I am using Spring MVC and this is my method:
/**
* Upload single file using Spring Controller.
*/
#RequestMapping(value = "/uploadFile", method = RequestMethod.POST)
public #ResponseBody ResponseEntity<GenericResponseVO<? extends IServiceVO>> uploadFileHandler(
#RequestParam("name") String name,
#RequestParam("file") MultipartFile file,
HttpServletRequest request,
HttpServletResponse response) {
if (!file.isEmpty()) {
try {
byte[] bytes = file.getBytes();
// Creating the directory to store file
String rootPath = System.getProperty("catalina.home");
File dir = new File(rootPath + File.separator + "tmpFiles");
if (!dir.exists()) {
dir.mkdirs();
}
// Create the file on server
File serverFile = new File(dir.getAbsolutePath() + File.separator + name);
BufferedOutputStream stream = new BufferedOutputStream(new FileOutputStream(serverFile));
stream.write(bytes);
stream.close();
System.out.println("Server File Location=" + serverFile.getAbsolutePath());
return null;
} catch (Exception e) {
return null;
}
}
}
I need to pass the session id in postman and also the file. How can I do that?
In postman, set method type to POST.
Then select
Body -> form-data -> Enter your parameter name (file according to your code)
On the right side of the Key field, while hovering your mouse over it, there is a dropdown menu to select between Text/File. Select File, then a "Select Files" button will appear in the Value field.
For rest of "text" based parameters, you can post it like normally you do with postman. Just enter parameter name and select "text" from that right side dropdown menu and enter any value for it, hit send button. Your controller method should get called.
The Missing Visual Guide
You must first find the nearly-invisible pale-grey-on-white dropdown for File which is the magic key that unlocks the Choose Files button.
After you choose POST, then choose Body->form-data, then find the File dropdown, and then choose 'File', only then will the 'Choose Files' button magically appear:
Maybe you could do it this way:
Like this :
Body -> form-data -> select file
You must write "file" instead of "name"
Also you can send JSON data from Body -> raw field. (Just paste JSON string)
I got confused after seeing all of the answers, I couldn't find any proper screenshot to bring the Content Type column. After some time, I found it by my own. Hope this will help somebody like me.
Here is the steps:
click on red marked area of postman.
Now check the green marked option (Content Type).
Now change the search content type, in the yellow marked area.
In my case:
invoice_id_ls (key) contains the json data.
documents contains the file data.
placed_amount contains normal text string.
Select [Content Type] from [SHOW COLUMNS] then set content-type of "application/json" to the parameter of json text.
Don't give any headers.
Put your json data inside a .json file.
Select your both files one is your .txt file and other is .json file
for your request param keys.
If somebody wants to send json data in form-data format just need to declare the variables like this
Postman:
As you see, the description parameter will be in basic json format, result of that:
{ description: { spanish: 'hola', english: 'hello' } }
Kindly follow steps from top to bottom as shown in below image.
At third step you will find dropdown of type selection as shown in below image
Body > binary > Select File
If you need like
Upload file in multipart using form data and send json data(Dto object) in same POST Request
Get yor JSON object as String in Controller and make it Deserialize by adding this line
ContactDto contactDto = new ObjectMapper().readValue(yourJSONString, ContactDto.class);
If somebody needed:
body -> form-data
Add field name as array
Use below code in spring rest side :
#PostMapping(value = Constant.API_INITIAL + "/uploadFile")
public UploadFileResponse uploadFile(#RequestParam("file") MultipartFile file,String jsonFileVo) {
FileUploadVo fileUploadVo = null;
try {
fileUploadVo = new ObjectMapper().readValue(jsonFileVo, FileUploadVo.class);
} catch (Exception e) {
e.printStackTrace();
}
If you want to make a PUT request, just do everything as a POST request but add _method => PUT to your form-data parameters.
The way to send mulitpart data which containts a file with the json data is the following, we need to set the content-type of the respective json key fields to 'application/json' in the postman body tab like the following:
You can send both Image and optional/mandatory parameters.
In postman, there is Params tab.
I needed to pass both: a file and an integer. I did it this way:
needed to pass a file to upload:
did it as per Sumit's answer.
Request type : POST
Body -> form-data
under the heading KEY, entered the name of the variable ('file' in my backend code).
in the backend:
file = request.files['file']
Next to 'file', there's a drop-down box which allows you to choose between 'File' or 'Text'. Chose 'File' and under the heading VALUE, 'Select files' appeared. Clicked on this which opened a window to select the file.
