[英]POST complex parameters to REST service what would request URL and Body look like
So I am standing up a rest service. 所以我要站起来休息。 It is going to request a post to a resource that is pretty complex.
它将请求发布到非常复杂的资源。 The actual backend requires multiple (19!) parameters.
实际的后端需要多个(19!)参数。 One of which is a byte array.
其中之一是字节数组。 It sounds like this is going to need to be serialized by the client and sent to my service.
听起来这需要客户端进行序列化并将其发送到我的服务。
I am trying to understand how I can right a method that will handle this. 我试图了解如何正确处理此问题的方法。 I am thinking something like this
我在想这样的事情
@POST
@Path("apath")
@Consumes(MediaType.APPLICATION_JSON, MediaType.TEXT_HTML)
public Response randomAPI(@Parameter apiID, WhatParamWouldIPutHere confused){
}
What parameter types would I do to capture that inbound (Serialized) Post data. 我将执行什么参数类型来捕获该入站(序列化)的Post数据。 And what would the client URI look like?
客户端URI是什么样的?
In order to get all the parameters of the request you can use @Context UriInfo
as parameter of your randomAPI
method. 为了获取请求的所有参数,您可以使用
@Context UriInfo
作为您的randomAPI
方法的参数。
Then use UriInfo#getQueryParameters()
to get the full MultivaluedMap
of parameters. 然后使用
UriInfo#getQueryParameters()
得到充分的MultivaluedMap
的参数。
if you wish to convert the MultivaluedMap
to a simple HashMap
I added the code for that too. 如果您希望将
MultivaluedMap
转换为简单的HashMap
我也为此添加了代码。
so your method will basically look like something like this: 因此您的方法基本上看起来像这样:
@POST
@Path("apath")
@Consumes(MediaType.APPLICATION_JSON, MediaType.TEXT_HTML)
public Response randomAPI(@Context UriInfo uriInfo){
Map params= (HashMap) convertMultiToHashMap(uriInfo.getQueryParameters());
return service.doWork(params);
}
public Map<String, String> convertMultiToHashMap(MultivaluedMap<String, String> m) {
Map<String, String> map = new HashMap<String, String>();
for (Map.Entry<String, List<String>> entry : m.entrySet()) {
StringBuilder sb = new StringBuilder();
for (String s : entry.getValue()) {
sb.append(s);
}
map.put(entry.getKey(), sb.toString());
}
return map;
}
Additional Info : 附加信息 :
The
@Context
annotation allows you to inject instances ofjavax.ws.rs.core.HttpHeaders
,javax.ws.rs.core.UriInfo
,javax.ws.rs.core.Request
,javax.servlet.HttpServletRequest
,javax.servlet.HttpServletResponse
,javax.servlet.ServletConfig
,javax.servlet.ServletContext
, andjavax.ws.rs.core.SecurityContext
objects.@Context
批注允许您注入javax.ws.rs.core.HttpHeaders
,javax.ws.rs.core.UriInfo
,javax.ws.rs.core.Request
,javax.servlet.HttpServletRequest
和javax.servlet.HttpServletResponse
实例javax.servlet.HttpServletResponse
,javax.servlet.ServletConfig
,javax.servlet.ServletContext
和javax.ws.rs.core.SecurityContext
对象。
What I am thinking is you can simply use the httpservlet request and all the parameters can be retrieved as below 我在想的是您可以简单地使用httpservlet请求,并且可以按以下方式检索所有参数
@RequestMapping(value = "/yourMapping", method = RequestMethod.POST)
public @ResponseBody String yourMethod(HttpServletRequest request) {
Map<String, String[]> params = request.getParameterMap();
//Loop through the parameter maps and get all the paramters one by one including byte array
for(String key : params){
if(key == "file"){ //This param is byte array/ file data, specially handle it
byte[] content = params.get(key);
//Do what ever you want with the byte array
else if(key == "any of your special params") {
//handle
}
else {
}
}
}
http://docs.oracle.com/cd/E17802_01/products/products/servlet/2.3/javadoc/javax/servlet/ServletRequest.html#getParameterMap%28%29 http://docs.oracle.com/cd/E17802_01/products/products/servlet/2.3/javadoc/javax/servlet/ServletRequest.html#getParameterMap%28%29
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