nested if else function in javascript

Question

I am trying to add a barcode verify function to an item picking webapp page and i have the javascript as this:

 function barcodeSubmit () {

 if (barcode1 != barcode || barcode2 != barcode) {
   if (PUT_LPN == "") {
  barcodeF.focus();
  return false;
  }  
 } else {
 if (barcode1 != barcode || barcode2 != barcode) {
 if (PUT_LPN != "") {
     barcodeF.focus();
     return false;
     }  
   }
 }
} else {
 if (barcode1 == barcode || barcode2 == barcode) {
 if (PUT_LPN == "") {
    PUT_LPN.focus();
    return false;
  }  
 }
}
 else {
  if (barcode1 == barcode || barcode2 == barcode) {
  if (PUT_LPN != "") {
    return true;
   }  
  }
 }
}

When I submit this form it does not submit and I can not get it to do the function its connected to an onclick input field . It does call when I make it simpler it will execute what I want. I also tried it without the else statements and came out with slightly different result but still not triggering event properly

"When I submit this form i ", here this refer to which form — sanjeev
– sanjeev, Commented Sep 8, 2015 at 5:00
You have one if with 4 else which is wrong. Use else if — Manwal
– Manwal, Commented Sep 8, 2015 at 5:00

Raimondas Kazlauskas · Accepted Answer · 2015-09-08 05:07:26Z

1

Use switch instead:

switch(true) {
    case ((barcode1 != barcode || barcode2 != barcode) && PUT_LPN == ""):
        barcodeF.focus();
        return false;
    case ((barcode1 != barcode || barcode2 != barcode) && PUT_LPN != ""):
        barcodeF.focus();
        return false;
    case ((barcode1 == barcode || barcode2 == barcode) && PUT_LPN == ""):
        PUT_LPN.focus();
        return false;
    case ((barcode1 == barcode || barcode2 == barcode) && PUT_LPN != ""):
        return true;
    default:
        return true;
}

You can remove default case if not needed

answered Sep 8, 2015 at 5:07

Raimondas Kazlauskas

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Comments

Manwal · Accepted Answer · 2015-09-08 05:04:25Z

1

Your code having lot of problems you are using one if and 4 else and some unused conditions also try following:

function barcodeSubmit() {

    if (barcode1 != barcode || barcode2 != barcode) {
        if (PUT_LPN == "") {
            barcodeF.focus();
            return false;
        } else if (PUT_LPN != "") {
            barcodeF.focus();
            return false;
        }
    }
    else if (barcode1 == barcode || barcode2 == barcode) {
        if (PUT_LPN == "") {
            PUT_LPN.focus();
            return false;
        } else if (PUT_LPN != "") {
            return true;
        }
    }
}

answered Sep 8, 2015 at 5:04

Manwal

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Comments

samumaretiya · Accepted Answer · 2015-09-08 05:07:34Z

1

Please try something like following

function barcodeSubmit () {

 if (barcode1 != barcode || barcode2 != barcode) {
	if (PUT_LPN == "") {
	  barcodeF.focus();
	  return false;
	}  
 } else if (barcode1 != barcode || barcode2 != barcode) {
	if (PUT_LPN != "") {
	  barcodeF.focus();
      return false;
    }   
} else if (barcode1 == barcode || barcode2 == barcode) {
	if (PUT_LPN == "") {
		PUT_LPN.focus();
		return false;
	}  
}else {
	if (barcode1 == barcode || barcode2 == barcode) {
		if (PUT_LPN != "") {
			return true;
		}  
	}
}

}

answered Sep 8, 2015 at 5:07

samumaretiya

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Comments

score 1 · Accepted Answer · 2015-09-08 05:25:09Z

1

Problem seems to me is PUT_LPN.focus() and PUT_LPN == "". I assume its and input field and in order to compare value of it you need to use PUT_LPN.value. Again the syntax of if...else is also not valid.

if(condition) {
  //
} else if(condition) {
  //
} else {
  //
}

Also there are lot many repetitive conditions which you need to take care of.

function barcodeSubmit() {
  if (barcode1 === barcode || barcode2 === barcode) {
    var flag = ('' === PUT_LPN.value);
    if (flag)
      PUT_LPN.focus();
    return !flag;
  }
  barcodeF.focus();
  return false;
}

edited Sep 8, 2015 at 5:25

answered Sep 8, 2015 at 5:09

user2575725

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