2.
needed to pass an integer:
went to:
Params
entered variable name (e.g.: id) under KEY and its value (e.g.: 1) under VALUE
in the backend:
id = request.args.get('id')
Worked!
For each form data key you can set Content-Type, there is a postman button on the right to add the Content-Type column, and you don't have to parse a json from a string inside your Controller.
first, set post in method and fill link API
Then select Body -> form-data -> Enter your parameter name (file according to your code)
If you are using cookies to keep session, you can use interceptor to share cookies from browser to postman.
Also to upload a file you can use form-data tab under body tab on postman, In which you can provide data in key-value format and for each key you can select the type of value text/file. when you select file type option appeared to upload the file.
If you want the Id and File in one object you can add your request object to a method as standard and then within Postman set the Body to form-data and prefix your keys with your request object name. e.g. request.SessionId and request.File.
The steps of uploading a file through postman along with passing some input data is very well discussed in below blog along with the screenshot. In this blog, the api code is written in node js. You can go through it once to have more clarity.
https://jksnu.blogspot.com/2021/09/how-to-create-post-request-with.html
At Back-end part
Rest service in Controller will have mixed #RequestPart and MultipartFile to serve such Multipart + JSON request.
#RequestMapping(value = "/executesampleservice", method = RequestMethod.POST,
consumes = {"multipart/form-data"})
#ResponseBody
public boolean yourEndpointMethod(
#RequestPart("properties") #Valid ConnectionProperties properties,
#RequestPart("file") #Valid #NotNull #NotBlank MultipartFile file) {
return projectService.executeSampleService(properties, file);
}
At front-end :
formData = new FormData();
formData.append("file", document.forms[formName].file.files[0]);
formData.append('properties', new Blob([JSON.stringify({
"name": "root",
"password": "root"
})], {
type: "application/json"
}));
See in the image (POSTMAN request):
Click to view Postman request in form data for both file and json
To send image along with json data in postman you just have to follow the below steps .
Make your method to post in postman
go to the body section and click on form-data
provide your field name select file from the dropdown list as shown below
you can also provide your other fields .
now just write your image storing code in your controller as shown below .
postman :
my controller :
public function sendImage(Request $request)
{
$image=new ImgUpload;
if($request->hasfile('image'))
{
$file=$request->file('image');
$extension=$file->getClientOriginalExtension();
$filename=time().'.'.$extension;
$file->move('public/upload/userimg/',$filename);
$image->image=$filename;
}
else
{
return $request;
$image->image='';
}
$image->save();
return response()->json(['response'=>['code'=>'200','message'=>'image uploaded successfull']]);
}
That's it hope it will help you
I have a lambda, written in Java, that accepts a Request Object of the structure
{
"id": "be1c320a-144f-464d-b32c-38ec7fb4445b",
"userId": "foobar"
}
When I call this Lambda through the test interface with such an object, it works fine.
I want to create an API where a GET request to
/users/foobar/items/be1c320a-144f-464d-b32c-38ec7fb4445b
i.e. of the form
/users/{userId}/items/{id}
calls this Lambda.
I have created the API resources /users, {userId}, items, and {id} appropriately.
And I have created the GET method (on /users/{userId}/items/{id})and associated it to the lambda.
When I test the API, it invokes the lambda, but with null values in the request. I can see it package the path as {"id":"be1c320a-144f-464d-b32c-38ec7fb4445b","userId": "foobar"} in the logs, but that's not being sent in the body.
I have tried creating a template map (and have tried RTFM), but cannot see how to map path parameters to a body.
How do I achieve this mapping?
I think your Request Object structure may not be properly configured. There may be a few ways to configure this. Here is some information that has helped me.
How to pass a querystring or route parameter to AWS Lambda from Amazon API Gateway - Demonstrates this mapping (albeit with python). However, taking the top response, if you enable "Use Lambda Proxy integration", you can similarily do this with Java as so:
#Override
public Object handleRequest(APIGatewayProxyRequestEvent input, Context context) {
Map<String, String> pathParameters = input.getPathParameters();
String id = pathParameters.get("id");
String userId = pathParameters.get("userId");
// Handle rest of request..
}
This is a tuturial using the serverless framework to create an Api with Java. This tutorial similarily accesses the pathParameters by parsing the input rather than using the APIGatewayProxyRequestEvent java class.
#Override
public Object handleRequest(Map<String, Object> input, Context context) {
try {
// get the 'pathParameters' from input
Map<String,String> pathParameters = (Map<String,String>)input.get("pathParameters");
String id = pathParameters.get("id");
String userId = pathParameters.get("userId");
} catch (Exception ex) {
logger.error("Error in retrieving product: " + ex);
}
}
Use a mapping template.
