错误：传递给persist的分离实体 - 尝试保留复杂数据（Play-Framework） [英] Error: detached entity passed to persist - try to persist complex data (Play-Framework)

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问题描述

我通过play-framework持续数据存在问题。也许它不可能实现这个结果，但如果它能够工作，它会非常好。



简单：我有一个复杂的模型（Shop with Addresses），我想一次性更改商店并将它们存储在同样的方式（shop.save（））。但是发生错误分离的实体传递给persist 。

更新历史
05.11


  • 05.11




    • 更新具有属性的模型商店mappedBy =shop

    • 更新指向google用户组的链接


  • 09.11




    • 找到一个解决方法，但它不是通用的


  • 16.11


    • 更新示例html表单，感谢@Pavel

    • 更新解决方法（更新09.11）一个通用的方法，感谢@ mericano1


  • 21.11


    • 我放弃了试图找到一个解决方案，并等待玩2.0 ...




Dateil ：
我尝试将问题降至最低：


模型：

  @Entity 
公共类商店扩展模型{
 
 @Required（message =Shopname is required） 
 public String shopname; 
 
 @OneToMany（cascade = CascadeType.ALL，fetch = FetchType.EAGER，mappedBy =shop）
 public List< Address>地址; 
 
} 
  


  @Entity 
 public class Address extends Model {
 
 @Required 
 public String location; 
 
 @ManyToOne 
公共商店; 
}

现在我的前端代码：

 ＃{extends'main.html'/} 
 
＃{form @save（shop？.id）} 
 
< input type =hiddenname =shop.idvalue =$ {shop？.id}/> 
 
＃{field'shop.shopname'} 
< label for =shopName>商店名称：< / label> 
 ＃{/ field} 
 
< legend> Addressen< / legend> 
＃{list items：shop.addresses，as：address} 
< input type =hiddenname =shop.addresses [$ {address_index  -  1}] .idvalue = $ {address.id}/> 
< label>位置< / label> 
< input name =shop.addresses [$ {address_index  -  1}]。locationtype =textvalue =$ {address.location}/> 
＃{/ list} 
 
< input type =submitclass =btn primaryvalue =保存更改/> 
＃{/ form}

我只有店铺本身的Id和店名通过POST提供，如：？shop.shopname = foo



interssting部分是地址列表，我有地址和地址，结果就像是：？shop.shopname = foo& shop.addresses [0] .id = 1& shop.addresses [0] .location =栏。


$ b

现在数据的 Controller 部分： 

 公共类商店扩展CRUD {
 
公共商店= Shop.findById（ID）; 
 render（shop）; 
} 
 render（）; 
 
 
 public static void save（Long id，Shop shop）{
 
 //手动设置所有者（不要从FE编辑）
用户用户= User.find（byEmail，Security.connected（））。first（）; 
 shop.owner = user; 
 
 //验证
 validation.valid（shop）; 
 if（validation.hasErrors（））
 render（@ form，shop）; 
 
 shop.save（）; 
 index（）; 
}

现在问题：当我更改地址数据的代码达到了 shop.save（）; 对象商店里充满了所有数据，一切看起来都很好，但是当hibernate尝试持久化数据时，错误传递给persist的分离实体发生:(

）

我试图改变fetch模式，cascadetype，我也尝试过：

  Shop shop1 = shop.merge（）; 
 shop1.save（）; 
  
 $ b 
不幸的是，没有任何工作，要么发生错误，要么没有地址数据将被存储。
有没有办法存储数据如果有什么不清楚的地方请写信给我，我会很乐意尽可能地提供更多的信息。
 
 
 
 更新1  
我也将问题放在谷歌用户组合 
 
 
 
 更新2 + 3  
在用户组的帮助下（感谢bryan w。）和一个来自mericano1的答案我在这里找到了一个通用的解决方法。  
 
 
首先，您必须从属性地址中删除 cascade = CascadeType.ALL  c $ c>在shop.class中。然后，您必须在shops.class中更改方法 save 。  
 
 
  public static void save（Long id，Shop shop）{
 
 //手动设置所有者不要从FE编辑）
 User user = User.find（byEmail，Security.connected（））。first（）; 
 shop.owner = user; 
 
 //在店内存储复杂数据
 storeData（shop.addresses，shop.addresses）; 
 storeData（shop.links，shop.links）; 
 
 //验证
 validation.valid（shop）; 
 if（validation.hasErrors（））
 render（@ form，shop）; 
 
 shop.save（）; 
 index（）; 
} 
  
存储数据的通用方法如下所示：
  private static< T extends Model> void storeData（List< T> list，String parameterName）{
 for（int i = 0; i< list.size（）; i ++）{
 T relation = list.get（i）; 
 
 if（relation == null）
 continue; 
 
 if（relation.id！= null）{
 relation =（T）Model.Manager.factoryFor（relation.getClass（））。findById（relation.id）; 
 StringBuffer buf = new StringBuffer（parameterName）; 
 buf.append（'['）。append（i）.append（']'）; 
 Binder.bind（relation，buf.toString（），request.params.all（））; 
} 
 
 //尝试设置bidiritional关系（您需要一个界面或不适用）
 //relation.shop = shop; 
 relation.save（）; 
 
 
 $ / code $ / pre 
 
我在Shop.class中添加了一个链接，但我不会更新其他代码片段，因此如果发生编译错误时会发出警告。 
解决方案
当您在Hibernate中更新一个复杂的实例时，您需要确保它来自数据库（先取得它，然后更新相同的实例），以避免这种'分离的实例'问题。
 
 
 
我通常更喜欢总是首先获取，然后只更新我期望从UI获得的特定字段。 
 
 
 
您可以使用 
 
 
 （ T）Model.Manager.factoryFor（relation.getClass（））findById（relation.id）。 
  
 
I have a problem with persisting data via play-framework. Maybe it's not possible to achive that result, but it would be really nice if it would work. 



