Hi I am using loadJSONObject function in Processing IDE to receive a JSON Object from a URL.
Following is my code:
JSONObject jsonUserLocations = loadJSONObject("http://smrt.utd.sg/eLocation/getLocs.php?userId="+usrID+"&sTime="+strtTime+"&eTime="+endTym);
This code works when the php returns some data (for some users).
The problem occurs when it doesn't return any data. (php doesn't return any data when there is no data, browser shows a blank page. this means user has no location data)
At this instance, the Processing IDE gives me an error saying;
a jsonobject text must begin with {
My question is how can I handle empty JSON Object in this type of situation? I need to skip if this is empty and request data for next user. Your help is much appreciated.
Thanks,
Hasala
Edit:
This is sample json object I receive when there is data.
{"locations":[{"latitude":"1.3809274","longitude":"103.7654596","startTime":"1421918587868","duration":"0","accuracy":"30"},{"latitude":"1.3805307","longitude":"103.7661015","startTime":"1421941711737","duration":"0","accuracy":"45"},{"latitude":"1.3805304","longitude":"103.7660959","startTime":"1421942011727","duration":"0","accuracy":"45"},{"latitude":"1.3799822","longitude":"103.7658037","startTime":"1421942311835","duration":"0","accuracy":"82.5"}],"success":1}
You have two options:
You could read the String form the URL first, and if it's blank, don't bother with the parsing.
Or you could just catch the exception that Processing throws:
try{
JSONObject jsonUserLocations = loadJSONObject("http://smrt.utd.sg/eLocation/getLocs.php?userId="+usrID+"&sTime="+strtTime+"&eTime="+endTym);
}
catch(JSONException e){
e.printStackTrace();
//json was blank, do something else
}
Related
I have a microservices based application and wish to create a service that captures all Fault events with their message payloads (as json) and stores them in a database for later analysis and potential resubmission. I have created a Fault consumer and can capture the Fault but am unable to generically extract the message payload as json.
public Task Consume(ConsumeContext<Fault> context)
{
if (context is PipeContext pipeContext)
{
var result = pipeContext.TryGetPayload(out ConsumeContext<Fault> payload2);
var serCont = context.SerializerContext;
}
Console.WriteLine($"A message faulted:{context.Message.FaultedMessageId} " +
$"{context.Message.Exceptions} " +
$"{context.ConversationId}"
);
return Task.CompletedTask;
}
I can see the full details I want in the context.SerializerContext._message but this is unaccessable.
context.SerializerContext._message
I saw you comment for a similar question:
If you did want to later get the actual fault message, you could use
consumeContext.TryGetMessage<Fault>(out var faultContext) and if it
was present you'd get it back.
I don't have "T" from every service and therefore want to handle all Faults a JSON.
Is there a way I can capture the full Fault with the message, ideally as json, without having access to every T across my system?
I am on MassTransit 8.
Thanks
If you have the message type (T), you can use TryGetMessage<Fault<T>> and it will return the message type deserialized.
If you don't, or if you want to deal with the JSON in a message directly, using V8 you can get the actual JsonElement from the deserializer and navigate the JSON yourself:
var jsonElement = context.TryGetMessage<JsonElement>()
Previous answer, but for Newtonsoft: https://stackoverflow.com/a/46779547/1882
I have a java spring integration project that is receving emails through the below code:
ClassPathXmlApplicationContext ac =
new ClassPathXmlApplicationContext(
"/integration/gmail-imap-idle-config.xml");
DirectChannel inputChannel = ac.getBean("receiveChannel", DirectChannel.class);
inputChannel.subscribe(message -> {
org.springframework.messaging.Message<MimeMailMessage> received =
(org.springframework.messaging.Message<MimeMailMessage>) message;
log.info("content" + message);
List<String> sentences = null;
try {
} catch (Exception e) {
}
I get the email, and I can get the subject, but I can never actually extract the message body. How do I do this?
Thank you!
You have to use this option on the channel adapter:
simple-content="true"
See its description:
When 'true', messages produced by the source will be rendered by 'MimeMessage.getContent()'
which is usually just the body for a simple text email. When false (default) the content
is rendered by the 'getContent()' method on the actual message returned by the underlying
javamail implementation.
For example, an IMAP message is rendered with some message headers.
This attribute is provided so that users can enable the previous behavior, which just
rendered the body.
But still it is doubtful, since I see in case of GMail message it is never simple. The content is a MimeMultipart and we need to read its parts to get access to the real body.
So, this is how you should change your code as well:
log.info("content" + ((MimeMultipart) ((MimeMessage) message.getPayload()).getContent()).getBodyPart(0).getContent());
I've a JSP app. It uploads a file, but to do so the user has to authenticate using a name and a password. So my JSP file starts with:
//0.2.- We get the password
String password = (String) request.getParameter("pass"); // -> This returns NULL
//0.3.- We get the "uvus"
String uvus = (String) request.getParameter("uvus"); //-> This also returns NULL
//More code
So I'm trying to know why am I getting null from those variables.
I went to the form I was uploading, and look for the data that was being sent. Using Firefox Debug Tools, I saw:
So in fact, it was being sent.
As additional info, I'm building the request like this:
var pUvus = document.getElementById("uvus").value;
var pPassword = document.getElementById("pass").value;
var file = document.getElementById("userFile");
var formData = new FormData();
formData.append("upload", file.files[0]);
formData.append("uvus", pUvus);
formData.append("pass", pPassword);
xmlhttp.open("POST","uploadFile.jsp",true);
xmlhttp.send(formData);
At last, I would like to say that I can get vars from application object in the same JSP with no errors, and have received in another pair of JSP files vars at request object without more problems, so I think my fault should be in the way I'm building the request in Ajax, but I've no more clue about that...
