Spring mongodb 模板保存在同一对象中

Spring mongodb template saves in the same object

我有如下模型

@CompoundIndexes(value = {
        @CompoundIndex(name = "catalog_idx", def = "{'code' : 1, 'brand' : 1}", unique = true) })
@Document(collection = Catalog.ENTITY)
public class Catalog extends AbstractModel<String> {

    private static final long serialVersionUID = 1L;

    public static final String ENTITY = "catalog";

    @NotNull(message = "Code is required")
    @Field("code")
    private String code;

    @NotNull(message = "Brand is required")
    @DBRef(lazy = true)
    @Field("brand")
    private Brand brand;
}

当我使用 mongoTemplate.save(object); 保存时,我只看到在数据库中创建了 2 个对象,而不是 6 个。就在保存要保存的对象的调试行之前。

Catalog [code=StagedCatalog, brand=Brand [code=Brand_3]]
Catalog [code=StagedCatalog, brand=Brand [code=Brand_2]]
Catalog [code=StagedCatalog, brand=Brand [code=Brand_1]]
Catalog [code=OnlineCatalog, brand=Brand [code=Brand_2]]
Catalog [code=OnlineCatalog, brand=Brand [code=Brand_1]]
Catalog [code=OnlineCatalog, brand=Brand [code=Brand_3]]

知道为什么吗?我觉得索引独特的东西不知何故不起作用。我希望 codebrandunique combination

public abstract class AbstractModel<ID extends Serializable> implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    private ID id;
}

您设置了唯一索引。这意味着您将无法拥有 2 个具有相同代码和品牌的文件。

现在您已将 ID 列设置为 ID 对象。您有 2 个插入而不是 6 个的事实意味着您对 3 个插入使用相同的 ID,例如:

for (code: {"StagedCatalog","OnlineCatalog"} ) {
    ID id=new ID(...);
    for (brand: {1, 2, 3}){
        Catalog cat=new Catalog();
        cat.setId(id);              // <<== this is wrong, you reuse the same id, you will insert first brand, then update to brand2 and brand3.
        cat.setCode(code);
        cat.setBrand(brand);
        mongoTemplate.persist(cat);
    }
}

为防止这种情况发生,您需要:

Catalog cat=new Catalog();
ID id=new ID(realUniqueId);  // RealuniqueId can be code+brand for instance
cat.setId(id); 
...

db.collection.save()

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