Simple: I have a complex Model (Shop with Addresses) and I want to change the shop with addresses at once and store them in the same way (shop.save()). But the error detached entity passed to persistoccurs. 



Udate History
05.11 



    

  • 05.11
    


      

    • update Model Shop with attribute mappedBy="shop"
      • 

    • update link to google user group
      • 

    • 

  • 09.11
    


      

    • find a workaround, but it's ot generic
      • 

    • 

  • 16.11
    


      

    • update example html form, thanks to @Pavel
      • 

    • update workaround (update 09.11) to a generic method, thanks to @mericano1
      • 

    • 

  • 21.11


      

    • I gave up trying to find a solution and waiting for play 2.0...
      • 

    • 




Dateil: 
I try to cut down the problem to a minimum: 


Model: 
@Entity
public class Shop extends Model {

    @Required(message = "Shopname is required")
    public String shopname;

    @OneToMany(cascade=CascadeType.ALL, fetch=FetchType.EAGER, mappedBy="shop")
    public List<Address> addresses;

}


@Entity
public class Address extends Model {

    @Required
    public String location;

    @ManyToOne
    public Shop shop;
}

now my Frontendcode: 
#{extends 'main.html' /}

#{form @save(shop?.id)}

    <input type="hidden" name="shop.id" value="${shop?.id}"/>

    #{field 'shop.shopname'}
        <label for="shopName">Shop name:</label>
        <input type="text" name="${field.name}" 
            value="${shop?.shopname}" class="${field.errorClass}" />
    #{/field}

    <legend>Addressen</legend>
    #{list items: shop.addresses, as: "address"}
        <input type="hidden" name="shop.addresses[${address_index - 1}].id" value="${address.id}"/>
        <label>Location</label>
        <input name="shop.addresses[${address_index - 1}].location" type="text" value="${address.location}"/>
    #{/list}

     <input type="submit" class="btn primary" value="Save changes" />
#{/form}

I have just the Id from the Shop itself and the shopname to deliver via POST like: ?shop.shopname=foo 



The interssting part is the list of addresses and there I have the Id and the location from the address and the result would be somthin like: ?shop.shopname=foo&shop.addresses[0].id=1&shop.addresses[0].location=bar. 



Now the Controller part for the data: 
public class Shops extends CRUD {

public static void form(Long id) {

    if (id != null) {
        Shop shop = Shop.findById(id);
        render(shop);
    }
    render();
}

public static void save(Long id, Shop shop) {

    // set owner manually (dont edit from FE)
    User user = User.find("byEmail", Security.connected()).first();
    shop.owner = user;

    // Validate
    validation.valid(shop);
    if (validation.hasErrors()) 
        render("@form", shop);

    shop.save();
    index();
}

Now the problem: When i change the address data the code reaches the shop.save(); the object shop is filled with all data and everything looks fine, but when hibernate tryes to persist the data, the error detached entity passed to persist occurs :( 



I tried to change the fetch mode, the cascadetype and i also tried: 
Shop shop1 = shop.merge();
shop1.save();

Unfortunately nothing worked, either the error occurs, or no address data will be stored. 
Is there a way to store the data in that way? 



If there is somthing not clear please write to me, I would be glad to give as much information as possible. 



Update 1
I also put the problem on the google user group



Update 2 + 3 
With the help of the user group (thanks to bryan w.) and an Answer from mericano1 here I found a generic workaround. 



First you have to remove cascade=CascadeType.ALL from attribute addresses in shop.class. Then you have to change the method save within shops.class. 
public static void save(Long id, Shop shop) {

    // set owner manually (dont edit from FE)
    User user = User.find("byEmail", Security.connected()).first();
    shop.owner = user;

    // store complex data within shop
    storeData(shop.addresses, "shop.addresses");
    storeData(shop.links, "shop.links");

    // Validate
    validation.valid(shop);
    if (validation.hasErrors()) 
        render("@form", shop);

    shop.save();
    index();
}

the generic method to store the data looks like that: 
private static <T extends Model> void  storeData(List<T> list, String parameterName) {
    for(int i=0; i<list.size(); i++) {
        T relation = list.get(i);

        if (relation == null)
            continue;

        if (relation.id != null) {
            relation = (T)Model.Manager.factoryFor(relation.getClass()).findById(relation.id);
            StringBuffer buf = new StringBuffer(parameterName);
            buf.append('[').append(i).append(']');
            Binder.bind(relation, buf.toString(), request.params.all());
        }

        // try to set bidiritional relation (you need an interface or smth)
        //relation.shop = shop;
        relation.save();
    }
}

I added in Shop.class a list of Links, but I won't update the other code snippets, so be warned if compiling errors occur. 
 解决方案 
When you update a complex instance in Hibernate you need to make sure it is coming from the database (fetch it first, then update that same instance) to avoid this 'detached instance' problem.



I generally prefer to always fetch first and then only update specific fields I'm expecting from the UI.



You can make your code a bit more generic using
(T)Model.Manager.factoryFor(relation.getClass()).findById(relation.id);



                        
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