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我花了半天的时间疯狂地让我的泽西服务接受并操纵JSON。将JSON发布到泽西岛请求
下面是我在做什么: 在PHP中使用Zend Framework:
$client = new Zend_Http_Client("http://localhost:8080/api/");
$data = array("city"=> "Paris", "zip" => "1111");
$json = json_encode($data);
$client->setHeaders("Content-type", "application/json");
$client->setRawData($json, 'application/json')->request("GET");
API方法:
@GET
@Produces(MediaType.APPLICATION_JSON)
@Consumes(MediaType.APPLICATION_JSON)
public Response getAPI(Address addr) {
JSONObject out = new JSONObject();
out.put("city test",addr.getCity());
Response response = null;
return response.ok(out.toString()).header("Accept", "application/json").build();
}
在一个单独的文件,我有我的注解类:
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
@XmlRootElement
public class Address
{
@XmlElement(name="city")
public String city;
@XmlElement(name="zip")
public String zip;
public String getCity() {
return city;
}
}
我收到不受支持的媒体类型错误:
Zend_Http_Response Object
(
[version:protected] => 1.1
[code:protected] => 415
[message:protected] => Unsupported Media Type
[headers:protected] => Array
(
[Server] => Apache-Coyote/1.1
[Content-type] => text/html;charset=utf-8
[Content-length] => 1117
[Date] => Tue, 29 May 2012 17:55:03 GMT
[Connection] => close
)
[body:protected] =>
我错过了什么?
谢谢大家, 丹尼尔
感谢您的回答,除了输出,我想我的问题是在输入中获取json。这正是我努力实现的目标! – Daniele
只要你的json输入结构正确,你应该全部设置。编组和解组是为你完成的。如果它不起作用,请尝试添加城市/邮编的设置方法,但我不确定是否需要它们。另外,我假设你在服务类中正确设置了@Path? –