我想知道是否有更好的方法来管理功能组件中对话框的打开和关闭?您可以在下面找到一个示例:
import React, { useState } from 'react';
import PropTypes from 'prop-types';
import EditDialog from './EditDialog';
import DeleteDialog from './DeleteDialog';
const ContactCard = ({ contact }) => {
const [editOpen, setEditOpen] = useState(false);
const [deleteOpen, setDeleteOpen] = useState(false);
const handleEditOpen = () => {
setEditOpen(true);
};
const handleEditClose = () => {
setEditOpen(false);
};
const handleDeleteOpen = () => {
setDeleteOpen(true);
};
const handleDeleteClose = () => {
setDeleteOpen(false);
};
const { type, firstName, lastName, phoneNumber, mail } = contact;
return (
<>
<div className={classes.main}>
{/* All my contact informations */}
</div>
<EditDialog handleClose={handleEditClose} open={editOpen} />
<DeleteDialog handleClose={handleDeleteClose} open={deleteOpen} />
</>
);
};
ContactCard.propTypes = {
contact: PropTypes.object.isRequired
};
export default ContactCard;
我认为这是超级多余的,但我找不到更好的方法来管理几个不同的对话框。
const handleEditOpen = () => {
setEditOpen(true);
};
const handleEditClose = () => {
setEditOpen(false);
};
const handleDeleteOpen = () => {
setDeleteOpen(true);
};
const handleDeleteClose = () => {
setDeleteOpen(false);
};
非常感谢您的时间和建议!
为了减少代码的一些冗余,您可以通过基本上切换当前状态来在一个函数中设置打开/关闭。我做了内联的,但你仍然可以创建一个handleEdit函数并在那里切换状态。
import React, {useState} from "react";
import ReactDOM from "react-dom";
function App() {
const [editCard, setEditCard] = useState(false)
return (
<div className="App">
<h1>Hello CodeSandbox</h1>
<h2>Start editing to see some magic happen!</h2>
<button onClick={() => setEditCard(!editCard)}>Toggle Edit</button>
{editCard && <div>Card is open for editing</div>}
</div>
);
}
const rootElement = document.getElementById("root");
ReactDOM.render(<App />, rootElement);
下面是代码的另一个示例。我没有运行它,但它应该看起来像这样。
import React, { useState } from 'react';
import PropTypes from 'prop-types';
import EditDialog from './EditDialog';
import DeleteDialog from './DeleteDialog';
const ContactCard = ({ contact }) => {
const [editOpen, setEditOpen] = useState(false);
const [deleteOpen, setDeleteOpen] = useState(false);
const handleEdit = () => {
setEditOpen(!editOpen);
};
const handleDelete = () => {
setDeleteOpen(!deleteOpen);
};
const { type, firstName, lastName, phoneNumber, mail } = contact;
return (
<>
<div className={classes.main}>
{/* All my contact informations */}
</div>
{
editOpen && <EditDialog handleEdit={handleEdit} />
}
{
deleteOpen && <DeleteDialog handleClose={handleClose} />
}
</>
);
};
ContactCard.propTypes = {
contact: PropTypes.object.isRequired
};
export default ContactCard;
打开对话框的责任应该是主要组成部分。这样,仅当状态属性为 true 时,才会呈现模态。 另一个提示是使用<React.Fragment>
而不是<>
import React, { useState } from 'react';
import PropTypes from 'prop-types';
import EditDialog from './EditDialog';
import DeleteDialog from './DeleteDialog';
const ContactCard = ({ contact }) => {
const [editOpen, setEditOpen] = useState(false);
const [deleteOpen, setDeleteOpen] = useState(false);
const handleEditOpen = () => {
setEditOpen(!editOpen);
};
const handleDeleteOpen = () => {
setDeleteOpen(!deleteOpen);
};
const { type, firstName, lastName, phoneNumber, mail } = contact;
return (
<React.Fragment>
<div className={classes.main}>
{/* All my contact informations */}
</div>
{
editOpen && <EditDialog handleClose={handleEditOpen} />
}
{
deleteOpen && <DeleteDialog handleClose={handleDeleteOpen} />
}
</React.Fragment>
);
};
ContactCard.propTypes = {
contact: PropTypes.object.isRequired
};
export default ContactCard;
为了封装更改对话框打开状态的逻辑,我建议创建单独的钩子:
const useToggle = (defaultValue) => {
return useReducer((value) => !value, !!defaultValue)
}
这个钩子基本上是useState
但 setState 函数不会等待参数更新状态,它用当前状态的反面更新状态。
这在使用对话框时可能很有用:
const ContactCard = () => {
const [editOpen, toggleEditOpen] = useToggle(false);
const [deleteOpen, toggleDeleteOpen] = usetoggle(false);
return (
<>
<div className={classes.main}>
{/* All my contact informations */}
</div>
{editOpen && <EditDialog handleEdit={toggleEditOpen} />}
{deleteOpen && <DeleteDialog handleClose={toggleDeleteOpen} />}
</>
);
};