Anyone can guide me?
Thanks for your help
Update: #rickz asked for how do I get the file and parse the request (what is done after my problem, trying to get the objects from the request scope):
List items;
items = servlet_up.parseRequest(request);
for(int i=0;i<items.size();i++)
{
FileItem item = (FileItem) items.get(i);
if (! item.isFormField())
{
request.getParameter() won't work for a multipart/form-data request.
If you are using org.apache.commons.fileupload then you should be using something like
if(item.isFormField()){
name = item.getFieldName();
...
}
I have a servicestack service which when called via the browser (restful) Url ex:http://localhost:1616/myproducts, it works fine.
The service method has RedisCaching enabled. So first time it hits the data repository and caches it for subsequent use.
My problem is when I try calling it from a c# client via Soap12ServiceClient. It returns the below error:
Error in line 1 position 183. Expecting element '<target response>'
from namespace 'http://schemas.datacontract.org/2004/07/<target namespace>'..
Encountered 'Element' with name 'base64Binary',
namespace 'http://schemas.microsoft.com/2003/10/Serialization/'.
Below is my Client code:
var endpointURI = "http://mydevelopmentapi.serverhostingservices.com:1616/";
using (IServiceClient client = new Soap12ServiceClient(endpointURI))
{
var request = new ProductRequest { Param1 = "xy23432"};
client.Send<ProductResponse>(request);
}
It seems that the soapwsdl used is giving the problem, but I appear to have used the defaults as generated by servicestack..
Any help will be much appreciated.
Update
I was able over come this error by changing the cache code at the service end:
Code that returned error at client end:
return RequestContext.ToOptimizedResultUsingCache(this.CacheClient, cacheKey,
() =>
new ProductResponse(){CreateDate = DateTime.UtcNow,
products = new productRepository().Getproducts(request)
});
Code that works now:
var result = this.CacheClient.Get<ProductResponse>(cacheKey);
if (result == null)
{
this.CacheClient.Set<ProductResponse>(cacheKey, productResult);
result = productResult;
}
return result;
But I am still curious to know why the first method (RequestContext.ToOptimizedResultUsingCache) returned error at c# client?
But I am still curious to know why the first method (RequestContext.ToOptimizedResultUsingCache) returned error at c# client?
From what I can tell, the ToOptimizedResultUsingCache is trying to pull a specific format (xml, html, json, etc) out of the cache based on the RequestContext's ResponseContentType (see code here and here). When using the Soap12ServiceClient the ResponseContentType is text/html (not sure if this is correct/intentional within ServiceStack). So what ToOptimizedResultUsingCache is pulling out of the cache is a string of html. The html string is being returned to the Soap12ServiceClient and causing an exception.
By pulling directly out of the cache you are bypassing ToOptimizedResultUsingCache's 'format check' and returning something the Soap12ServiceClient can handle.
** If you are using Redis and creating your key with UrnId.Create method you should see a key like urn:ProductResponse:{yourkey}.html
Thanks for your response paaschpa.
I revisited the code and I was able to fix it. Since your response gave me the direction, I have accepted your answer. Below is my fix.
I moved the return statement from RequestContext to the response DTO.
Code which throws error when used via c# client (code was returning entire requestcontext):
return RequestContext.ToOptimizedResultUsingCache(this.CacheClient, cacheKey,
() =>
new ProductResponse(){CreateDate = DateTime.UtcNow,
products = new productRepository().Getproducts(request)
});
Fixed Code (return moved to response DTO):
RequestContext.ToOptimizedResultUsingCache(this.CacheClient, cacheKey,
() => {
return new ProductResponse(){CreateDate = DateTime.UtcNow,
products = new productRepository().Getproducts(request)
}
});
Actually my question is short.
How can I get a HttpPostedFile from a ASP.NET Web API POST or PUT?
I did see that I can get various information from the Request like Request.Header, Request.Content, Request.Properties. Where in there can I find the file I passed and how can I create a HttpPostedFile from it?
Thanks in advance!
Check out the great article from Henrik Nielsen to post multi-part content (i.e posting a form with file)
UPDATE: Add simple code for a controller to receive a file without multipart content
If you only need your controller to receive a file (i.e. no multipart content), you could do something like the above. The request only contains the file binary and the filename is passed within the URL.
public Task<HttpResponseMessage> Post([FromUri]string filename)
{
Guid uploadedFile = Guid.NewGuid();
Task<HttpResponseMessage> task = Request.Content.ReadAsStreamAsync().ContinueWith<HttpResponseMessage>(t =>
{
if (t.IsFaulted || t.IsCanceled)
throw new HttpResponseException(HttpStatusCode.InternalServerError);
try
{
using (Stream stream = t.Result)
{
//TODO: Write the stream to file system / db as you need
}
}
catch (Exception e)
{
Object o = e;
return Request.CreateResponse(HttpStatusCode.InternalServerError, e.GetBaseException().Message);
}
return Request.CreateResponse(HttpStatusCode.Created, uploadedFile.ToString());
});
return task;
}
Your short question does not have a short answer I am afraid.
ASP.NET Web API exposes you to the wonders of HTTP while ASP.NET MVC abstracted some of it - in this case for HttpPostedFile.
So a bit of background:
HTTP posts where a file is involved usually has multipart form data content. This means that you are mixing different kind of content-type: your normal form data will be sent using formurlencoded while the files will be sent application/octent-stream.
So in Web API, all you have to do is to say
var contents = message.Content.ReadAsMultipartAsync(); // message is HttpRequestMessage
One of the contents will contain your file.