Edittext inputtype constant value doesnot match

问题:

In android xml file im using editext as

<EditText
  android:id="@+id/email"
  android:layout_width="fill_parent"
  android:layout_height="33dp"
  android:inputType="textEmailAddress"
  android:hint="Enter your mail id" />

In java file while validating that editext.

if(editextobj.getInputType()==InputType.TYPE_TEXT_VARIATION_EMAIL_ADDRESS){

}

or

if(getInputType()==(InputType.TYPE_CLASS_TEXT | InputType.TYPE_TEXT_VARIATION_EMAIL_ADDRESS)){

}

this condition is not working since editextobj.getInputType() returns 33 whereas developer document gives TYPE_TEXT_VARIATION_EMAIL_ADDRESS constant value as 32

How to validate inputype programatically?

回答1:

There is nothing wrong with your code. 32 stays for TYPE_TEXT_VARIATION_EMAIL_ADDRESS. Also it's a flag, so you should test it like this. See InputType example(at the top under class overview) for more details.

if(editextobj.getInputType() & InputType.TYPE_TEXT_VARIATION_EMAIL_ADDRESS == 1){

}

回答2:

You need to test each flag separately, e.g.:

if ( ( editextobj.getInputType() & InputType.TYPE_TEXT_VARIATION_EMAIL_ADDRESS) != 0)
{
     // This is an email address!
}

回答3:

The following is happening:
InputType.TYPE_TEXT_VARIATION_EMAIL_ADDRESS value is: 32.
InputType.TYPE_CLASS_TEXT value is: 1.

the (|) is a bitwise OR operation. It's doing modification at the binary level:

Decimal: 32|1 results in 33
Binary: 100000|1 results in 100001 which is 33 in decimal.

editextobj.getInputType() value is 33

回答4:

Try this:

if(editextobj.getInputType() == (InputType.TYPE_TEXT_VARIATION_EMAIL_ADDRESS + 1)) {....}