Reactjs React Formik-仅在表单提交时触发验证
我在React本机应用程序中使用Formik。在登录表单上,我有两个字段:email和password,它们都是必需的 我写过这样的验证规则:Reactjs React Formik-仅在表单提交时触发验证,reactjs,validation,react-native,formik,yup,Reactjs,Validation,React Native,Formik,Yup,我在React本机应用程序中使用Formik。在登录表单上,我有两个字段:email和password,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() .mi
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'),
});
我只需要在表单提交时触发验证,并显示一个错误弹出窗口。我已经阅读了文档,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;
const Login=({navigation})=>{
const[isLoading,setIsLoading]=useState(true);
const[isVisible,setIsVisible]=useState(false);
useffect(()=>{
//稍后检查令牌
const tokenIsStored=true;
如果(存储令牌){
设置加载(假);
}
});
const onLogin=值=>{
log(值为“登录时”);
//将值传递到端点
导航。导航(“仪表板”);
};
const onModalClose=()=>{
setIsVisible(假);
};
console.log(isVisible);
如果(!isLoading){
返回(
onModalClose()}
/>
onLogin(值)}
>
{props=>(
{props.errors.email&&setIsVisible(true)}
)}
);
}
返回;
};
导出默认登录;
检查
您可以根据需要更改
和/或
道具的值来控制Formik何时运行验证。默认情况下,Formik将按如下方式运行验证方法:
将道具
validateOnChange={false}
和validateOnBlur={false}
传递给你的Formik
是的。你可以这样做
<Formik
initialValues={initialValues}
validationSchema={validationSchema}
validateOnChange={false}
validateOnBlur={false}
onSubmit={(values, { validate }) => {
validate(values);
}}
>
{
验证(数值);
}}
>
这对v1也是一样吗?@NaveenVignesh是的,这对我不起作用。