JPA entity with column of type long that have a Null value - jpa

Question

I'm new to JPA . I'm using spring data and I want to create an entity "File" (a folder is a File ) . each File can have a parent (or not).
this is my table's columns :id ,name,path ,mtime ,parentid.
I used this in my entities
#ManyToOne
#JoinColumn(name = "parentid")
private File parent;
public File() {}
public File(String name, String path,File parent) {
this.name = name;
this.path=path;
this.parent=parent;
}
#Override
public String toString() {
return String.format(
"File[ id=%d,name='%s', path='%s',parentid=%d]",
id,name, path,parent.id);
}
in my test I did this :
fileRepository.save(new File("file1","file1",null));
File file1=fileRepository.findOne((long) 1);
fileRepository.save(new File("file2","file2",file1));
and it inserted the first line with parentid NULL and a second line with parentid 1 (the first file).( I confirmed that on phpmyadmin)
I wanted to show the lines so i did this :
for (Object element : Lists.newArrayList(fileRepository.findAll())) {
System.out.println(element);
}
but it doesn't work .
when I removed parentid from my toString() function I get the correct result :
File[ id=1,name='file1', path='file1']
File[ id=2,name='file2', path='file2']
I get the same problem if I add a new column of type Long and when one of the lines have a NULL value in that column .
how Can I fix that ?

Answer 1

Your toString() method "doesn't work" because you're trying to get the id of a null parent, which obviously causes a NullPointerException. You need to test if the parent is null before trying to get and print its ID. Reading the message, type and stack trace of the exception would have allowed you to find this out. Using a debugger as well.
A variable of type long can't hold null. Primitive types are not nullable. So you need to use the type java.lang.Long for this field.