I am new to Angular JS and trying my hands on File upload. My requirement is to submit the multipart data on button click.
I read on ng-model not working on type="file", so I got to know the work around and i copied the directive. after working through that directive, while sending data there is no Content-disposition data set. I mean file name, content type etc which I want to read at server side.
that is why I am getting null at headerOfFilePart.getFileName()
what I am doing wrong. what is the right way to achieve things i described above in Angular JS.
<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>
And this is my JS part
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();
}