I am using an Azure Function that has a HTTP trigger with a route parameter {id} which is the fileId of the JSON file I want to read.
I am using a Blob Input Binding to bind where my JSON files are stored. The JSON files are stored in a container called "conversations" and then in a folder called "Conversation".
An example of a file route is "https://<STORAGE_ACCOUNT_NAME>/conversations/Conversation/8da3d7ad3e35273-1aWpKU4rVghHiTaYkjOjVC-eu%7C0000000.json"
Below is my code.
public static class GetConvo
{
[FunctionName("GetConvo")]
public static async Task<IActionResult> Run(
[HttpTrigger(AuthorizationLevel.Anonymous, "get", "post", Route = "getConvo/{id}")] HttpRequest req,
[Blob("conversations/{id}", FileAccess.Read, Connection = "AzureWebJobsStorage")] string json,
ILogger log, string id)
{
log.LogInformation($"File name: {id}");
if (json == null)
{
log.LogInformation($"File {id} not found");
return new NotFoundResult();
}
else
{
log.LogInformation($"Content: {json}");
}
return new OkObjectResult(JsonConvert.DeserializeObject<Message>(json));
The above code works if I move a JSON file to outside the "Conversation" folder, I can access it and receive a 200OK code.
I have tried changing the Blob input binding path to "conversations/Conversation/{id}" as below but that returns a 404 code.
[FunctionName("GetConvo")]
public static async Task<IActionResult> Run(
[HttpTrigger(AuthorizationLevel.Anonymous, "get", "post", Route = "getConvo/{id}")] HttpRequest req,
[Blob("conversations/Conversation/{id}", FileAccess.Read, Connection = "AzureWebJobsStorage")] string json,
ILogger log, string id)
Is this a blob input path problem?
How would I read JSON files that are in a folder in a blob container using an azure function?
#AjgB, yes the Blob path is incorrect. You need to provide the file extension.
Lets say the file is placed directly in your 'conversations' folder. Then your BLOB input bindings should be -
[Blob("conversations/{id}.json", FileAccess.Read, Connection = "AzureWebJobsStorage")] string json
Note the .json in the blob path
I found out what my error was.
The blob input path was correct. It was a URL encoding problem for:
https://<STORAGE_ACCOUNT_NAME>/conversations/Conversation/8da3d7ad3e35273-1aWpKU4rVghHiTaYkjOjVC-eu%7C0000000.json
The % was not recognised and required; putting 25 after the % resolved this error:
https://<STORAGE_ACCOUNT_NAME>/conversations/Conversation/8da3d7ad3e35273-1aWpKU4rVghHiTaYkjOjVC-eu%257C0000000.json
Sending a POST request (Apache httpclient, here Kotlin source code):
val httpPost = HttpPost("http://localhost:8000")
val builder = MultipartEntityBuilder.create()
builder.addBinaryBody("file", File("testFile.zip"),
ContentType.APPLICATION_OCTET_STREAM, "file.ext")
val multipart = builder.build()
httpPost.entity = multipart
val r = httpClient.execute(httpPost)
r.close()
I receive the request in my post handler as a via spark-java Request-object. How do I retrieve the original file (plus the file name as a bonus) from the post request? The request.bodyAsBytes() method seems to add some bytes because the body is larger than the original file.
Thanks, Jörg
Near the bottom of Spark's Documentation page there is a section "Examples and FAQ". The first example is "How do I upload something?".
From there, it links further to an example on GitHub.
In short:
post("/yourUploadPath", (request, response) -> {
request.attribute("org.eclipse.jetty.multipartConfig", new MultipartConfigElement("/temp"));
try (InputStream is = request.raw().getPart("file").getInputStream()) {
// Use the input stream to create a file
}
return "File uploaded";
});
To access the original file name:
request.raw().getPart("file").getSubmittedFileName()
To handle multiple files or parts, I usually have code similar to the following (assuming only files are included in the multi-part encoded upload):
for (Part part : req.raw().getParts()) {
try (InputStream stream = part.getInputStream()) {
String filename = part.getSubmittedFileName();
// save the input stream to the filesystem, and the filename to a database
}
}
I would like to know what can I do to upload attachments in CouchDB using the update function.
here you will find an example of my update function to add documents:
function(doc, req){
if (!doc) {
if (!req.form._id) {
req.form._id = req.uuid;
}
req.form['|edited_by'] = req.userCtx.name
req.form['|edited_on'] = new Date();
return [req.form, JSON.stringify(req.form)];
}
else {
return [null, "Use POST to add a document."]
}
}
example for remove documents:
function(doc, req){
if (doc) {
for (var i in req.form) {
doc[i] = req.form[i];
}
doc['|edited_by'] = req.userCtx.name
doc['|edited_on'] = new Date();
doc._deleted = true;
return [doc, JSON.stringify(doc)];
}
else {
return [null, "Document does not exist."]
