我正在解雇一个方法:
this.transitionTo('route', {},{search: this.state.search, type: this.state.type});
传递网址http://...route?search=%param10&type=param2
是否存在传递这些参数而不在URL中打印的等效方法?
答案 0 :(得分:6)
试试这个:
this.transitionTo("route", { search: this.state.search, type: this.state.type });