Restlet使用json接收和响应实现post
首先,我想知道的是我正在做的是正确的方法 我有一个场景,我将收到一个json请求,我必须用它更新数据库,一旦数据库被更新,我必须用json确认回复 到目前为止,我所做的是创建类扩展应用程序,如下所示:Restlet使用json接收和响应实现post,json,postgresql,rest,restlet,Json,Postgresql,Rest,Restlet,首先,我想知道的是我正在做的是正确的方法 我有一个场景,我将收到一个json请求,我必须用它更新数据库,一旦数据库被更新,我必须用json确认回复 到目前为止,我所做的是创建类扩展应用程序,如下所示: @Override public Restlet createRoot() { // Create a router Restlet that routes each call to a // new instance of Sc
@Override
public Restlet createRoot() {
// Create a router Restlet that routes each call to a
// new instance of ScanRequestResource.
Router router = new Router(getContext());
// Defines only one route
router.attach("/request", RequestResource.class);
return router;
}
我的资源类正在扩展ServerResource,我的资源类中有以下方法
@Post("json")
public Representation post() throws ResourceException {
try {
Representation entity = getRequestEntity();
JsonRepresentation represent = new JsonRepresentation(entity);
JSONObject jsonobject = represent.toJsonObject();
JSONObject json = jsonobject.getJSONObject("request");
getResponse().setStatus(Status.SUCCESS_ACCEPTED);
StringBuffer sb = new StringBuffer();
ScanRequestAck ack = new ScanRequestAck();
ack.statusURL = "http://localhost:8080/status/2713";
Representation rep = new JsonRepresentation(ack.asJSON());
return rep;
} catch (Exception e) {
getResponse().setStatus(Status.SERVER_ERROR_INTERNAL);
}
我首先关心的是我在实体中接收到的对象是inputrepresentation,因此当我从创建的jsonrepresentation获取jsonobject时,我总是得到空/空对象
我已经尝试用以下代码和附加的客户端传递json请求
function submitjson(){
alert("Alert 1");
$.ajax({
type: "POST",
url: "http://localhost:8080/thoughtclicksWeb/request",
contentType: "application/json; charset=utf-8",
data: "{request{id:1, request-url:http://thoughtclicks.com/status}}",
dataType: "json",
success: function(msg){
//alert("testing alert");
alert(msg);
}
});
};
客户过去常打电话来
ClientResource requestResource = new ClientResource("http://localhost:8080/thoughtclicksWeb/request");
Representation rep = new JsonRepresentation(new JSONObject(jsonstring));
rep.setMediaType(MediaType.APPLICATION_JSON);
Representation reply = requestResource.post(rep);
非常感谢您提供的任何帮助或线索
谢谢,
Rahul当我使用以下JSON作为请求时,它可以工作:
{"request": {"id": "1", "request-url": "http://thoughtclicks.com/status"}}
请注意示例中没有的双引号和附加冒号。仅使用1个JARjse-x.y.z/lib/org.restlet.JAR,您就可以在客户端手动构建JSON,以满足简单请求:
ClientResource res = new ClientResource("http://localhost:9191/something/other");
StringRepresentation s = new StringRepresentation("" +
"{\n" +
"\t\"name\" : \"bank1\"\n" +
"}");
res.post(s).write(System.out);
在服务器端,仅使用两个jar-gson-x.y.z.jar和jse-x.y.z/lib/org.restlet.jar:
public class BankResource extends ServerResource {
@Get("json")
public String listBanks() {
JsonArray banksArray = new JsonArray();
for (String s : names) {
banksArray.add(new JsonPrimitive(s));
}
JsonObject j = new JsonObject();
j.add("banks", banksArray);
return j.toString();
}
@Post
public Representation createBank(Representation r) throws IOException {
String s = r.getText();
JsonObject j = new JsonParser().parse(s).getAsJsonObject();
JsonElement name = j.get("name");
.. (more) .. ..
//Send list on creation.
return new StringRepresentation(listBanks(), MediaType.TEXT_PLAIN);
}
}
在官方REST讨论论坛上问这个问题: