input field validation

[Former Member]

Hi

i have a input field with label PIN code which should not be NULL and should accept exactly six digits only and should not even accept Spaces in between, can u tell the required code

Thanks

Durga

[venkatakalyan_karanam]

Hi

First check the null

if not call the below method

Use the following method

public boolean isAllNumeric( java.lang.String stringValue )

{

//@@begin isAllNumeric()

boolean isNumeric = false;

isNumeric = Pattern.matches("0-9+", stringValue);

return isNumeric;

//@@end

}

which gives you whether the value is numeric or not.if not throw an error

if yes find the length of the string if not equals 6 throw an error.

Also you can restrict the input field to 6 digits by writing this code in init method.

use try catch block

ISimpleTypeModifiable homeAddrPostalCode = wdContext.wdGetAPI().getModifiableTypeOf ("HomeAddData<nodename>.PostalCode<attribute>");

homeAddrPostalCode.setMaxExternalLength(6);

Regards

Kalyan

[nikhil_bose]

make an attribute of type long and name it as pin. I am creating it in

in wdDoModifyView()

the Root context

    IWDAttributeInfo info = wdContext.getNodeInfo().getAttribute("Pin");
    info.getModifiableSimpleType().setFormat("######");

in the OnAction() where you are validating, put this code

if(wdContext.currentContextElement().getPin() == 0 ) {
 //display empty error message
}
else {
String temp = Long.toString(wdContext.currentContextElement().getPin());
if(temp.length >6 )
// display limit exceeds error message
} // if (temp)
}//else

nikhil

[Former Member]

HI

Try this to validate Pin Number Input.

String pin = wdContext.current<node>Element().getPinNo() ;
if(pin.length != 6 ){
wdComponentAPI.getMessageManager().reportException("Invalid Pin No." , true);
} 

Mandeep Virk

[Former Member]

Hi,

The following code should help you with your requirements.

if(wdContext.currentContextElement().getMobileNo().length()==6)//For calculating length.

{

int valid_flag=1;

for(int i=0;i<6;i++)

{

int j = wdContext.currentContextElement().getMobileNo().charAt(i);

if(j >=48 && j<=57)

{

valid_flag=1;

}

else

{

valid_flag=2 ;

break;

}

}

if(valid_flag==2)

{

wdComponentAPI.getMessageManager().reportSuccess("Please Enter a valid password");

}

else

{

wdComponentAPI.getMessageManager().reportSuccess("Valid password");

}

}

else

{

wdComponentAPI.getMessageManager().reportSuccess("Please Enter a valid password");

}

Regards,

Sudeep

[Former Member]

Hi sudeep

when i enter null value iam geeting null pointer exception

can u tell how to write validation for null value

thanks

durga

[Former Member]

Hi Durga,

The input field for which you are entering null, try to read the value form the context of the corresponding field.

After reading this value store this in any String variable and validate,i.e:-

String strVal = wdontext.currentContextElement.getName();

IWDMessageManager messageManager = wdComponentAPI.getMessageManager();

if(strVal.equalsIgnoreCase(null)){

/// print the error msg

messageManager .reportException(("error msg",true);

}

else{

///----needed code

}

Hope that may help you.

Regards,

Deepak

[Former Member]

hi

i think this will help u