我使用的是KeyboardAvoidingView,我想把道具的“行为”传递给它,以防平台是IOS平台,而不是安卓平台。我不想为相同的内容编写两个单独的JSX组件声明。有没有办法决定通过道具。Based on the platform the component should be declared the following way
<KeyboardAvoidingViewmodalContainer} >
{this.props.children} //I've lots of child