如何获取 Jersey JaxRS 中的所有查询参数？

我正在构建一个通用的 Web 服务，需要将所有的查询参数集成到一个字符串中，以便稍后进行解析。我怎么能这么做？

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最佳答案

You can access a single param via @QueryParam("name") or all of the params via the context:

@POST
public Response postSomething(@QueryParam("name") String name, @Context UriInfo uriInfo, String content) {
MultivaluedMap<String, String> queryParams = uriInfo.getQueryParameters();
String nameParam = queryParams.getFirst("name");
}

The key is the @Context jax-rs annotation, which can be used to access:

UriInfo, Request, HttpHeaders, SecurityContext, Providers

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The unparsed query part of the request URI can be obtained from the UriInfo object:

@GET
public Representation get(@Context UriInfo uriInfo) {
String query = uriInfo.getRequestUri().getQuery();
...
}

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Adding a bit more to the accepted answer. It is also possible to get all the query parameters in the following way without adding an additional parameter to the method which maybe useful when maintaining swagger documentation.

@Context
private UriInfo uriInfo;


@POST
public Response postSomething(@QueryParam("name") String name) {
MultivaluedMap<String, String> queryParams = uriInfo.getQueryParameters();
String nameParam = queryParams.getFirst("name");
}

ref