edittext visible means how can i check the if condition in android

问题

I have checked my edittext is visible or invisible in android.

Now i have to check the condition is.,

If my edittext is visible means how can i insert the data.
If my edittext is gone means how can i insert on another data.

This is my code for if i have to check the checkbox means the edittext is invisible otherwise the edittext is visible .:

 chkIos = (CheckBox) findViewById(R.id.addresscheckbox);

    chkIos.setOnClickListener(new View.OnClickListener() {
      public void onClick(View v) {
        if (((CheckBox) v).isChecked())
        {
             S_address = (TextView)findViewById(R.id.address1);  
             S_address.setVisibility(View.GONE);  

        Saddress = (EditText)findViewById(R.id.tf_address1);  
        Saddress.setVisibility(View.GONE);  
      }
        else
        {
             S_address = (TextView)findViewById(R.id.address1);  
             S_address.setVisibility(View.VISIBLE);  

       Saddress = (EditText)findViewById(R.id.tf_address1);  
        Saddress.setVisibility(View.VISIBLE);
        if(!(Saddress.getText().toString().length() == 0)){

                showAlertbox(" Shipping Address is Required!!"); 
            }
      }

The below code is ., if my edittext is visible means insert the saddr value.otherwise insert the baddr value.how can i check the condition.

Here below error is showing: VISIBLE cannot be resolved to a variable.

      if(Saddress== VISIBLE)
    {
        PropertyInfo saddress =new PropertyInfo();
        saddress.setName("Saddress");//Define the variable name in the web service method
        saddress.setValue(saddr);//Define value for fname variable
        saddress.setType(String.class);//Define the type of the variable
        request.addProperty(saddress);//Pass properties to the variable

    }
    else
    {
    PropertyInfo saddress =new PropertyInfo();
    saddress.setName("Saddress");//Define the variable name in the web service method
    saddress.setValue(baddr);//Define value for fname variable
    saddress.setType(String.class);//Define the type of the variable
    request.addProperty(saddress);//Pass properties to the variable
    }

please see the full source code here:full code

EDIT:

In my code i have to check the checkbox means the Saddress is invisible know.that time i have to click the button means the baddr value is inserted...If i have to uncheck the checkbox means the Saddress value is visible.that time i have to insert the saddr value.

Here i have to run the app means the baddr value is insered both Saddress==visible and Saddess==invisible case.how can i write the condition for these.

回答1:

use

if(Saddress.getVisibility() == View.VISIBLE){
  //.. your code here
}

instead of

if(Saddress== VISIBLE){
//.. your code here
}

because VISIBLE,GONE and INVISIBLE is part of View class instead of Activity

回答2:

EditText editText = (EditText)findViewById(R.id.editText1);
editText.getVisibility();

this will provide the edittext is VISIBLE, INVISIBLE, or GONE.

if(editText.getVisibility() == View.VISIBLE){
   //.. your actions are performed here
}

Refer This. You can find some interesting details about VISIBLE ,GONE and INVISIBLE

来源：https://stackoverflow.com/questions/14555787/edittext-visible-means-how-can-i-check-the-if-condition-in-android

标签

java

android

if-statement

checkbox

android-edittext