2015-01-13 31 views
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?

+0

更好地檢查這個http://stackoverflow.com/questions/27891667/react-js-pass-parameters-with-router-not-in-url – user2360560

回答

6

試試這個:

this.transitionTo("route", { search: this.state.search, type: this.state.type });