How to Check Whether an Integer Is Null or Zero in Java

Checking Integer Wrapper against NULL as well as primitive value 0

Replace or by and :

if( statusId != null && statusId != 0 ) 

It will work because only if statusId is not null :

statusId != null

you will try to unbox statusId to int:

statusId != 0 

And in the case of statusId is null, the short-circuiting && operator will prevent to throw a NullPointerException as statusId != 0 will not be evaluated.

How can I tell if a Java integer is null?

parseInt() is just going to throw an exception if the parsing can't complete successfully. You can instead use Integers, the corresponding object type, which makes things a little bit cleaner. So you probably want something closer to:

Integer s = null;

try {
s = Integer.valueOf(startField.getText());
catch (NumberFormatException e) {
// ...

if (s != null) { ... }

Beware if you do decide to use parseInt()! parseInt() doesn't support good internationalization, so you have to jump through even more hoops:

try {
NumberFormat nf = NumberFormat.getIntegerInstance(locale);
nf.setMaximumIntegerDigits(9); // Or whatever you'd like to max out at.

// Start parsing from the beginning.
ParsePosition p = new ParsePosition(0);

int val = format.parse(str, p).intValue();
if (p.getIndex() != str.length()) {
// There's some stuff after all the digits are done being processed.

// Work with the processed value here.
} catch (java.text.ParseFormatException exc) {
// Something blew up in the parsing.

Check if the value is null and assign it 0

You can use ternary operator to check a condition and set the value like 0

String num1 = editTextnum1.getText().toString();
num1 = num1.isEmpty() ? "0" : num1;

Integer variable is zero when is it null in database

You have to handle that case yourself. The ResultSet documentation clearly states (emphasis mine):

getInt int getInt(int columnIndex)

Retrieves the value of the designated column in the current row of this ResultSet object as an int in the
Java programming language.


* columnIndex - the first column is 1, the second is 2, ...


the column value; if the value
is SQL NULL, the value returned is 0

You have to call wasNull() after reading a column value and set your field to null if that was the case. So something like the following:

Integer projectId = rs.getInt("PROJECT_ID");
if (rs.wasNull()) projectId = null;
return new Project (rs.getInt("ID"), projectId), ...);

For sanity reasons it's probably nicer if you move that code into a new method:

* Retrieves the value from the designated column as an {@link Integer}
* object.
* @param rs
* The ResultSet to read from.
* @param columnName
* The column name to read.
* @return {@code null} if the column value was SQL NULL; its value as an
* integer otherwise.
public static Integer getInteger(ResultSet rs, String columnName) {
int v = rs.getInt(columnName);
return rs.wasNull() ? null : v;

Check if a Integer is empty

Class Integer is just an wrapper on top of primitive int type. So it can either be null or store a valid integer value. There is no obvious "empty" definition for it.

If you just compare Integer against empty String, you''ll get false as a result. Always. See Integer.equals(Object o) implementation:

public boolean equals(Object obj) {
if (obj instanceof Integer) {
return value == ((Integer)obj).intValue();
return false;

First of all, you can get a NumberFormatException during parsing integer in the line:

Integer varsta = Integer.parseInt(request.getParameter("varsta"));

And you are getting it, since For input string: "" looks like a NumberFormatExpection message.

In your example you should either check whether the "varsta" attribute value is a number (assume it's a string) and then parse it, or parse it as is and catch NumberFormatException that Integer.parseInt() throws on incorrect argument.

First solution:

Integer varsta = null;
String varstaStr = request.getParameter("varsta"); // read string 'varsta' field
if (varstaStr != null && varstaStr.matches("\\d+")) { // null-check and regex check to make sure the string contains only digits
varsta = Integer.parseInt(varstaStr);

Second solution:

Integer varsta = null;
String varstaStr = request.getParameter("varsta"); // read string 'varsta' field
try {
varsta = Integer.parseInt(varsta);
} catch (NumberFormatException e) {
// handle error

After this, you have one more problem in the line:

if(varsta == null && "".equals(varsta)){

The varsta reference has type Integer here, so "".equals(varsta) will always return false:

(varsta == null && "".equals(varsta)) = [assume varsta is null] =
((null) == null && "".equals(null)) = (true && false) = false


if(varsta == null && "".equals(varsta)){


if(varsta == null){

This should help you.

P.S. If you use Java of version 7 or higher, consider use try-with-resources to manage Connection and PreparedStatement.

Compare nullable Integer to 0

x is a field in your class, so when you create it without making it to reference to any Integer object (Integer x = new Integer(7) for example), the compiler gives it a null for you (the default values for Object references). It seems like you have hence: Integer x = null;

So to compare it just use the equals() method that is implemented by Integer wrapper class.

new Integer(0).equals(x)

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