[英]Automatic Hibernate index creation too long
我回來了,現在遇到了陽光下的錯誤/問題。 通常,我在H2DB上測試本地開發和更改,但是據我所知,它也必須在Oracle和MSSQL上工作。 現在再次在oracle上進行測試會出現此問題:密鑰COR_VIEWSETTINGSCOR_USERSETTINGS_FK0和COR_VIEWSETTINGSCOR_USERSETTINGS_FK1是自動生成的,對於oracle數據庫來說太長了。 要知道如何創建這些鍵,我現在將向您顯示實體UserSettings和UserViewSettings。 提示:您可以忽略實體,如果它們使您感到困惑,則可以繼續進行編輯。 也許您仍然可以幫助我。
用戶設置
/**
The Class UserSettings.
*/
@org.hibernate.envers.Audited
@DataObject( value = UserSettings.DATA_OBJECT_NAME )
@CRUDDefinition( supportsRead = true, supportsCreate = true, supportsUpdate = true, supportsDelete = true )
@Entity( name = UserSettings.DATA_OBJECT_NAME )
@NamedQuery( name = UserSettings.DATA_OBJECT_NAME, query = "from userSettings e where e.name = :name" )
@javax.persistence.Inheritance( strategy = javax.persistence.InheritanceType.TABLE_PER_CLASS )
@AttributeOverrides( { @AttributeOverride( name = "id", column = @Column( name = "USERSETTINGS_ID" ) )
} )
@Table( name = "COR_USERSETTINGS", indexes = {
@javax.persistence.Index( name="COR_USERSETTINGS_FK0", columnList = "SETTINGSTYPE_ID" ),
@javax.persistence.Index( name="COR_USERSETTINGS_FK1", columnList = "USER_ID" ),
}
)
public class UserSettings extends NamedRevisionEntity implements NameSettingsType, NameSettings
{
/** The Constant serialVersionUID. */
private static final long serialVersionUID = 1L;
/** The Constant DATA_OBJECT_NAME. */
public static final String DATA_OBJECT_NAME = "userSettings";
@javax.persistence.Basic( fetch = javax.persistence.FetchType.EAGER, optional = false )
@Column( name = "SETTINGS", nullable = false, unique = false, insertable = true, updatable = true )
@javax.persistence.Lob
private java.lang.String settings;
@javax.persistence.ManyToOne( fetch = javax.persistence.FetchType.EAGER, optional = false )
@javax.persistence.JoinColumn( name = "SETTINGSTYPE_ID", nullable = false, unique = false, insertable = true, updatable = true )
private SettingsType settingsType;
@javax.persistence.ManyToOne( fetch = javax.persistence.FetchType.EAGER, optional = true )
@javax.persistence.JoinColumn( name = "USER_ID", nullable = true, unique = false, insertable = true, updatable = true )
private User user;
public SettingsType getSettingsType()
{
return settingsType;
}
public void setSettingsType( SettingsType settingsType )
{
this.settingsType = settingsType;
}
public User getUser()
{
return user;
}
public void setUser( User user )
{
this.user = user;
}
public java.lang.String getSettings()
{
return settings;
}
public void setSettings( java.lang.String settings )
{
this.settings = settings;
}
@Override
public String getDataObjectName()
{
return DATA_OBJECT_NAME;
}
@Override
public String toString()
{
StringBuilder builder = new StringBuilder( super.toString() );
builder.append( ", " );
try
{
builder.append( ToStringUtils.referenceToString( "settingsType", "SettingsType", this.settingsType ) );
}
catch( Exception ex )
{
builder.append( ex.getClass().getName() );
builder.append( ": " );
builder.append( ex.getMessage() );
}
builder.append( ", " );
try
{
builder.append( ToStringUtils.referenceToString( "user", "User", this.user ) );
}
catch( Exception ex )
{
builder.append( ex.getClass().getName() );
builder.append( ": " );
builder.append( ex.getMessage() );
}
builder.append( "]" );
return builder.toString();
}
}
UserViewSettings
/**
The Class UserViewSettings.
