Java 检查Edittext是否为空,如果不是,则执行特定的任务
我正在制作一个应用程序,但我遇到了一个问题。我想知道我能做些什么来实现这一点: 我想检查编辑文本是否为空,如果为空,则执行特定的任务,如果为空,则祝酒,祝酒词为“请输入姓名” 这是我的密码:`Java 检查Edittext是否为空,如果不是,则执行特定的任务,java,android,Java,Android,我正在制作一个应用程序,但我遇到了一个问题。我想知道我能做些什么来实现这一点: 我想检查编辑文本是否为空,如果为空,则执行特定的任务,如果为空,则祝酒,祝酒词为“请输入姓名” 这是我的密码:` final TextView percentage = (TextView) findViewById(R.id.percentage_text); Button calculateButton = (Button) findViewById(R.id.calculate_button); ca
final TextView percentage = (TextView) findViewById(R.id.percentage_text);
Button calculateButton = (Button) findViewById(R.id.calculate_button);
calculateButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
Random noGen = new Random();
int number = noGen.nextInt(101);
EditText firestName = (EditText) findViewById(R.id.first_name);
String firstNameString = firestName.getEditableText().toString();
EditText secName = (EditText) findViewById(R.id.sec_name);
String secNameString = secName.getEditableText().toString();
percentage.setText(Integer.toString(number));
Toast.makeText(getApplicationContext(), "The love score between " + firstNameString +" & " + secNameString + " is " + number + "%", Toast.LENGTH_SHORT ).show();
}
});
`
我知道我可以很容易地用if-else语句做到这一点,但我不能做到
提前感谢 试试这个:
if(firestName.getText()!=null &&
!firestName.getText().equals("") && secName.getText()!=null &&
!secName.getText().equals("")){
// do your task
}else{
// show msg
}
可以检查字符串变量是否为空。您可以参考下面的代码。您可以在Internet上找到更多字符串操作
Button calculateButton = (Button) findViewById(R.id.calculate_button);
EditText firestName = (EditText) findViewById(R.id.first_name);
EditText secName = (EditText) findViewById(R.id.sec_name);
calculateButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
Random noGen = new Random();
int number = noGen.nextInt(101);
String firstNameString = firestName.getEditableText().toString();
String secNameString = secName.getEditableText().toString();
if(!firstNameString.isEmpty() && !secNameString.isEmpty()){
//Perform your task
}
percentage.setText(Integer.toString(number));
Toast.makeText(getApplicationContext(), "The love score between " + firstNameString +" & " + secNameString + " is " + number + "%", Toast.LENGTH_SHORT ).show();
}
});
试试下面的代码
EditText firestName = (EditText) findViewById(R.id.first_name);
String firstNameString = firestName.getEditableText().toString();
EditText secName = (EditText) findViewById(R.id.sec_name);
String secNameString = secName.getEditableText().toString();
if(firstNameString.matches("") && secNameString.matches("")){
//Do your code here
}
}else{
Toast.makeText(getApplicationContext(),“请输入”,Toast.LENGTH_SHORT).show();
}
您可以做到这一点。试试看!我做了,但是我不能使用String。isEmpty()
比String.matches(“”)更容易阅读。格式化答案会更好。
if(firestName.getText().toString().equals("")){
//do some stuff