Reactjs 隐藏和取消隐藏“切换”按钮上的所有记录单击
我在数组中有3条记录。单击“切换”按钮时,我需要分别隐藏和取消隐藏每个记录。
我的问题是,每次单击特定的切换按钮时,它不会隐藏并取消隐藏该记录,而是会 隐藏和取消隐藏所有三个记录。
我想我需要为每个切换按钮设置一个defmacro print_block(_msg, do: code_block) do
IO.inspect(code_block)
code_block
end
。有人可以帮我吗
{post.id}
答案 0 :(得分:0)
一种方法是,向数据中的每个项目添加一个标志,然后切换该标志:
class AutoButton extends React.Component {
constructor(props) {
super(props);
this.state = {
data: [],
shown: true,
};
}
componentDidMount() {
this.setState({
data: [
{ id: "1", title: "my first title", visible: true },
{ id: "2", title: "my second title", visible: true },
{ id: "3", title: "my third title", visible: true }
]
});
}
toggle(id) {
const newData = this.state.data.map(item => {
if(item.id === id) {
return { ...item, visible: !item.visible};
}
return item;
})
this.setState({
data: newData
});
}
render() {
return (
<div>
<label>
<ul>
{this.state.data.map((post, i) => (
<li key={i}>
<h2 style={{ display: post.visible ? "block" : "none"}}> {post.title} --{post.id} </h2>
<button onClick={this.toggle.bind(this, post.id)}>Toggle</button>
<br />
</li>
))}
</ul>
</label>
</div>
);
}
}
另一种方法是跟踪隐藏的项目,例如:
class AutoButton extends React.Component {
constructor(props) {
super(props);
this.state = {
data: [],
shown: true,
hiddenItems: []
};
}
componentDidMount() {
this.setState({
data: [
{ id: "1", title: "my first title" },
{ id: "2", title: "my second title" },
{ id: "3", title: "my third title" }
]
});
}
toggle(id) {
if(this.isHidden(id)) {
const newHiddenItems = this.state.hiddenItems.filter(item => item !== id);
this.setState({
hiddenItems: newHiddenItems
});
} else {
this.setState((state) => ({
hiddenItems: state.hiddenItems.concat(id)
}));
}
}
isHidden(id) {
return this.state.hiddenItems.includes(id);
}
render() {
return (
<div>
<label>
<ul>
{this.state.data.map((post, i) => (
<li key={i}>
<h2 style={{ display: this.isHidden(post.id) ? "none" : "block"}}> {post.title} --{post.id} </h2>
<button onClick={this.toggle.bind(this, post.id)}>Toggle</button>
<br />
</li>
))}
</ul>
</label>
</div>
);
}
}