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Home/ Questions/Q 223857
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T19:15:40+00:00 2026-05-11T19:15:40+00:00

Because calling a flush() to get every entities persist from memory to database. So

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Because calling a flush() to get every entities persist from memory to database. So if I use call too much unnecessary flush(), it could take much time therefore not a good choice for the performance. Here is a scenario that I don’t know when to call a flush()?

//Order and Item have Bidirectional Relationships
Order ord = New ord("my first order");
Item item = New Item("tv",10);

//...process item and ord object

em.persist(ord);//em is an instance of EntityManager
em.flush();// No.1 flush()

item.setOrder(ord);
em.persist(item);

Set<Item> items= new HashSet<Item>();
items.add(item);
ord.setItems(items);

em.flush();// No.2 flush()

My question is: calling of the No.1 flush could be avoid or not?

The things I worried is: in order to do the item.setOrder(ord), we need an database id of ord. And calling only em.persist(ord) cannot generate an database id, so I have to call the em.flush() before item.setOrder(ord). So what’s your opinion guys?

Thanks in advance.

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  1. Editorial Team
    Editorial Team
    2026-05-11T19:15:41+00:00Added an answer on May 11, 2026 at 7:15 pm

    i should first construct the structure, and after that persist everything.

    Order ord = New ord("my first order");
    Item item = New Item("tv",10);
    
    item.setOrder(ord);
    
    Set<Item> items= new HashSet<Item>();
    items.add(item);
    ord.setItems(items);
    
    em.persist(ord);
    

    In this way, you persist the whole tree in one call and is flush not needed.

    In good object design, you should use the way duffymo described to wire your objects.

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