directive('fileModel', ['$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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