[英]edittext visible means how can i check the if condition in android
I have checked my edittext is visible or invisible in android. 我检查了我的edittext在android中可见还是不可见。
Now i have to check the condition is., 现在我要检查条件是。,
This is my code for if i have to check the checkbox means the edittext is invisible otherwise the edittext is visible .: 这是我的代码,如果我必须选中此复选框,则表示edittext不可见,否则edittext可见。
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. 下面的代码是。,如果我的edittext是可见的意味着插入saddr值。否则插入baddr值。我该如何检查条件。
Here below error is showing: VISIBLE cannot be resolved to a variable. 下面的错误显示:VISIBLE无法解析为变量。
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 请在此处查看完整的源代码: 完整代码
EDIT: 编辑:
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. 在我的代码中,我必须选中复选框表示Saddress不可见知道。那时候我必须单击按钮意味着就插入了baddr值...如果我必须取消选中复选框则意味着Saddress值是可见的。必须插入saddr值。
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. 在这里我必须运行该应用程序,这意味着将在Saddress == visible和Saddess == invisible case.s中插入baddr值。我该如何为它们写条件。
use 采用
if(Saddress.getVisibility() == View.VISIBLE){
//.. your code here
}
instead of 代替
if(Saddress== VISIBLE){
//.. your code here
}
because VISIBLE
, GONE
and INVISIBLE
is part of View class instead of Activity 因为
VISIBLE
, GONE
和INVISIBLE
是View类而不是Activity的一部分
EditText editText = (EditText)findViewById(R.id.editText1);
editText.getVisibility();
this will provide the edittext is VISIBLE, INVISIBLE, or GONE. 这将提供edittext是VISIBLE,INVISIBLE或GONE。
if(editText.getVisibility() == View.VISIBLE){
//.. your actions are performed here
}
Refer This . 请参阅此 。 You can find some interesting details about VISIBLE , GONE and INVISIBLE
您可以找到有关VISIBLE , GONE和INVISIBLE的一些有趣的详细信息
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