}
}
thanks for your help,
It is possible to add attachments to a document using an update function by modifying the document's _attachments property. Here's an example of an update function which will add an attachment to an existing document:
function (doc, req) {
// skipping the create document case for simplicity
if (!doc) {
return [null, "update only"];
}
// ensure that the required form parameters are present
if (!req.form || !req.form.name || !req.form.data) {
return [null, "missing required post fields"];
}
// if there isn't an _attachments property on the doc already, create one
if (!doc._attachments) {
doc._attachments = {};
}
// create the attachment using the form data POSTed by the client
doc._attachments[req.form.name] = {
content_type: req.form.content_type || 'application/octet-stream',
data: req.form.data
};
return [doc, "saved attachment"];
}
For each attachment, you need a name, a content type, and body data encoded as base64. The example function above requires that the client sends an HTTP POST in application/x-www-form-urlencoded format with at least two parameters: name and data (a content_type parameter will be used if provided):
name=logo.png&content_type=image/png&data=iVBORw0KGgoA...
To test the update function:
Find a small image and base64 encode it:
$ base64 logo.png | sed 's/+/%2b/g' > post.txt
The sed script encodes + characters so they don't get converted to spaces.
Edit post.txt and add name=logo.png&content_type=image/png&data= to the top of the document.
Create a new document in CouchDB using Futon.
Use curl to call the update function with the post.txt file as the body, substituting in the ID of the document you just created.
curl -X POST -d #post.txt http://127.0.0.1:5984/mydb/_design/myddoc/_update/upload/193ecff8618678f96d83770cea002910
This was tested on CouchDB 1.6.1 running on OSX.
Update: #janl was kind enough to provide some details on why this answer can lead to performance and scaling issues. Uploading attachments via an upload handler has two main problems:
The upload handlers are written in JavaScript, so the CouchDB server may have to fork() a couchjs process to handle the upload. Even if a couchjs process is already running, the server has to stream the entire HTTP request to the external process over stdin. For large attachments, the transfer of the request can take significant time and system resources. For each concurrent request to an update function like this, CouchDB will have to fork a new couchjs process. Since the process runtime will be rather long because of what is explained next, you can easily run out of RAM, CPU or the ability to handle more concurrent requests.
After the _attachments property is populated by the upload handler and streamed back to the CouchDB server (!), the server must parse the response JSON, decode the base64-encoded attachment body, and write the binary body to disk. The standard method of adding an attachment to a document -- PUT /db/docid/attachmentname -- streams the binary request body directly to disk and does not require the two processing steps.
The function above will work, but there are non-trivial issues to consider before using it in a highly-scalable system.
I got a Web API that performs a function and posts a JSON response back to a calling page.
This is standard Web API behaviour and works beautifully.
Now I want to modify the controller so that in addition to the post back the user is redirected back to the page on the calling web site where the result of the Web API call can be displayed (in JSON).
So basically I want to:
(1) Server side post back the results in JSON to a page and redirect to the same page from the Web API
(2) On the caller's site, I want to display the JSON that was posted back.
How do I do this?
I already tried for many hours ...
e.g.:
using (WebClient client = new WebClient())
{
client.Headers.Add("Content-Type", "text/json");
client.Headers.Add("Accept", "text/json");
try
{
ErrorText = client.UploadString(redirectURL, "POST", JsonConvert.SerializeObject(orderresponse));
Response.Redirect(redirectURL);
}
catch (WebException err)
{
ErrorText = err.Message; //Todo - write to logfile
}
}
Instead of doing the redirect on the server, instruct the client to do it by using the appropriate HTTP status code. For example:
public HttpResponseMessage Post(MyModel model)
{
// handle the post
MyResult result = ...;
// redirect
var response = Request.CreateResponse<MyResult>(HttpStatusCode.Moved, result);
response.Headers.Location = new Uri("http://www.yourdomain.com/redirectURI");
return response;
}
In an application I am developing RESTful API and we want the client to send data as JSON. Part of this application requires the client to upload a file (usually an image) as well as information about the image.
I'm having a hard time tracking down how this happens in a single request. Is it possible to Base64 the file data into a JSON string? Am I going to need to perform 2 posts to the server? Should I not be using JSON for this?
As a side note, we're using Grails on the backend and these services are accessed by native mobile clients (iPhone, Android, etc), if any of that makes a difference.
I asked a similar question here:
How do I upload a file with metadata using a REST web service?
You basically have three choices:
Base64 encode the file, at the expense of increasing the data size by around 33%, and add processing overhead in both the server and the client for encoding/decoding.
Send the file first in a multipart/form-data POST, and return an ID to the client. The client then sends the metadata with the ID, and the server re-associates the file and the metadata.
