在我使用 Formik 的 React/TypeScript 应用程序中，出现错误

Object is possibly 'null'.  TS2531

    125 |             <Modal.Footer>
  > 126 |                 <Button variant="primary" type="submit" form="nicknameForm" disabled={!(formRef.current.isValid && formRef.current.dirty)}>Apply</Button>
        |                                                                                            ^
    127 |             </Modal.Footer>
    128 |         </Modal>
    129 |     )

尝试将 formRef.current.isValid 更改为 formRef!.current.isValid 并将 formRef.current.dirty 更改为 formRef! .current.dirty 但错误仍然存​​在。

为什么会这样，我们如何解决这个错误？谢谢!

import React, { useState, useEffect, useRef } from 'react';
import { Button, Modal, Form } from 'react-bootstrap';
import { Formik } from 'formik';

interface IModal {
    show: boolean;
    handleClose: () => void;
}

export function NicknameModal({show, handleClose}: IModal) {

    const formRef = useRef(null);

    return (
        <Modal show={show} onHide={handleClose}>
            <Modal.Header closeButton>
                <Modal.Title>My Title</Modal.Title>
            </Modal.Header>
            <Modal.Body>
                <Formik
                    initialValues={{
                        nickname: '',
                    }}
                    innerRef={formRef}
                    onSubmit={(
                        values,
                        { setSubmitting }
                    ) => {
                        setSubmitting(true);
                        handleClose();
                        setSubmitting(false);
                    }}
                >
                    {({values, errors, touched, handleChange, handleBlur, handleSubmit, isSubmitting, setFieldValue }) => (
                        <Form id="nicknameForm" onSubmit={handleSubmit}>
                            <Form.Group controlId="formNickname">
                                <Form.Label>Nickname</Form.Label>
                                <Form.Control type="text" name="nickname" onChange={handleChange} onBlur={handleBlur} value={values.nickname} />
                            </Form.Group>
                        </Form>  
                    )}
                </Formik>
            </Modal.Body>
            <Modal.Footer>
                <Button variant="primary" type="submit" 
                 disabled={!(formRef.current.isValid && formRef.current.dirty)}
                 form="nicknameForm">Apply</Button>
            </Modal.Footer>
        </Modal>
    )
}

更新:

如果 const formRef = useRef(null); 更改为 const formRef = useRef();，我们现在会遇到不同的错误:

Type 'MutableRefObject<undefined>' is not assignable to type '((instance: FormikProps<{ nickname: string; }> | null) => void) & MutableRefObject<undefined>'.
  Type 'MutableRefObject<undefined>' is not assignable to type '(instance: FormikProps<{ nickname: string; }> | null) => void'.
    Type 'MutableRefObject<undefined>' provides no match for the signature '(instance: FormikProps<{ nickname: string; }> | null): void'.  TS2322

    71 |                         nickName: '',
    72 |                     }}
  > 73 |                     innerRef={formRef}
       |                     ^
    74 |                     onSubmit={(
    75 |                         values: Values,
    76 |                         { setSubmitting }: FormikHelpers<Values>

最佳答案

你需要为 useRef 设置类型，其中 FormValues 是你的表单值

type FormValues = {};
useRef<FormikProps<FormValues>>(null);

https://github.com/formium/formik/issues/2600#issuecomment-693479057

关于javascript - 对象可能是 'null' : TypeScript, React useRef & Formik innerRef，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/66145167/