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how can i call web services with curl with json param?

I'm programming in Java.
I developed a test web service in which I pass a random json and the service returns me hello
Here the web service code:
#Path("test")
#GET
#Consumes(MediaType.APPLICATION_JSON)
public String test(#HeaderParam("token") String token, #QueryParam("array")String array)
{
return "hello";
}
I call the service with curl
curl -v -H 'token:aaaa' 'http://140.145.1.2/clayapi/restservices/test?array=[{"name":"george","lastname":"ronney"}]';
message of error:
curl: (3) [globbing] illegal character in range specification at pos 60
I tried to add -g but it doesn't work .. how should I do?

Use -G along with --data-urlencode:
curl -v -G 'http://example.org' \
-H 'header:value' \
--data-urlencode 'array=[{"name":"george","lastname":"ronney"}]'
From the documentation:
-G, --get
When used, this option will make all data specified with -d, --data, --data-binary or --data-urlencode to be used in an HTTP GET request instead of the POST request that otherwise would be used. The data will be appended to the URL with a ? separator. [...]
--data-urlencode <data>
(HTTP) This posts data, similar to the other -d, --data options with the exception that this performs URL-encoding. [...]

To accept -d #json , How put method should be written in rest api

In PUT rest call , Using curl I would like to use -d option , what should be method signature . I am using java.
curl -X PUT -u testUser 'http://hostname:port/method' -d #test.json -H "Content-Type: application/json"
This should be my curl when I fire .
I have written a method signature which accepts string as input
#PUT #Path ("/test")
#Produces("application/json")
public Response method(String input) throws Exception {...}
When I run above command , Having above method signature is throwing error .
Error processing command: java.lang.NullPointerException
String input needs data in json format .
{
"name" : "test",
"host" : "10.xx.xx.xx",
"port" : 22
.
.
} //around 15 values
having these in a single url like this
"http://hostname:port/method?name=test&host=10.xx.xx.xx&port=22........."
makes url grow pretty long but its working though.
I would like to use -d option to send json input through a file.
What should be the method signature for curl to work with -d #test.json in java ?
--Thanks in Advance.

wiremock how to get post data

i'm trying to get data from the post does any one try
{
"name":"sasha",
"age":"15",
"country":"italy"
}
I will like to get the value sash from the post, any ideas?
only find to reply...
http://wiremock.org/docs/request-matching/
buy I don't want to reply, I want the data.
I would like something linke this:
String name = wiremock.withRequestBody(name());
print name -> sasha
something like this:
Body postBody = aResponse().withBody(matching("[?(#.name]"));
I'm trying to get data from the POST like curl -i -H "Accept: application/json" -H "Content-Type: application/json" -X POST -d "{'json':{'name':'sasha'}}" 127.0.0.1:8080/to/post I want to get sasha

Convert Python request with image to cURL

I found this example of a API request. Unfortunately I didn't find any other example how to upload an image to the API.
As I'm not familiar with Python I'm trying to understand how to do the same in a cURL command.
import requests
auth_headers = {
'app_id': 'your_app_id',
'app_key': 'your_app_key'
}
url = 'https://XXXXXXX'
files = {
'source': open('media/test.jpg')
}
data = {
'timeout': 60
}
response = requests.post(url, files=files, data=data, headers=auth_headers)
I tried to convert it by trying out a cURL to python converter, but I don't know how to build it with the files.
In the end I want to do the request in JAVA, but I think if I would know the request in cURL I can figure it out.
Hope anyone can help me with that.

This will do it:
#!/bin/bash
args=(
-H 'app_id: your_app_id'
-H 'app_key: your_app_key'
-F 'source=#/path/to/file'
-F 'timeout=60'
'http://httpbin.org/post'
)
curl "${args[#]}"
or, as a one-liner:
curl -H 'app_id: your_app_id' -H 'app_key: your_app_key' -F 'source=#/path/to/file' -F 'timeout=60' 'http://httpbin.org/post'
Use -H to specify header fields (repeat for every field) and -F to specify form fields - either as key=value pairs, or filename=#path pairs. When -F is used, POST method is the default, and Content-Type is multipart/form-data (but that too can be overridden).

Spring REST controller post request

I have this controller in spring
#RestController
public class GreetingController {
#RequestMapping(value = "/greeting", method = RequestMethod.POST)
public String greeting(#RequestParam("uouo") String uouo) {
return uouo;
}
}
and when I testing it
curl -k -i -X POST -H "Content-Type:application/json" -d uouo=test http://192.168.1.104:8080/api/greeting
the result of the testing
HTTP Status 400 - Required String parameter 'uouo' is not present
I tried may thing, but I think #RequestParam can't use for POST it always passed the parameter in URL using GET, I use post only if I had object JSON as parameter using #RequestBody, is there any way to make string parameter send using POST?

The Servlet container will only provide parameters from the body for POST requests if the content type is application/x-www-form-urlencoded. It will ignore the body if the content type is anything else. This is specified in the Servlet Specification Chapter 3.1.1 When Parameters Are Available
The following are the conditions that must be met before post form
data will be populated to the parameter set:
The request is an HTTP or HTTPS request.
The HTTP method is POST.
The content type is application/x-www-form-urlencoded.
The servlet has made an initial call of any of the getParameter family of methods on the request object.
If the conditions are not met and the post form data is not included
in the parameter set, the post data must still be available to the
servlet via the request object’s input stream. If the conditions are
met, post form data will no longer be available for reading directly
from the request object’s input stream.
Since you aren't sending any JSON, just set the appropriate content type
curl -k -i -X POST -H "Content-Type:application/x-www-form-urlencoded" -d uouo=test http://192.168.1.104:8080/api/greeting
or let curl infer it
curl -k -i -X POST -d uouo=test http://192.168.1.104:8080/api/greeting?uouo=test
Note that you can still pass query parameters in the URL
curl -k -i -X POST -H "Content-Type:application/json" http://192.168.1.104:8080/api/greeting?uouo=test