inheritance_mapping.po

hibernate 开源框架的代码 jar包希望大家能喜欢

#, fuzzy
msgid ""
msgstr ""
"PO-Revision-Date: YEAR-MO-DA HO:MI+ZONE\n"
"Last-Translator: FULL NAME <EMAIL@ADDRESS>\n"
"Content-Type: text/plain; charset=utf-8\n"

#: index.docbook:5
msgid "Inheritance Mapping"
msgstr "Mapeo de Herencia"

#: index.docbook:8
msgid "The Three Strategies"
msgstr "Las Tres Estrategias"

#: index.docbook:10
msgid "Hibernate supports the three basic inheritance mapping strategies:"
msgstr ""
"Hibernate soporta las tres estrategias b&#x00e1;sicas de mapeo de herencia:"

#: index.docbook:16
msgid "table per class hierarchy"
msgstr "<para>tabla por jerarqu&#x00ed;a de clases</para>"

#: index.docbook:21
msgid "<para>table per subclass</para>"
msgstr "<para>tabla por subclase</para>"

#: index.docbook:26
msgid "table per concrete class"
msgstr "tabla por clase concreta"

#: index.docbook:32
msgid ""
"In addition, Hibernate supports a fourth, slightly different kind of "
"polymorphism:"
msgstr ""
"En adici&#x00f3;n, Hibernate soporta un cuarto, ligeramente diferente tipo "
"de polimorfismo:"

#: index.docbook:39
msgid "implicit polymorphism"
msgstr "polimorfismo impl&#x00ed;cito"

#: index.docbook:45
msgid ""
"It is possible to use different mapping strategies for different branches of "
"the same inheritance hierarchy, and then make use of implicit polymorphism "
"to achieve polymorphism across the whole hierarchy. However, Hibernate does "
"not support mixing <literal>&lt;subclass&gt;</literal>, and <literal>&lt;"
"joined-subclass&gt;</literal> and <literal>&lt;union-subclass&gt;</literal> "
"mappings under the same root <literal>&lt;class&gt;</literal> element. It is "
"possible to mix together the table per hierarchy and table per subclass "
"strategies, under the the same <literal>&lt;class&gt;</literal> element, by "
"combining the <literal>&lt;subclass&gt;</literal> and <literal>&lt;join&gt;</"
"literal> elements (see below)."
msgstr ""
"UNTRANSLATED! It is possible to define <literal>subclass</literal>, "
"<literal>union-subclass</literal>, and <literal>joined-subclass</literal> "
"mappings in separate mapping documents, directly beneath <literal>hibernate-"
"mapping</literal>. This allows you to extend a class hierachy just by adding "
"a new mapping file. You must specify an <literal>extends</literal> attribute "
"in the subclass mapping, naming a previously mapped superclass. Note: "
"Previously this feature made the ordering of the mapping documents "
"important. Since Hibernate3, the ordering of mapping files does not matter "
"when using the extends keyword. The ordering inside a single mapping file "
"still needs to be defined as superclasses before subclasses."

#: index.docbook:59
msgid ""
"It is possible to define <literal>subclass</literal>, <literal>union-"
"subclass</literal>, and <literal>joined-subclass</literal> mappings in "
"separate mapping documents, directly beneath <literal>hibernate-mapping</"
"literal>. This allows you to extend a class hierachy just by adding a new "
"mapping file. You must specify an <literal>extends</literal> attribute in "
"the subclass mapping, naming a previously mapped superclass. Note: "
"Previously this feature made the ordering of the mapping documents "
"important. Since Hibernate3, the ordering of mapping files does not matter "
"when using the extends keyword. The ordering inside a single mapping file "
"still needs to be defined as superclasses before subclasses."
msgstr ""
"<![CDATA[\n"
"                                 <hibernate-mapping>\n"
"                                         <subclass name=\"DomesticCat\" "
"extends=\"Cat\" discriminator-value=\"D\">\n"
"                                                 <property name=\"name\" "
"type=\"string\"/>\n"
"                                         </subclass>\n"
" </hibernate-mapping>]]>"

