Question

我尝试使用输入参数Diagnosticcentreid调用API并将错误视为

java.lang.IllegalArgumentException：未知参数名称：Diagnosticcentreid

我的错误可以帮助我吗？

public List<DiagnosticDoctor> getDiagnosticdoctor(int Diagnosticcentreid){
    Session session = SessionUtil.getSession();
    Query query = session.createSQLQuery("SELECT d.diagnosticcentreid, d.doctorname, d.specializationname, d.tagline, d.clinicname, d.daysname, d.time\n" +
            " FROM diagnosticdoctormapping ddm, doctor d where d.diagnosticcentreid = :ddm.Diagnosticcentreid");
    query.setInteger("Diagnosticcentreid", Diagnosticcentreid);
    List<Object[]> dataList =  query.list();
    List<DiagnosticDoctor> diagnosticDoctorList =new ArrayList<DiagnosticDoctor>();

    for(Object[] obj:dataList){
        DiagnosticDoctor dt = new DiagnosticDoctor();
        dt.setDiagnosticcentreid(Integer.parseInt(String.valueOf(obj[0])));
        dt.setDoctorname(obj[1].toString());
        dt.setSpecializationname(obj[2].toString());
        dt.setTagline(obj[3].toString());
        dt.setClinicname(obj[4].toString());
        dt.setDaysname(obj[5].toString());
        dt.setTime(obj[6].toString());

        diagnosticDoctorList.add(dt);
    }
    session.close();
    return diagnosticDoctorList;
}

这是我的资源类：

  @Path("/getdoctor/{Diagnosticcentreid}")
@Produces("application/json")
public Response getDiagnosticdoctor(@PathParam("Diagnosticcentreid") int Diagnosticcentreid) {
    DiagnosticDoctorMappingDAO dao = new DiagnosticDoctorMappingDAO();
    List dtm = dao.getDiagnosticdoctor(Diagnosticcentreid);
    String json = new Gson().toJson(dtm);
    return Response.ok().entity(json.toString()).build();
}

Answer 1

将您的sql更改为

Query query = session.createSQLQuery("SELECT d.diagnosticcentreid, d.doctorname, d.specializationname, d.tagline, d.clinicname, d.daysname, d.time\n" +
        " FROM diagnosticdoctormapping ddm, doctor d where d.diagnosticcentreid = :Diagnosticcentreid");

注意删除＆＃34; ddm。＆＃34;

java.lang.IllegalArgumentException：未知参数名称：Diagnosticcentreid

1 个答案: