onchange事件处理程序的刷新问题

mznpcxlj  于 2021-09-13  发布在  Java
关注(0)|答案(1)|浏览(482)

我有一个带有onchange事件处理程序的输入,每当我尝试在其中键入字符时,它都会刷新屏幕。我想知道你们是否能帮我解决这个问题。
这是我的密码:

  1. const HeaderUser: React.FC<TabWrapper> = () => {
  2. const [drawerEmpreendimentosVisible, setDrawerEmpreendimentosVisible] = useState(false);
  3. const intl = useIntl().formatMessage;
  4. const [currentEmpreendimento] = useLocalStorage().createState('currentEmpreendimento');
  5. const [displayRoadmapScreen, setDisplayRoadmapScreen] = useState(false)
  6. const [displayRegisterScreen, setDisplayRegisterScreen] = useState(false);
  7. const [inspector, setInspector] = useState('');
  8. const [roadmap, setRoadmap] = useState('');
  9. const RegisterScreen = () => {
  10. const handleOk = () => {
  11. setDisplayRegisterScreen(false);
  12. };
  13. const handleCancel = () => {
  14. setDisplayRegisterScreen(false);
  15. };
  16. return (
  17. <div>
  18. <Modal
  19. title="Cadastrar novo roteiro"
  20. visible={displayRegisterScreen}
  21. onOk={handleOk}
  22. onCancel={handleCancel}>
  23. <b>Roteiro:</b>
  24. <br />
  25. //Refreshing issue <Input type="text" placeholder="Roteiro" onChange={e =>
  26. setRoadmap(e.target.value)} /><br />
  27. <b>Inspetor:</b>
  28. <br />
  29. <Input type="text" name="inspector" placeholder="Inspetor" onChange={(e) => console.log(e.target.value)} />
  30. </Modal>
  31. </div>
  32. )
  33. };
wgxvkvu9

wgxvkvu91#

可能是因为您在registerscreen组件之外定义了自己的状态?尝试在组件中定义它。

  1. const RegisterScreen = () => {
  2. const [roadmap, setRoadmap] = useState('');
  3. //rest of code...

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