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
}
My goal is to retrieve JSON data from "The virus tracker" API and parse it into a label. This is for my IB personal project so I just started learning c#. Fount this old code and tried fixing it, it worked with basic GET API's but it doesn't work with the one I'm using. (English not my main language)
POSTMAN RESPONSE
{
"results": [
{
"total_cases": 5954607,
"total_recovered": 2622507,
"total_unresolved": 2255875,
"total_deaths": 363208,
"total_new_cases_today": 53700,
"total_new_deaths_today": 1659,
"total_active_cases": 45257,
"total_serious_cases": 2698001,
"total_affected_countries": 213,
"source": {
"url": "https://thevirustracker.com/"
}
}
],
"stat": "ok"
}
C# CODE
//Creating Client connection
RestClient restClient = new RestClient("https://thevirustracker.com/free-api?global=stats");
//Creating request to get data from server
RestRequest restRequest = new RestRequest("total_cases", Method.GET);
// Executing request to server and checking server response to the it
IRestResponse restResponse = restClient.Execute(restRequest);
// Extracting output data from received response
string response = restResponse.Content;
// Parsing JSON content into element-node JObject
var jObject = JObject.Parse(restResponse.Content);
//Extracting Node element using Getvalue method
string cases = jObject.GetValue("total_cases").ToString();
label1.Text = (cases);
ERROR
An unhandled exception of type 'Newtonsoft.Json.JsonReaderException' occurred in Newtonsoft.Json.dll
Unexpected character encountered while parsing value: <. Path '', line 0, position 0.
My final goal with this code is that label1 says "5954607"
Remember that I'm really new to this so if you can explain the changes you made to the code ill really appreciate it.
You're not using RestRequest correctly or initialized the RestClient wrongly.
If that url you used in the RestClient is the whole url then you don't need a "resource" in the RestRequest.
Due to that mistake you got an HTML response indicated by the error
Unexpected character encountered while parsing value: <. Path '', line 0, position 0.
The value of restResponse.Content probably starts with <html and that is not valid JSON.
The Json payload is a bit more complex then you coded for. The results object holds an array with an object that has your stat property.
Putting these fixes together gives you this code.
RestClient restClient = new RestClient("https://thevirustracker.com/free-api?global=stats");
//Creating request to get data from server
RestRequest restRequest = new RestRequest(null, DataFormat.Json);
// Executing request to server and checking server response to the it
IRestResponse restResponse = restClient.Execute(restRequest);
// Extracting output data from received response
string response = restResponse.Content;
// Parsing JSON content into element-node JObject
var jObject = JObject.Parse(restResponse.Content);
//Extracting
// the object has a results property
// that is an array with one item (zero-based)
// on index 0 there is an object
// that has a property total_cases
string cases = (string) jObject["results"][0]["total_cases"];
label1.Text = (cases);
I need to be able to append an encoded object to a URI to pass it to a Web API endpoint.
In this post, the author is creating an object:
var request = new Object();
request.SearchWindowStart = start.toISOString();
request.SearchWindowEnd = end.toISOString();
request.ServiceId = "5f3b6e7f-48c0-e511-80d7-d89d67631c44";
request.Direction = '0';
request.NumberOfResults = 10;
request.UserTimeZoneCode = 1;
Then they are appending it to a URL:
var req = new XMLHttpRequest()
req.open("GET", clientUrl + "/api/data/v8.0/Search(AppointmentRequest=#request)?#request=" + JSON.stringify(request) , true);
I actually cannot modify the C sharp code however I have two options. The first option is to add the parameters into the URL I actually cannot modify the c# code however I have two options. The first option is to add the parameters into the URL and the other option would be to add a body to the request with my intended object.
If I know the structure of the object ahead of time how do I include it with my request?
You can do it in a two ways.
Just add every propery of the object with the value to the url eg. /search?property1=value1&property2=value2
Of course each value should be url encoded.
Serialize the whole object into json and send it via post or get. Look at the https://www.newtonsoft.com/json how to do it. Sending request could be done by simple WebClient class.
Based on the code snippet you need to serialize the object to JSON. You can use Json.Net as already linked in the other answer.
Using OP as an example...
var request = new {
SearchWindowStart = "some_start_value",
SearchWindowEnd = "some_end_value",
ServiceId = "5f3b6e7f-48c0-e511-80d7-d89d67631c44",
Direction = '0',
NumberOfResults = 10,
UserTimeZoneCode = 1
};
//JSON.stringify(request)
var json = JsonConvert.SerializeObject(request);
var url = clientUrl + "/api/data/v8.0/Search(AppointmentRequest=#request)?#request=" + json;
From there you should be able to use the URL as desired.
let encodedObj = encodeURIComponent(JSON.stringify(yourObject))
Then you could just.
req.open("GET", clientUrl +
"/api/data/v8.0/Search(AppointmentRequest=#request)?#request=" + encodedObj, true);
Although according to: https://javascript.info/url
The encode functions are based on the obsolete version RFC2396, and advises you to use these classes instead:
URL and URLSearchParams to build a URL.
