我想从我的数据库中选择查询。我需要从列表中查找,因为它在同一查询中有许多搜索项。所以我想得到清单来做到这一点。添加Model.addattribute
时,选择查询的效果非常好。但我不希望这样。我想从Ajax做到这一点。所以我想将列表传递给另一个为ajax编写的控件。所以请有人帮助我这样做。
这是我的控制器类,它可以获取我的页面
@RequestMapping(path = "/applicationManage")
public String viewApplicationPage(Model model) {
Object principal = SecurityContextHolder.getContext().getAuthentication().getPrincipal();
long userId = 0;
String agency = "";
String units = "";
List<String> unitsSLSI;
if (principal != null && principal instanceof AuthenticatedUser) {
AuthenticatedUser auth = (AuthenticatedUser) principal;
userId = auth.getUserId();
agency = auth.getAgency();
unitsSLSI = auth.getJobUnits();
}
return "applicationManage";
}
我想将unitsSLSI
列表传递给另一个控制器?其余服务(ajax)控制器是
@RequestMapping(value = "/SLSIApp", produces = {MediaType.APPLICATION_JSON_VALUE}, method = RequestMethod.GET)
@JsonIgnore
public ResponseEntity<List<SLSNotification>> listAllSLSI(List<String> unitList) {
List<SLSNotification> appbyUserID = jservice.getApplicationsByUnit(unitList);
if (appbyUserID.isEmpty()) {
return new ResponseEntity<List<SLSNotification>>(HttpStatus.NO_CONTENT);//You many decide to return HttpStatus.NOT_FOUND
}
System.out.println("hibernate query " + new ResponseEntity<List<SLSNotification>>(appbyUserID, HttpStatus.OK));
return new ResponseEntity<List<SLSNotification>>(appbyUserID, HttpStatus.OK);
}
当我添加此电源线时
List<SLSNotification> appbyUserID = jservice.getApplicationsByUnit(unitList);
我想从我之前的控制器获取unitList
,因为它有我的搜索列表。我怎么能这样做?
答案 0 :(得分:2)
在您收到反应时删除@JsonIgnore
...