#: index.docbook:70
msgid ""
"<![CDATA[\n"
" <hibernate-mapping>\n"
"     <subclass name=\"DomesticCat\" extends=\"Cat\" discriminator-value=\"D"
"\">\n"
"          <property name=\"name\" type=\"string\"/>\n"
"     </subclass>\n"
" </hibernate-mapping>]]>"
msgstr ""
"Es posible usar estrategias de mapeo diferentes para diferentes "
"ramificaciones de la misma jerarqu&#x00ed;a de herencia, y entonces usar "
"polimorfismo impl&#x00ed;cito para conseguir polimorfismo a trav&#x00e9;s de "
"toda la jerarqu&#x00ed;a. Sin embargo, Hibernate no soporta la mezcla de "
"mapeos <literal>&lt;subclass&gt;</literal>, y <literal>&lt;joined-"
"subclass&gt;</literal> y <literal>&lt;union-subclass&gt;</literal> bajo el "
"mismo elemento <literal>&lt;class&gt;</literal> ra&#x00ed;z. Es posible "
"mezclar juntas las estrategias de tabla por jerarqu&#x00ed;a y tabla por "
"subclase, bajo el mismo elemento <literal>&lt;class&gt;</literal>, "
"combinando los elementos <literal>&lt;subclass&gt;</literal> y <literal>&lt;"
"join&gt;</literal> (ver debajo)."

#: index.docbook:74
msgid "Table per class hierarchy"
msgstr "Tabla por jerarqu&#x00ed;a de clases"

#: index.docbook:76
msgid ""
"Suppose we have an interface <literal>Payment</literal>, with implementors "
"<literal>CreditCardPayment</literal>, <literal>CashPayment</literal>, "
"<literal>ChequePayment</literal>. The table per hierarchy mapping would look "
"like:"
msgstr ""
"Sup&#x00f3;n que tenemos una interface <literal>Payment</literal>, con los "
"implementadores <literal>CreditCardPayment</literal>, <literal>CashPayment</"
"literal>, <literal>ChequePayment</literal>. El mapeo de tabla por "
"jerarqu&#x00ed;a se ver&#x00ed;a as&#x00ed;:"

#: index.docbook:83
msgid ""
"<![CDATA[<class name=\"Payment\" table=\"PAYMENT\">\n"
"    <id name=\"id\" type=\"long\" column=\"PAYMENT_ID\">\n"
"        <generator class=\"native\"/>\n"
"    </id>\n"
"    <discriminator column=\"PAYMENT_TYPE\" type=\"string\"/>\n"
"    <property name=\"amount\" column=\"AMOUNT\"/>\n"
"    ...\n"
"    <subclass name=\"CreditCardPayment\" discriminator-value=\"CREDIT\">\n"
"        <property name=\"creditCardType\" column=\"CCTYPE\"/>\n"
"        ...\n"
"    </subclass>\n"
"    <subclass name=\"CashPayment\" discriminator-value=\"CASH\">\n"
"        ...\n"
"    </subclass>\n"
"    <subclass name=\"ChequePayment\" discriminator-value=\"CHEQUE\">\n"
"        ...\n"
"    </subclass>\n"
"</class>]]>"
msgstr ""
"<![CDATA[<class name=\"Payment\" table=\"PAYMENT\">\n"
"    <id name=\"id\" type=\"long\" column=\"PAYMENT_ID\">\n"
"        <generator class=\"native\"/>\n"
"    </id>\n"
"    <discriminator column=\"PAYMENT_TYPE\" type=\"string\"/>\n"
"    <property name=\"amount\" column=\"AMOUNT\"/>\n"
"    ...\n"
"    <subclass name=\"CreditCardPayment\" discriminator-value=\"CREDIT\">\n"
"        <property name=\"creditCardType\" column=\"CCTYPE\"/>\n"
"        ...\n"
"    </subclass>\n"
"    <subclass name=\"CashPayment\" discriminator-value=\"CASH\">\n"
"        ...\n"
"    </subclass>\n"
"    <subclass name=\"ChequePayment\" discriminator-value=\"CHEQUE\">\n"
"        ...\n"
"    </subclass>\n"
"</class>]]>"

#: index.docbook:85
msgid ""
"Exactly one table is required. There is one big limitation of this mapping "
"strategy: columns declared by the subclasses, such as <literal>CCTYPE</"
"literal>, may not have <literal>NOT NULL</literal> constraints."
msgstr ""
"Se requiere exactamente una tabla. Hay una gran limitaci&#x00f3;n de esta "
"estrategia de mapeo: las columnas declaradas por las subclases, como "
"<literal>CCTYPE</literal>, no pueden tener restricciones <literal>NOT NULL</"
"literal>."

