通过从输入文本框中传递参数，通过 jQuery AJAX 调用 Java EE REST 服务答案

【问题标题】：Calling Java EE REST service through jQuery AJAX by passing the parameters from input text box通过从输入文本框中传递参数，通过 jQuery AJAX 调用 Java EE REST 服务
【发布时间】：2019-01-23 03:30:27
【问题描述】：

我有一个 html 代码来创建一个接受输入表单用户的输入文本框。并且参数必须与 url 一起传递给休息服务。这是我的ajax调用代码：

$(function() {

var empid = document.getElementById("ManagerId").value;
$('#submit').click(function(){ 
$.ajax({ 
    crossDomain : true,
     type: "GET",
     dataType: "json",
    url: "http://localhost:8088/JirasTrackingApp/reporter/Reportees?empid="+empid,

     success: function(result){        
        console.log(result);
        document.write(empid.value);
     }
 });
});

这是我的服务：

@Path("/Reportees")
public class ReporteesService {
    ReporteeList   reportee = new ReporteeList();

    @GET
    @Produces(MediaType.APPLICATION_JSON)
    public Map<Object, Object> getList(String empid) throws Exception {
        System.out.println("id is"+empid);  //when I try to print the empid,it  displays nothing
        Map<Object, Object> map=reportee.getReportees(empid);  
        return map;             
    }
});

这是我在 ReporteeList 类中的 getReportees()

public class ReporteeList {

    public Map<Object, Object> getReportees(String idOfEmp) throws Exception {
        System.out.println(idOfEmp);
        String msg = "error";
        String api = "https://connect.ucern.com/api/core/v3/people/";
        String id = idOfEmp;
        String ext = "/@reports";
        String url = api + id + ext;
        String name = "*********";
        String password = "*********";
        String authString = name + ":" + password;
        String authStringEnc = new BASE64Encoder().encode(authString.getBytes());
        System.out.println("Base64 encoded auth string: " + authStringEnc);
        Client restClient = Client.create();
        WebResource webResource = restClient.resource(url);
        ClientResponse resp = webResource.accept("application/json")
                                         .header("Authorization", "Basic " + authStringEnc)
                                         .get(ClientResponse.class);
        if (resp.getStatus() != 200) {
            System.err.println("Unable to connect to the server");
        }
        String output = resp.getEntity(String.class);

        // JSONParser reads the data from string object and break each data into key
        // value pairs
        JSONParser parse = new JSONParser();
        // Type caste the parsed json data in json object
        JSONObject jobj = (JSONObject) parse.parse(output);
        // Store the JSON object in JSON array as objects (For level 1 array element i.e list)

        JSONArray jsonarr_s = (JSONArray) jobj.get("list");
        Map<Object, Object> map = new HashMap<Object, Object>(); //error in this line 

        if (jsonarr_s.size() > 0) {

            // Get data for List array
            for (int i = 0; i < jsonarr_s.size(); i++) {
                JSONObject jsonobj_1 = (JSONObject) jsonarr_s.get(i);
                JSONObject jive = (JSONObject) jsonobj_1.get("jive");
                Object names = jsonobj_1.get("displayName");
                Object userid = jive.get("username");
                map.put(names, userid);                  
            }

            return map;
        } else {
            map.put("errorcheck", msg);
        }
        return map;
    }
}

服务未使用来自 ajax 调用的值 empid。请告诉我如何从 url 中获取参数并传递给其他服务。

【问题讨论】：

标签： java ajax rest web-services jakarta-ee

【解决方案1】：

您还必须为您的 getList 方法指定 @QueryParam 注释：

@GET
@Produces(MediaType.APPLICATION_JSON)
public Map<Object, Object> getList(@QueryParam("empId") String empid) throws Exception {
    System.out.println("id is"+empid);  //when I try to print the empid,it  displays nothing
    Map<Object, Object> map=reportee.getReportees(empid);  
    return map;             
}

【讨论】：