Send the metadata first, and return an ID to the client. The client then sends the file with the ID, and the server re-associates the file and the metadata.
You can send the file and data over in one request using the multipart/form-data content type:
In many applications, it is possible for a user to be presented with
a form. The user will fill out the form, including information that
is typed, generated by user input, or included from files that the
user has selected. When the form is filled out, the data from the
form is sent from the user to the receiving application.
The definition of MultiPart/Form-Data is derived from one of those
applications...
From http://www.faqs.org/rfcs/rfc2388.html:
"multipart/form-data" contains a series of parts. Each part is
expected to contain a content-disposition header [RFC 2183] where the
disposition type is "form-data", and where the disposition contains
an (additional) parameter of "name", where the value of that
parameter is the original field name in the form. For example, a part
might contain a header:
Content-Disposition: form-data; name="user"
with the value corresponding to the entry of the "user" field.
You can include file information or field information within each section between boundaries. I've successfully implemented a RESTful service that required the user to submit both data and a form, and multipart/form-data worked perfectly. The service was built using Java/Spring, and the client was using C#, so unfortunately I don't have any Grails examples to give you concerning how to set up the service. You don't need to use JSON in this case since each "form-data" section provides you a place to specify the name of the parameter and its value.
The good thing about using multipart/form-data is that you're using HTTP-defined headers, so you're sticking with the REST philosophy of using existing HTTP tools to create your service.
I know that this thread is quite old, however, I am missing here one option. If you have metadata (in any format) that you want to send along with the data to upload, you can make a single multipart/related request.
The Multipart/Related media type is intended for compound objects consisting of several inter-related body parts.
You can check RFC 2387 specification for more in-depth details.
Basically each part of such a request can have content with different type and all parts are somehow related (e.g. an image and it metadata). The parts are identified by a boundary string, and the final boundary string is followed by two hyphens.
Example:
POST /upload HTTP/1.1
Host: www.hostname.com
Content-Type: multipart/related; boundary=xyz
Content-Length: [actual-content-length]
--xyz
Content-Type: application/json; charset=UTF-8
{
"name": "Sample image",
"desc": "...",
...
}
--xyz
Content-Type: image/jpeg
[image data]
[image data]
[image data]
...
--foo_bar_baz--
Here is my approach API (i use example) - as you can see, you I don't use any file_id (uploaded file identifier to the server) in API:
Create photo object on server:
POST: /projects/{project_id}/photos
body: { name: "some_schema.jpg", comment: "blah"}
response: photo_id
Upload file (note that file is in singular form because it is only one per photo):
POST: /projects/{project_id}/photos/{photo_id}/file
body: file to upload
response: -
And then for instance:
Read photos list
GET: /projects/{project_id}/photos
response: [ photo, photo, photo, ... ] (array of objects)
Read some photo details
GET: /projects/{project_id}/photos/{photo_id}
response: { id: 666, name: 'some_schema.jpg', comment:'blah'} (photo object)
Read photo file
GET: /projects/{project_id}/photos/{photo_id}/file
response: file content
So the conclusion is that, first you create an object (photo) by POST, and then you send second request with the file (again POST). To not have problems with CACHE in this approach we assume that we can only delete old photos and add new - no update binary photo files (because new binary file is in fact... NEW photo). However if you need to be able to update binary files and cache them, then in point 4 return also fileId and change 5 to GET: /projects/{project_id}/photos/{photo_id}/files/{fileId}.
I know this question is old, but in the last days I had searched whole web to solution this same question. I have grails REST webservices and iPhone Client that send pictures, title and description.
I don't know if my approach is the best, but is so easy and simple.
I take a picture using the UIImagePickerController and send to server the NSData using the header tags of request to send the picture's data.
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:#"myServerAddress"]];
[request setHTTPMethod:#"POST"];
[request setHTTPBody:UIImageJPEGRepresentation(picture, 0.5)];
[request setValue:#"image/jpeg" forHTTPHeaderField:#"Content-Type"];
[request setValue:#"myPhotoTitle" forHTTPHeaderField:#"Photo-Title"];
[request setValue:#"myPhotoDescription" forHTTPHeaderField:#"Photo-Description"];
NSURLResponse *response;
NSError *error;
[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
At the server side, I receive the photo using the code:
InputStream is = request.inputStream
def receivedPhotoFile = (IOUtils.toByteArray(is))
def photo = new Photo()
photo.photoFile = receivedPhotoFile //photoFile is a transient attribute
photo.title = request.getHeader("Photo-Title")
photo.description = request.getHeader("Photo-Description")
photo.imageURL = "temp"
if (photo.save()) {
File saveLocation = grailsAttributes.getApplicationContext().getResource(File.separator + "images").getFile()
saveLocation.mkdirs()
File tempFile = File.createTempFile("photo", ".jpg", saveLocation)
photo.imageURL = saveLocation.getName() + "/" + tempFile.getName()
tempFile.append(photo.photoFile);
} else {
println("Error")
}
I don't know if I have problems in future, but now is working fine in production environment.
