这个问题与this非常相似,但是,出于某种原因,除了返回键(回车键)外,每个键都可以正常工作。如果密码正确,我想要的是让用户进入下一页。任何帮助将非常感激
//代码
<TextInput
style={styles.txtfield}
placeholder="Password"
placeholderTextColor = 'rgba(249, 129, 37, 1)'
secureTextEntry={true}
onChangeText={ password => this.setState({ password })}
keyboardType="default"
returnKeyType="next"
onKeyPress={ (event) => {
if(event.nativeEvent.key == "Enter"){
alert(event.nativeEvent.key) // doesn't output anything nor execute the signin function
// this.signIn();
}
else {
alert('Something else Pressed') // show a valid alert with the key info
}
}}
/>
最佳答案
您将获得 onPress
事件 输入 key 仅当存在 multiline
时TextInput
.
单线用TextInput
,您将在 onSubmitEditing
中获得“Enter”或“Submit”按键事件方法。
<TextInput
style={styles.txtfield}
placeholder="Password"
placeholderTextColor = 'rgba(249, 129, 37, 1)'
secureTextEntry={true}
onChangeText={ password => this.setState({ password })}
keyboardType="default"
returnKeyType="next"
onSubmitEditing={()=>{
alert('on submit') // called only when multiline is false
}}
onKeyPress={ (event) => {
if(event.nativeEvent.key == "Enter"){
alert(event.nativeEvent.key) //called when multiline is true
// this.signIn();
}
else {
alert('Something else Pressed')
}
}}
/>
关于react-native - react native : e. nativeEvent.key == 'Enter' 不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53508294/