I am trying to get data from a URL that contains JSON values. These values are to be used in my windows 8 desktop app. The code I have tried is seen below.
HttpClient client = new HttpClient();
HttpResponseMessage response = await client.GetAsync(App.DataServiceUrl + "/productcategory");
var Groups = new List<GroupList>();
if (response.IsSuccessStatusCode)
{
var content = await response.Content.ReadAsStringAsync();
var prods = JsonConvert.DeserializeObject<dynamic>(content);
foreach (var data in prods)
{
var dataGroup = new GroupList
(
data.term_id,
data.name,
data.slug,
data.description,
data.taxonomy
);
Groups.Add(dataGroup);
}
}
The URL I am accessing is actually a page in a website where I call some fucntions to get some data in php and then use json_encode($all); to return json data. When I access the url, I can see all the JSON data.
One example of the data from URL is:
[{"term_id":"64","name":"Argentina","slug":"argentina","term_group":"0","term_taxonomy_id":"64","taxonomy":"product_cat","description":"","parent":"13","count":"20","meta_id":"154","woocommerce_term_id":"64","meta_key":"order","meta_value":"0","cat_ID":"64","category_count":"20","category_description":"","cat_name":"Argentina","category_nicename":"argentina","category_parent":"13"},...]
But the application doesn't recognize the json values when I try to access them such as term_id, name etc. (as seen in code above)
Instead when I run the application I get this error:
Unexpected character encountered while parsing value: <. Path", line 4, position 2
How do I get and use the values from the url in my C# code?
You could use the DynamicJson class to parse the JSON, as far as I know it's the best option.
Your page produces some random text in front of JSON. You can take any HTTP debugger (my favorite is Fiddler, but anything would do) to confirm. Here are first several lines of the response from http://cbbnideas.com/brydens-website/api/ - note that JSON starts at 8th line:
37
<!--All Categories (And Sub Categories)-->
6bd8
[{"term_id":"64","name":"Argentina","slug":"argentina","term_gr...
I am trying to use a HTTP request to return the contents of a json string from google shopping api. What I have so far seems to be working but I would like to know how I can use the contents of the object to display the data on a page.
public string HttpGet(string url)
{
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
HttpWebResponse response = (HttpWebResponse)request.GetResponse();
try
{
using (Stream stream = response.GetResponseStream())
{
StreamReader reader = new StreamReader(stream);
return reader.ReadToEnd();
}
}
finally
{
response.Close();
}
}
protected void submit_Click(object sender, EventArgs e)
{
string json = HttpGet("https://www.googleapis.com/shopping/search/v1/public/products?key={KEY}4&country=UK&q=BX80637153570K&rankBy=price:ascending&maxResults=1&alt=json");
dynamic obj = JsonObject.Parse(json);
}
Ok looking at the responses it looks as though I need a C# class for the data returned in json. I have created a classes using json2csharp.com. This is the data I need to return from the Json and display on the page. Maybe it will help explain my problem better.
https://www.googleapis.com/shopping/search/v1/public/products?key=AIzaSyCC0j59RBeGNtf2W2ft6avhfoTdJ1FQ2c4&country=UK&q=BX80637153570K&rankBy=price:ascending&maxResults=1&alt=json
Can anyone advise how I can use this information on my website. I am a little lost now as I'm new to all this and have tried several different methods.I don't need all of the returned data if that makes a difference? Just price and link.
Supposing dynamic obj is rightly filled you can use it's content in web page
For examples, if json is:
{
"error" : {
"code": 400,
"message": "Bad Request"
}
}
You can access object properties with code like this:
Response.Write("Error code is" + obj.error.code);
Of course Response.Write is only a sample on how you can send retrieved data to the page.
Edit 1:
It seems json converter used in question is not working or not working right.
In many cases, it's overkill to create a concrete class only to parse a json, expecially since C# 4 that can use ExpandoObject
This is a sample on how you can deserialize in a dynamic object without the need to create a concrete object
var url = "http://www.google.com/ig/calculator?hl=en&q=100USD=?EUR";
string json = HttpGet(url);
//this is json string:
//{lhs: "100 U.S. dollars",rhs: "78.1799703 Euros",error: "",icc: true}
//now convert in a dynamic object
var jss = new DynamicJsonConverter();
var serializer = new JavaScriptSerializer();
serializer.RegisterConverters(new[] { new DynamicJsonConverter() });
dynamic obj = serializer.Deserialize(json, typeof(object));
//now you have access to json content
string text1 = obj.lhs;
string text2 = obj.rhs;
DynamicJsonConverter class can be created using code you can find here:
Deserialize JSON into C# dynamic object?
I think you need to deserialize the response
JavaScriptSerializer json_serializer = new JavaScriptSerializer();
You will need to create a class with the property resulted by the response.
Say test class
class Test
{
//define a similar property here
// which you suppose it will be return in the response
}
Then
Test routes_list = (Test)json_serializer.DeserializeObject(yourjson);
Edit 1
Or you can go through this link
How to Convert JSON object to Custom C# object?