FormData Objects: Upload Files Using Ajax
XMLHttpRequest Level 2 adds support for the new FormData interface.
FormData objects provide a way to easily construct a set of key/value pairs representing form fields and their values, which can then be easily sent using the XMLHttpRequest send() method.
function AjaxFileUpload() {
var file = document.getElementById("files");
//var file = fileInput;
var fd = new FormData();
fd.append("imageFileData", file);
var xhr = new XMLHttpRequest();
xhr.open("POST", '/ws/fileUpload.do');
xhr.onreadystatechange = function () {
if (xhr.readyState == 4) {
alert('success');
}
else if (uploadResult == 'success')
alert('error');
};
xhr.send(fd);
}
https://developer.mozilla.org/en-US/docs/Web/API/FormData
Since the only missing example is the ANDROID example, I'll add it.
This technique uses a custom AsyncTask that should be declared inside your Activity class.
private class UploadFile extends AsyncTask<Void, Integer, String> {
#Override
protected void onPreExecute() {
// set a status bar or show a dialog to the user here
super.onPreExecute();
}
#Override
protected void onProgressUpdate(Integer... progress) {
// progress[0] is the current status (e.g. 10%)
// here you can update the user interface with the current status
}
#Override
protected String doInBackground(Void... params) {
return uploadFile();
}
private String uploadFile() {
String responseString = null;
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://example.com/upload-file");
try {
AndroidMultiPartEntity ampEntity = new AndroidMultiPartEntity(
new ProgressListener() {
#Override
public void transferred(long num) {
// this trigger the progressUpdate event
publishProgress((int) ((num / (float) totalSize) * 100));
}
});
File myFile = new File("/my/image/path/example.jpg");
ampEntity.addPart("fileFieldName", new FileBody(myFile));
totalSize = ampEntity.getContentLength();
httpPost.setEntity(ampEntity);
// Making server call
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
int statusCode = httpResponse.getStatusLine().getStatusCode();
if (statusCode == 200) {
responseString = EntityUtils.toString(httpEntity);
} else {
responseString = "Error, http status: "
+ statusCode;
}
} catch (Exception e) {
responseString = e.getMessage();
}
return responseString;
}
#Override
protected void onPostExecute(String result) {
// if you want update the user interface with upload result
super.onPostExecute(result);
}
}
So, when you want to upload your file just call:
new UploadFile().execute();
I wanted send some strings to backend server. I didnt use json with multipart, I have used request params.
#RequestMapping(value = "/upload", method = RequestMethod.POST)
public void uploadFile(HttpServletRequest request,
HttpServletResponse response, #RequestParam("uuid") String uuid,
#RequestParam("type") DocType type,
#RequestParam("file") MultipartFile uploadfile)
Url would look like
http://localhost:8080/file/upload?uuid=46f073d0&type=PASSPORT
I am passing two params (uuid and type) along with file upload.
Hope this will help who don't have the complex json data to send.
You could try using https://square.github.io/okhttp/ library.
You can set the request body to multipart and then add the file and json objects separately like so:
MultipartBody requestBody = new MultipartBody.Builder()
.setType(MultipartBody.FORM)
.addFormDataPart("uploadFile", uploadFile.getName(), okhttp3.RequestBody.create(uploadFile, MediaType.parse("image/png")))
.addFormDataPart("file metadata", json)
.build();
Request request = new Request.Builder()
.url("https://uploadurl.com/uploadFile")
.post(requestBody)
.build();
try (Response response = client.newCall(request).execute()) {
if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);
logger.info(response.body().string());
#RequestMapping(value = "/uploadImageJson", method = RequestMethod.POST)
public #ResponseBody Object jsongStrImage(#RequestParam(value="image") MultipartFile image, #RequestParam String jsonStr) {
-- use com.fasterxml.jackson.databind.ObjectMapper convert Json String to Object
}
Please ensure that you have following import. Ofcourse other standard imports
import org.springframework.core.io.FileSystemResource
void uploadzipFiles(String token) {
RestBuilder rest = new RestBuilder(connectTimeout:10000, readTimeout:20000)
def zipFile = new File("testdata.zip")
def Id = "001G00000"
MultiValueMap<String, String> form = new LinkedMultiValueMap<String, String>()
form.add("id", id)
form.add('file',new FileSystemResource(zipFile))
def urld ='''http://URL''';
def resp = rest.post(urld) {
header('X-Auth-Token', clientSecret)
contentType "multipart/form-data"
body(form)
}
println "resp::"+resp
println "resp::"+resp.text
println "resp::"+resp.headers
println "resp::"+resp.body
println "resp::"+resp.status
}