React Formik - 仅在表单提交时触发验证

Question

我在我的 React Native 应用程序中使用 Formik。在登录表单上，我有两个字段：电子邮件和密码，它们都是必需的。

我写过这样的验证规则：

export const LoginSchema = Yup.object().shape({
  email: Yup.string()
    .email('The email is invalid')
    .required('This field is required'),
  password: Yup.string()
    .min(6, 'The password is too short')
    .max(12, 'The password is too long')
    .required('This field is required'),
});

我只需要在提交表单时触发验证并显示错误弹出窗口。我已阅读文档，但找不到解决方案，因为验证会触发 onBlur。如何做到这一点？

谢谢！

const Login = ({ navigation }) => {
  const [isLoading, setIsLoading] = useState(true);
  const [isVisible, setIsVisible] = useState(false);

  useEffect(() => {
    // Later check for token
    const tokenIsStored = true;

    if (tokenIsStored) {
      setIsLoading(false);
    }
  });

  const onLogin = values => {
    console.log(values, 'on login');

    // Pass value to BE endpoint
    navigation.navigate('Dashboard');
  };

  const onModalClose = () => {
    setIsVisible(false);
  };

  console.log(isVisible);

  if (!isLoading) {
    return (
      <ScrollContainer keyboardShouldPersistTaps="handled">
        <ThemedStatusBar />

        <ThemedModal
          isVisible={isVisible}
          primaryMessage="Log In Failed"
          secondaryMessage="Please check your password"
          btnTitle="OK"
          btnPress={() => onModalClose()}
        />

        <Formik
          initialValues={{ email: '', password: '' }}
          validationSchema={LoginSchema}
          onSubmit={values => onLogin(values)}
        >
          {props => (
            <View>
              <ScrollContainer BackgroundColor={Colors.greyColor} Padding="0px" style={styles.loginForm}>
                <ThemedInput
                  onChangeText={props.handleChange('email')}
                  onBlur={props.handleBlur('email')}
                  value={props.values.email}
                  placeholder="Email"
                  keyboardType="email-address"
                />
                <ThemedInput
                  onChangeText={props.handleChange('password')}
                  onBlur={props.handleBlur('password')}
                  value={props.values.password}
                  placeholder="Password"
                  overrideStyles={styles.loginInputBottom}
                  secureTextEntry
                />
                {props.errors.email && setIsVisible(true)}
              </ScrollContainer>
              <ThemedButton onPress={props.handleSubmit} title="Log In" />
            </View>
          )}
        </Formik>
      </ScrollContainer>
    );
  }
  return <ThemedLoader isLoading />;
};

export default Login;

Answer 1

勾选docs

You can control when Formik runs validation by changing the values of <Formik validateOnChange> and/or <Formik validateOnBlur> props depending on your needs. By default, Formik will run validation methods as follows:

将道具 validateOnChange={false} 和 validateOnBlur={false}

传递给您的 Formik

Answer 2

是的。你可以这样做。

<Formik
        initialValues={initialValues}
        validationSchema={validationSchema}
        validateOnChange={false}
        validateOnBlur={false}
        onSubmit={(values, { validate }) => {
            validate(values);
        }}
    >

Answer 3

如果使用useFormik hook，需要添加下一个配置

  const formik = useFormik({
    initialValues,
    validationSchema,
    validateOnChange: false, // this one
    validateOnBlur: false, // and this one
    onSubmit: (values) => {
     do something on submit
    },
  });

React Formik - 仅在表单提交时触发验证

React Formik - Trigger validation only on form submit

validation

reactjs

react-native

yup

formik