3
我燒成的方法隱藏查詢: this.transitionTo('route', {},{search: this.state.search, type: this.state.type});
反應路由器transitionTo(routeNameOrPath,[PARAMS [,查詢]])從URL
其中通過在URL http://...route?search=%param10&type=param2
查詢參數是否有等效方法,該方法傳遞這些參數而不在URL中打印?
更好地檢查這個http://stackoverflow.com/questions/27891667/react-js-pass-parameters-with-router-not-in-url – user2360560