javaweb作为服务端，怎么配置服务器，能让小程序访问到 - WXOPEN Club

yanjin 1楼•5 年前

[@Controller](/user/Controller)

//比如：影院列表：/cinema/queryCinema.action

[@RequestMapping](/user/RequestMapping)("/cinema")

public class CinemaController {

[@Autowired](/user/Autowired)

private CinemaService cinemaService;

[@RequestMapping](/user/RequestMapping)("/queryCinema")

public [@ResponseBody](/user/ResponseBody) Map<String, Cinema[]> queryCinema(Cinema cinema) throws Exception {

List<Cinema> cinemaList=cinemaService.findCinemaList(cinema);

Cinema[] cl = (Cinema[]) cinemaList.toArray(new Cinema[cinemaList.size()]);

Map<String, Cinema[]> map = new HashMap<String, Cinema[]>();

map.put("cinemas", cl);

return map;

}

这是后台controller代码用注解@ResponseBody直接可以返回json格式的数据

小程序可以直接访问这个链接访问到数据

rding 2楼•5 年前

如果正式发布了，怎么办呢

li93 3楼•5 年前

不用的

把最下面的打钩就不检测域名

mingfeng 4楼•5 年前

小程序的代码是什么

dengxiulan 5楼•5 年前

我没用wx.socket，把Javaweb工程发布到Tomcat，直接一个链接访问Tomcat你的项目，http://localhost:8080/wxTicket/cinema/queryCinema.action?lat=37.5285652317&lng=121.3655765023

libai 6楼•5 年前

我写成这样，不对

var connectionbol=false;
//input.js
Page({
data: {

},
onLoad:function(){
     wx.connectSocket({
       url: ‘ws://192.168.0.220:8080/WebScada/getPowerCurveFromApp.action’,
       data: {

       },
       header: {
         ‘content-type’: ‘application/json’
       },
       method: ‘POST’,
       success: function (msg) {
         console.log(msg);

       }

     })
     wx.onSocketOpen(function () {
       connectionbol=true;
      })
     wx.onSocketMessage(function () {
       data: “”
     });
},
myquery:function(){
    if (connectionbol){
      wx.sendSocketMessage({
        data: {
          “paramid”: “M_20161017003”,
          “thetype”: “t_moniterCenter”,
          “starttime”: “2016-11-15”,
          “datatype”: “realtime”
        }
      });
    }
}
})