First, in the Method Request section, you should see userId and id as Request Paths
Then, in the Integration Request, do not choose Proxy Integration.
Then in the Mapping Templates section, add a new mapping template for application/json of the form
{
"id" : "$method.request.path.id",
"userId" : "$method.request.path.user_id"
}
I have a client application requesting a list of channels from a webservice. Is it possible to take the "response" from the web service and store it in an ArrayList?
Meaning if I wanted to store a list of channels for example, it would normally come from the web service as a response, typically from ResponseBuilder.
And I want to store it in an ArrayList from the client, like List.
How would I go about doing that?
You can use TypeReference to instantiate your Channel object list, here is an example:
import com.fasterxml.jackson.core.type.TypeReference;
public class ChannelClient {
public void getChannels() {
Response serviceResponse = client.target("http://your_service_url/channels/").
request(MediaType.APPLICATION_JSON).get(Response.class);
String responseString = serviceResponse.readEntity(String.class);
List<Channel> list = new ObjectMapper().readerFor(new TypeReference<List<Channel>>() {
}).readValue(responseString);
}
}
Make sure to have Jersey JSON Jackson jar in your dependencies, you can get it from here
https://mvnrepository.com/artifact/org.glassfish.jersey.media/jersey-media-json-jackson/2.26-b07
EDIT: In case you want to consume MediaType.TEXT_PLAIN response, you will just change the request method argument to your specified type like this:
Response serviceResponse = client.target("http://your_service_url/channels/").
request(MediaType.TEXT_PLAIN).get(Response.class);
I am moving from SQL to Couch DB from my web application, my very first application.
While i can not say why I do not like SQL queries, not sure that i don not, the idea of making CURL requests to access my database sound must better than using PHPs PDO .
I have spent a little over a day and a half trying to acquaint myself with the couch DB HTTP API. I can not claim I have throughly read the API , but who thoroughly reads an API before beginning to code. So my, possibly silly, question is - how do I pass an variable other than doc to a map function while making a http request to the view. The API clearly says that a map function only takes a single parameter which is "doc", in which case the function below itself is wrong but I can't find any section in the API that lets me query a database using end-user provided input.
my map function is
function(doc, pid2){
if (doc.pid === pid2)
{
emit(doc._id, doc) ;
}
}
pid2 is a number that will be provided by a front end user.
<?php
$pid2 = file_get_contents(facebook graphi api call_returns a Profile ID) ;
$user_exists = HTTP request to couch DB view to return
in JSON format the list of JSON documents with pid = $pid2
?>
Let your view emit the documents with doc.pid as the key
function(doc) {
emit(doc.pid, doc);
}
and use the key parameter to retrieve the right document:
http://localhost:5984/<database>/_design/<designdoc>/_view/<viewname>?key=<pid2>
This should return all documents with doc.pid === pid2.
I'm trying to do integration with Salesforce using their REST API and CF8.
I got the OAuth bit working, getting data etc but now I'm trying to update some records in Contact table.
First I tought about doing it the "proper" way as their docs say -
Update a record using HTTP PATCH.
But CFHTTP doesn't support PATCH method.
So then I tried running a SOQL query:
UPDATE Contact SET MailingStreet = 'Blah Blah' WHERE Id = '003A000000Zp4ObIAJ'
but here I'm getting
{"message":"unexpected token: UPDATE","errorCode":"MALFORMED_QUERY"}
Does anyone have an idea how to do it?
You can create your own PATCH method if your client supports it, but there is an easier way. From the Force.com REST API Developer's Guide:
If you use an HTTP library that doesn't allow overriding or setting an
arbitrary HTTP method name, you can send a POST request and provide an
override to the HTTP method via the query string parameter
_HttpMethod. In the PATCH example, you can replace the PostMethod line
with one that doesn't use override:
PostMethod m = new PostMethod(url + "?_HttpMethod=PATCH");
In CF9 CFScript, using the method that Paddyslacker already suggested for adding _HttpMethod=PATCH to the URL:
private boolean function patchObject(required string sfid, required string type, required struct obj) {
local.url = variables.salesforceInstance & '/services/data/v' & variables.apiVersion &'/sobjects/' & arguments.type & '/' & arguments.sfid &'?_HttpMethod=PATCH';
local.http = new Http(url=local.url,method='post');
//... convert obj to a json string, add to local.http ...
local.httpSendResult = local.http.send().getPrefix();
}
We have a CF9 CFC that we wrote that wraps most of the REST API that we will be open sourcing soon. I'll come back and link to it when we do.