#: index.docbook:94
msgid "Table per subclass"
msgstr "Tabla por subclase"

#: index.docbook:96
msgid "A table per subclass mapping would look like:"
msgstr "Un mapeo de tabla por sublclase se ver&#x00ed;a as&#x00ed;:"

#: index.docbook:100
msgid ""
"<![CDATA[<class name=\"Payment\" table=\"PAYMENT\">\n"
"    <id name=\"id\" type=\"long\" column=\"PAYMENT_ID\">\n"
"        <generator class=\"native\"/>\n"
"    </id>\n"
"    <property name=\"amount\" column=\"AMOUNT\"/>\n"
"    ...\n"
"    <joined-subclass name=\"CreditCardPayment\" table=\"CREDIT_PAYMENT\">\n"
"        <key column=\"PAYMENT_ID\"/>\n"
"        <property name=\"creditCardType\" column=\"CCTYPE\"/>\n"
"        ...\n"
"    </joined-subclass>\n"
"    <joined-subclass name=\"CashPayment\" table=\"CASH_PAYMENT\">\n"
"        <key column=\"PAYMENT_ID\"/>\n"
"        ...\n"
"    </joined-subclass>\n"
"    <joined-subclass name=\"ChequePayment\" table=\"CHEQUE_PAYMENT\">\n"
"        <key column=\"PAYMENT_ID\"/>\n"
"        ...\n"
"    </joined-subclass>\n"
"</class>]]>"
msgstr ""
"<![CDATA[<class name=\"Payment\" table=\"PAYMENT\">\n"
"    <id name=\"id\" type=\"long\" column=\"PAYMENT_ID\">\n"
"        <generator class=\"native\"/>\n"
"    </id>\n"
"    <property name=\"amount\" column=\"AMOUNT\"/>\n"
"    ...\n"
"    <joined-subclass name=\"CreditCardPayment\" table=\"CREDIT_PAYMENT\">\n"
"        <key column=\"PAYMENT_ID\"/>\n"
"        <property name=\"creditCardType\" column=\"CCTYPE\"/>\n"
"        ...\n"
"    </joined-subclass>\n"
"    <joined-subclass name=\"CashPayment\" table=\"CASH_PAYMENT\">\n"
"        <key column=\"PAYMENT_ID\"/>\n"
"        ...\n"
"    </joined-subclass>\n"
"    <joined-subclass name=\"ChequePayment\" table=\"CHEQUE_PAYMENT\">\n"
"        <key column=\"PAYMENT_ID\"/>\n"
"        ...\n"
"    </joined-subclass>\n"
"</class>]]>"

#: index.docbook:102
msgid ""
"Four tables are required. The three subclass tables have primary key "
"associations to the superclass table (so the relational model is actually a "
"one-to-one association)."
msgstr ""
"Se requieren cuatro tablas. Las tres tablas de subclase tienen asociaciones "
"de clave primaria a la tabla de superclase (de modo que en el modelo "
"relacional es realmente una asociaci&#x00f3;n uno-a-uno)."

#: index.docbook:111
msgid "Table per subclass, using a discriminator"
msgstr "Tabla por subclase, usando un discriminador"

#: index.docbook:113
msgid ""
"Note that Hibernate's implementation of table per subclass requires no "
"discriminator column. Other object/relational mappers use a different "
"implementation of table per subclass which requires a type discriminator "
"column in the superclass table. The approach taken by Hibernate is much more "
"difficult to implement but arguably more correct from a relational point of "
"view. If you would like to use a discriminator column with the table per "
"subclass strategy, you may combine the use of <literal>&lt;subclass&gt;</"
"literal> and <literal>&lt;join&gt;</literal>, as follow:"
msgstr ""
"Observa que la implementaci&#x00f3;n de Hibernate de tabla por subclase no "
"requiere ninguna columna discriminadora. Otros mapeadores objeto/relacional "
"usan una implementaci&#x00f3;n diferente de tabla por subclase que requiere "
"una columna discriminadora de tipo en la tabla de superclase. Este enfoque "
"es mucho m&#x00e1;s dif&#x00ed;cil de implementar pero discutiblemente "
"m&#x00e1;s correcto desde un punto de vista relacional. Si quisieras usar "
"una columna discriminadora con la estrategia de tabla por subclase, puedes "
"combinar el uso de <literal>&lt;subclass&gt;</literal> y <literal>&lt;"
"join&gt;</literal>, como sigue:"

#: index.docbook:125
msgid ""
"<![CDATA[<class name=\"Payment\" table=\"PAYMENT\">\n"
"    <id name=\"id\" type=\"long\" column=\"PAYMENT_ID\">\n"
"        <generator class=\"native\"/>\n"
"    </id>\n"
"    <discriminator column=\"PAYMENT_TYPE\" type=\"string\"/>\n"
"    <property name=\"amount\" column=\"AMOUNT\"/>\n"
"    ...\n"
"    <subclass name=\"CreditCardPayment\" discriminator-value=\"CREDIT\">\n"
"        <join table=\"CREDIT_PAYMENT\">\n"
"            <key column=\"PAYMENT_ID\"/>\n"
"            <property name=\"creditCardType\" column=\"CCTYPE\"/>\n"
"            ...\n"
"        </join>\n"
"    </subclass>\n"