[英]React/redux + bootstrap, make modal show unique for component
I have a component that shows a modal to pop up some content in my map. 我有一个组件,显示一个模式来弹出我的地图中的一些内容。 I have a pretty straight forward set up :
我有一个很直接的设置:
The JSX looks like this : JSX看起来像这样:
<Modal show={this.props.results.showPreviewModal} >
{myPreviewContent}
</Modal>
2 action-creators to open, close, and set the current item : 2个动作创建者来打开,关闭和设置当前项目:
export function previewAsset(result) {
return {
currentResult: result,
type: actions.PREVIEW_ASSET
};
}
export function closePreviewModal() {
return {
type: actions.CLOSE_PREVIEW_MODAL
};
}
And their reducers : 他们的减速器:
case actions.PREVIEW_ASSET:
return state.set('currentPreview', action.currentResult).set('showPreviewModal', true);
case actions.CLOSE_PREVIEW_MODAL:
return state.set('showPreviewModal', false);
Now, this seems to work fine. 现在,这似乎工作正常。 However, the issue is that the component that has the modal inside of it is inside a map, as it is a singular search result (each result component has a some functionality so it is it's own component that is mapped over with results).
但是,问题在于其中包含模态的组件位于地图内部,因为它是一个单一的搜索结果(每个结果组件都有一些功能,因此它是自己的组件,与结果一起映射)。 The issue is that if I have 10 results, this modal opens 10 times when I click the button that fires the
previewAsset
action. 问题是如果我有10个结果,当我点击触发
previewAsset
动作的按钮时,这个模态会打开10次。
This makes sense, because the showPreviewModal
is accessible by all components, but what I am wondering is if there is a way to make then unique for each component individually, so only the 1 modal opens, not all 10. Unsure how to approach this within react/redux, would very much appreciate any advice, thanks! 这是有道理的,因为所有组件都可以访问
showPreviewModal
,但我想知道的是,是否有一种方法可以单独为每个组件制作唯一的,所以只打开1个模态,而不是全部10.不确定如何在react / redux,非常感谢任何建议,谢谢!
An approach I've used successfully is to pull the Modal component out of the loop (or map()
in this case), and have a reducer for currentItem
or something similar, which gets set when an item is selected (you could also use currentItemIndex
, and then select the current item based on that in your connect()
call). 我成功使用的一种方法是将Modal组件拉出循环(或者在这种情况下为
map()
),并为currentItem
或类似的东西提供一个reducer,当选择一个项目时它会被设置(你也可以使用) currentItemIndex
,然后根据connect()
调用中的项选择当前项。
In the parent component, you'd have the Modal as a child, and only display it if that currentItem
is not null. 在父组件中,您将Modal作为子项,并且只有在
currentItem
不为null时才显示它。
Here's a quick JSBin example to show you what I mean: 这是一个快速的JSBin示例,向您展示我的意思:
http://jsbin.com/fefoxoy/edit?html,js,console,output http://jsbin.com/fefoxoy/edit?html,js,console,output
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