java - JPA query using domain object Spring mvc jpa rest -

i want know if possible build jpa query using domain object.

for example:

       @entity        public class user {        private string firstname;        private string lastname;        private string email;     } 

in real object has more fields. receive valid json string of object. not fields filled.

for example filled firstname:

    {"firstname":"peter","lastname":"","email":""} 

this json gets deserialized user object

and want make search using object received parameter. result should user firstname peter.

the question is possible give object query?

thanks inputs

---

edit 1

thanks until now. found solution search contracts using entity. in solution post updated values of contract , other methods.

the json this:

contract = {fromdate:moment($('#datepickervon').val(), 'dd-mm-yyyy').format('dd-mm-yyyy'), enddate:moment($('#datepickervon').val(), 'dd-mm-yyyy').format('dd-mm-yyyy'), season: {           season: "so14"}, name: {name: "peter"}, category:{category:"somestring"}};            console.log(contract);       $.ajax({           url:'/contracts/search/',           datatype: "json",           type: "post",           mimetype: 'application/json',           contenttype: "application/json",           data: json.stringify(contract),           success: function(data) {               console.log(data);           }           });  

the controller receives this:

@requestmapping(value = "/search/", method = requestmethod.post, headers = "accept=application/json") public responseentity<string> getvertagfromsearch(@requestbody string json, uricomponentsbuilder uribuilder){     contract contract = contract.fromjsontocontract(json);     list<contract> contractlist = contract.findallcontractspercontract(contract);     return new responseentity<string>(contract.tojsonarray(contractlist), headers, httpstatus.ok); } 

the deserialization here:

public static contract fromjsontocontract(string json) {     return new jsondeserializer<contract>().use(calendar.class, new calendartransformer("dd-mm-yyyy hh:mm")).use(null, contract.class).deserialize(json); } 

and here transaction method.

 @transactional public static list<contract> findallcontractspercontract(contract contract) {     entitymanager e = entitymanager().getentitymanagerfactory().createentitymanager();     session session = e.unwrap(session.class);     example contractexample = example.create(contract);     criteria c = session.createcriteria(contract.class).add(contractexample);     list<contract> list = c.list();     return list; } 

with example results. question now, possible add criteria sql "bigger or less" or "like". because @ moment seems criteria searched "equals".

thanks

yes wouldnt result. happen jpa provider (in case hibernate) detect type entity object , comparison based on @id of entity supplying, guess in case have null id.

you have like:

 @entity    @namedquery(name="user.findbyfirstname", query="select u user u u.firstname = :firstname")    public class user {        private string firstname;        private string lastname;        private string email;     } 

and on session bean

public class userbean{      public list<user> findusers(user user){         return getentitymanager().createnamedquery("user.findbyfirstname")                 .setparameter("firstname", user.getfirstname())                 .getresultlist();      }    }