DTD与XSD定义的XML语言的范围-ScopeofXMLlanguagesdefinedbyDTDvsXSD

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An interesting question; well asked!

一个有趣的问题;好问!

The answer is "no", in both directions.

两个方向的答案都是“不”。

Here is a DTD which has no equivalent in XSD:

这是一个在XSD中没有等效的DTD:

The set of character sequences accepted by this DTD includes both and &egbdf;, but not &beadgcf;.

该DTD接受的字符序列集包括 和 &egbdf; ,但不包括 &beadgcf; 。

Since XSD validation operates on an information set in which entities have all already been expanded, no XSD schema can distinguish the third case from the second.

由于XSD验证在一个实体已经扩展的信息集上运行,因此没有XSD架构可以区分第三种情况和第二种情况。

A second area where DTDs can express constraints not expressible in XSD involves NOTATION types. I won't give an example; the details are too complicated for me to remember them correctly without looking them up, and not interesting enough to make me want to do so.

DTD可以表达在XSD中不可表达的约束的第二个区域涉及NOTATION类型。我不会举一个例子;细节太复杂了,我不能正确地记住它们,而且没有足够的兴趣让我想要这么做。

A third area: DTDs treat namespace attributes (aka namespace declarations) and general attributes in the same way; a DTD can therefore constrain the appearance of namespace declarations in documents. An XSD schema cannot. The same applies to attributes in the xsi namespace.

第三个方面:DTD以相同的方式处理命名空间属性(也称为命名空间声明)和一般属性;因此,DTD可以约束文档中名称空间声明的外观。 XSD架构不能。这同样适用于xsi名称空间中的属性。

If we ignore all of those issues, and formulate the question with respect only to character sequences containing no references to named entities other than the pre-defined entities lt, gt, etc., then the answer changes: for every DTD not involving NOTATION declarations, there is an XSD schema that accepts precisely the same set of documents after entity expansion and with 'same' defined in a way that ignores namespace attributes and attributes in the xsi namespace.

如果我们忽略所有这些问题,并且只针对不包含对预定义实体lt,gt等以外的命名实体的引用的字符序列来表达问题,则答案会发生变化:对于每个不涉及NOTATION声明的DTD ,有一个XSD架构在实体扩展后接受完全相同的文档集,并且以忽略xsi命名空间中的命名空间属性和属性的方式定义“相同”。

In the other direction, the areas of difference include these:

在另一个方向,差异的领域包括:

• XSD is namespace aware: the following XSD schema accepts any instance of element e in the specified target namespace, regardless of what prefix is bound to that namespace in the document instance.

  XSD是名称空间感知:以下XSD架构接受指定目标名称空间中元素e的任何实例,而不管文档实例中哪个前缀绑定到该名称空间。

  No DTD can successfully accept all and only the e elements in the given namespace.

  没有DTD可以成功接受给定命名空间中的所有和仅e元素。

• XSD has a richer set of datatypes and can use datatypes to constrain elements as well as attributes. The following XSD schema has no equivalent DTD:

  XSD具有更丰富的数据类型集,可以使用数据类型来约束元素和属性。以下XSD架构没有等效的DTD:

  This schema accepts the document 42 but not the document 42d Street. No DTD can make that distinction, because DTDs have no mechanism for constraining #PCDATA content. The closest DTD would be , which accepts both sample documents.

  此架构接受文档 42 ,但不接受文档 42d Street 。没有DTD可以做出这种区分,因为DTD没有约束#PCDATA内容的机制。最接近的DTD是 ,它接受两个样本文件。

• XSD's xsi:type attribute allows in-document modifications of content models. The XSD schema described by the following schema document has no equivalent DTD:

  XSD的xsi:type属性允许对内容模型进行文档内修改。以下架构文档描述的XSD架构没有等效的DTD:

  This schema accepts the document and rejects the document . DTDs have no mechanism for making content models depend on an attribute value given in the document instance.

  此架构接受文档 并拒绝文档 。 DTD没有使内容模型依赖于文档实例中给出的属性值的机制。

• XSD wildcards allow the inclusion of arbitrary well-formed XML among the children of specified elements; the closest one can come to that with a DTD is to use an element declaration of the form , which is not the same because it requires declarations for all the elements which in fact appear.

  XSD通配符允许在指定元素的子元素中包含任意格式良好的XML;使用DTD最接近的是使用 形式的元素声明,这是不一样的,因为它需要声明实际出现的所有元素。

• XSD 1.1 provides assertions and conditional type assignment, which have no analogues in DTDs.

  XSD 1.1提供断言和条件类型赋值,它们在DTD中没有类似物。

There are probably other ways in which the expressive power of XSD exceeds that of DTDs, but I think the point has been illustrated adequately.

XSD的表现力可能超过DTD的其他方式,但我认为这一点已得到充分说明。

I think a fair summary would be: XSD can express everything DTDs can express, with the exception of entity declarations and special cases like namespace declarations and xsi:* attributes, because XSD was designed to be able to do so. So the loss of information when translating a DTD to an XSD schema document is relatively modest, well understood, and mostly involves things most vocabulary designers regard as DTD artefacts not of substantive interest.

我认为一个公平的总结是:XSD可以表达DTD可以表达的所有内容,但实体声明和名称空间声明和xsi:*属性等特殊情况除外,因为XSD旨在能够这样做。因此,在将DTD翻译成XSD模式文档时丢失信息是相对适度的,易于理解的,并且主要涉及大多数词汇设计者认为不具有实质意义的DTD文物的事物。

XSD can express more than DTDs can, again because XSD was designed to do so. In the general case, translation from XSD to DTD necessarily involves loss of information (the set of documents accepted may need to be larger, or smaller, or to be an overlapping set). Different choices can be made about how to manage the loss of information, which gives the question "How does one best translate an XSD into DTD form?" a certain theoretical interest. (Very few people, however, seem to find it an interesting question in practice.)

XSD可以表达超过DTD的能力,因为XSD的设计也是如此。在一般情况下,从XSD到DTD的转换必然涉及信息丢失(接受的文档集可能需要更大或更小,或者是重叠集)。可以对如何管理信息丢失做出不同的选择,这就提出了“如何最好地将XSD转换为DTD形式?”的问题。一定的理论兴趣。 (然而,很少有人在实践中发现这是一个有趣的问题。)

All of this focuses, as did your question, on documents as character sequences, on languages as document sets, and on schema languages as generators of languages in that sense. Issues of maintainability and information present in the schema that does not turn into differences in the extension of document sets (e.g. the treatment of class hierarchies in the document model) is left out of account.

所有这一切都集中在您的问题上,作为字符序列的文档,作为文档集的语言,以及作为这种意义上的语言生成器的模式语言。在模式中存在的可维护性和信息的问题不会变成文档集的扩展中的差异(例如,文档模型中的类层次结构的处理)。

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Without qualifiers, the answer is no.

没有限定词,答案是否定的。

You have to define what is it you call a "language". In my mind, these you refer to are languages meant to define document schemata. A schemata defines constraints on the document structure and content. The constraints expressible by XSD are far more powerful than DTD. So no, they wouldn't be the same.

你必须定义你称之为“语言”的东西。在我看来,你所指的是用于定义文档模式的语言。模式定义了对文档结构和内容的约束。 XSD可以表达的约束比DTD强大得多。所以不,他们不会是一样的。

A comparison of DTD vs. XSD might help you understand why not.

DTD与XSD的比较可能有助于您理解为什么不这样做。