*/
@org.hibernate.envers.Audited
@DataObject( value = UserViewSettings.DATA_OBJECT_NAME )
@CRUDDefinition( supportsRead = true, supportsCreate = true, supportsUpdate = true, supportsDelete = true )
@Entity( name = UserViewSettings.DATA_OBJECT_NAME )
@AttributeOverrides( { @AttributeOverride( name = "id", column = @Column( name = "VIEWSETTINGS_ID" ) )
} )
@Table( name = "COR_VIEWSETTINGS", uniqueConstraints = {
@javax.persistence.UniqueConstraint( name="COR_VIEWSETTINGS_UNQ1", columnNames = { "NAME", "SETTINGSTYPE_ID", "VIEW_NAME", "VIEWTYPE_ID" } ),
}
, indexes = {
@javax.persistence.Index( name="COR_VIEWSETTINGS_FK0", columnList = "VIEWTYPE_ID" ),
}
)
public class UserViewSettings extends UserSettings implements NameViewName, NameViewType
{
/** The Constant serialVersionUID. */
private static final long serialVersionUID = 1L;
/** The Constant DATA_OBJECT_NAME. */
public static final String DATA_OBJECT_NAME = "userViewSettings";
@javax.persistence.Basic( fetch = javax.persistence.FetchType.EAGER, optional = false )
@Column( name = "VIEW_NAME", nullable = false, unique = false, insertable = true, updatable = true )
private java.lang.String viewName;
@javax.persistence.ManyToOne( fetch = javax.persistence.FetchType.EAGER, optional = true )
@javax.persistence.JoinColumn( name = "VIEWTYPE_ID", nullable = true, unique = false, insertable = true, updatable = true )
private ViewType viewType;
public java.lang.String getViewName()
{
return viewName;
}
public void setViewName( java.lang.String viewName )
{
this.viewName = viewName;
}
public ViewType getViewType()
{
return viewType;
}
public void setViewType( ViewType viewType )
{
this.viewType = viewType;
}
@Override
public String getDataObjectName()
{
return DATA_OBJECT_NAME;
}
@Override
public String toString()
{
StringBuilder builder = new StringBuilder( super.toString() );
builder.append( ", " );
builder.append( "viewName" );
builder.append( "=" );
builder.append( this.viewName );
builder.append( ", " );
try
{
builder.append( ToStringUtils.referenceToString( "viewType", "ViewType", this.viewType ) );
}
catch( Exception ex )
{
builder.append( ex.getClass().getName() );
builder.append( ": " );
builder.append( ex.getMessage() );
}
builder.append( "]" );
return builder.toString();
}
}
從Hibernate 5.2和Oracle 11數據庫啟動Wildfly 10.0.0會導致錯誤,即自動生成的密鑰COR_VIEWSETTINGSCOR_USERSETTINGS_FK0和COR_VIEWSETTINGSCOR_USERSETTINGS_FK1對於數據庫而言太長了。
我看了一下Hibernate的NamingStrategies,甚至嘗試了一些,但是它們並沒有為我改變錯誤。
我如何影響這些密鑰的生成?
編輯:
所以打開調試給了我這個:
2016-11-29 09:22:03,190 DEBUG [org.hibernate.SQL] (ServerService Thread Pool -- 58) create index COR_USERSETTINGS_FK0 on COR_USERSETTINGS (SETTINGSTYPE_ID)
2016-11-29 09:22:03,190 DEBUG [org.hibernate.SQL] (ServerService Thread Pool -- 58) create index COR_USERSETTINGS_FK1 on COR_USERSETTINGS (USER_ID)
2016-11-29 09:22:03,190 DEBUG [org.hibernate.SQL] (ServerService Thread Pool -- 58) create index COR_VIEWSETTINGSCOR_USERSETTINGS_FK0 on COR_VIEWSETT INGS(SETTINGSTYPE_ID)
2016-11-29 09:22:03,190 DEBUG [org.hibernate.SQL] (ServerService Thread Pool -- 58) create index COR_VIEWSETTINGSCOR_USERSETTINGS_FK1 on COR_VIEWSETTINGS (USER_ID)
2016-11-29 09:22:03,190 DEBUG [org.hibernate.SQL] (ServerService Thread Pool -- 58) create index COR_VIEWSETTINGS_FK0 on COR_VIEWSETTINGS (VIEWTYPE_ID)
現在,我在org.hibernate.boot.model.naming包中找到了ImplicitIndexNameSource類,但是Internet並沒有給出我可以使用的示例,而且很長一段時間以來,它似乎一直是一個空類。
編輯2:
先前的編輯似乎是錯誤的路徑。 我找到了創建這些日志的地方。 它是從SchemaCreatorImpl調用的StandardIndexExporter 。 所以我需要更深入地研究框架,但是如果有人看到這一點。 這是正確的道路嗎? 我可以修改代碼以便他做我想做的事情嗎? 罪魁禍首似乎是hbm2ddl ,因為在以下幾行中在StandardIndexExport中創建了索引get:
final String indexNameForCreation;
if ( dialect.qualifyIndexName() ) {
indexNameForCreation = jdbcEnvironment.getQualifiedObjectNameFormatter().format(
new QualifiedNameImpl(
index.getTable().getQualifiedTableName().getCatalogName(),
index.getTable().getQualifiedTableName().getSchemaName(),
jdbcEnvironment.getIdentifierHelper().toIdentifier( index.getName() )
),
jdbcEnvironment.getDialect()
);
}
else {
indexNameForCreation = index.getName();
}
final StringBuilder buf = new StringBuilder()
.append( "create index " )
.append( indexNameForCreation )
.append( " on " )
.append( tableName )
.append( " (" );
boolean first = true;
Iterator<Column> columnItr = index.getColumnIterator();
while ( columnItr.hasNext() ) {
final Column column = columnItr.next();
if ( first ) {
first = false;
}
else {
buf.append( ", " );
}
buf.append( ( column.getQuotedName( dialect ) ) );
}
buf.append( ")" );
return new String[] { buf.toString() };
我將不勝感激。 這真令人沮喪
這樣我就可以了。 回答可能會發現此問題並且有相同問題的未來人們。
索引鍵由hibernate所引用的oracle方言創建。 因此,要做的是實現一個自定義OracleDialect,該自定義OracleDialect覆蓋方法getIndexExporter並指向自定義IndexExporter。 然后,您可以在此IndexExporter中修改鍵的創建方式。 就我而言,我固定了這樣的解決方案:
/**
* Gets the correct index name if it is a index for a TABLE_PER_CLASS inheritance and longer than
* 30 chars.
*
* @param index the index to decide for
* @return the correct index name
*/
private String getCorrectIndexName( Index index )
{
if ( index.getTable() instanceof DenormalizedTable && index.getName().length() > 30 )
{
String prefixedTable = index.getTable().getName();
String tableName = prefixedTable.substring( prefixedTable.indexOf( '_' ) + 1, prefixedTable.length() );
tableName = shortenName( tableName );
Iterator<Column> columnItr = index.getColumnIterator();
String reference;
if ( columnItr.hasNext() )
{
reference = extractReference( columnItr.next() );
}
else
{
/** backup strategy to prevent exceptions */
reference = shortenName( NamingHelper.INSTANCE.hashedName( index.getName() ) );
}
return tableName + "_" + reference;
}
return index.getName();
}
/**
* Extract the reference column of the index and hash the full name before shortening it with
* shortenName().
*
* @param index the index to extract the reference from.
* @return the reference with an appended _FK(hashedReference).
*/
private String extractReference( Column column )
{
String reference = column.getQuotedName( dialect );
String md5Hash = NamingHelper.INSTANCE.hashedName( reference );
md5Hash = md5Hash.substring( md5Hash.length() - 4, md5Hash.length() );
reference = shortenName( reference );
return reference + "_FK" + md5Hash;
}
/**
* Shorten the name to a maximum of 11 chars if it's longer.
*
* @param reference the reference to shorten
* @return the shortened string
*/
private static String shortenName( String reference )
{
if ( reference.length() > 11 )
{
return reference.substring( 0, 11 );
}
return reference;
}
這必須在Overriden函數getSqlCreateStrings中調用 。 更改后的行如下所示:
String indexName = getCorrectIndexName( index );
indexNameForCreation = jdbcEnvironment.getQualifiedObjectNameFormatter()
.format(
new QualifiedNameImpl( index.getTable().getQualifiedTableName().getCatalogName(),
index.getTable().getQualifiedTableName().getSchemaName(), jdbcEnvironment.getIdentifierHelper().toIdentifier( indexName ) ),
jdbcEnvironment.getDialect() );
希望對您有所幫助。
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