xml XSD all 和 XSD 序列之间的中间方式
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Middle way between XSD all and XSD sequence
提问by Bart van Heukelom
I'm defining a user element with XSD. For this example, a user has a name, email and one or more nationalities. I've tried:
我正在用 XSD 定义一个用户元素。对于此示例，用户具有姓名、电子邮件和一个或多个国籍。我试过了：
<xs:all>
<xs:element name="name" blabla />
<xs:element name="email" blabla />
<xs:element name="nationality" minOccurs="1" maxOccurs="unbounded" />
</xs:all>
However, that is illegal. Apparently elements inside an "All" can only occur one time (or not at all). I could fix this by changing the All to a Sequence, but then people would have to enter the properties in the exact order, which I actually don't care about.
然而，这是非法的。显然，“全部”中的元素只能出现一次（或根本不出现）。我可以通过将 All 更改为 Sequence 来解决这个问题，但是人们必须按照确切的顺序输入属性，而我实际上并不关心这些。
Is there a combination of these two available? Not according to http://www.w3schools.com/Schema/schema_complex_indicators.asp, but maybe it's hidden (or my inexperienced eyes don't see it).
有没有这两者的组合？不是根据http://www.w3schools.com/Schema/schema_complex_indicators.asp，但也许它是隐藏的（或者我没有经验的眼睛看不到它）。
By intuition, I also tried:
凭直觉，我也试过：
<xs:all>
<xs:element name="name" blabla />
<xs:element name="email" blabla />
<xs:sequence>
<xs:element name="nationality" minOccurs="1" maxOccurs="unbounded" />
</xs:sequence>
</xs:all>
But that's unfortunately invalid.
但不幸的是，这是无效的。
Here is the current, real, piece of XSD:
这是当前真实的 XSD 片段：
<!-- user -->
<xs:complexType name="user">
<xs:sequence>
<xs:element name="firstname" type="xs:string" minOccurs="1" maxOccurs="1" />
<xs:element name="appendix" type="xs:string" minOccurs="0" maxOccurs="1" />
<xs:element name="lastname" type="xs:string" minOccurs="1" maxOccurs="1" />
<xs:element name="address" type="xs:string" minOccurs="1" maxOccurs="1" />
<xs:element name="zipcode" type="xs:string" minOccurs="1" maxOccurs="1" />
<xs:element name="city" type="xs:string" minOccurs="1" maxOccurs="1"/>
<xs:element name="username" type="xs:string" minOccurs="1" maxOccurs="1"/>
<xs:element name="email" type="xs:string" minOccurs="1" maxOccurs="1"/>
<xs:element name="country" type="country" minOccurs="1" maxOccurs="1"/>
<xs:element name="nationality" type="xs:string" minOccurs="1" maxOccurs="unbounded"/>
</xs:sequence>
</xs:complexType>
采纳答案by jelovirt
Your code should be valid in XSD 1.1. For XSD 1.0 you have to find a workaround.
您的代码应该在 XSD 1.1 中有效。对于 XSD 1.0，您必须找到一种解决方法。
回答by marc_s
Could you just turn your "nationality" thingie into its own complexType and then use that new complex type inside your xs:all?
你能不能把你的“国籍”东西变成它自己的 complexType，然后在你的 xs:all 中使用那个新的复杂类型？
<xs:complexType name="NationalityType">
<xs:sequence>
<xs:element name="nationality" minOccurs="1" maxOccurs="unbounded" />
</xs:sequence>
</xs:complexType>
<xs:all>
<xs:element name="name" blabla />
<xs:element name="email" blabla />
<xs:element name="nationalities" type="NationalityType" />
</xs:all>
I don't have anything at hand to test this, so this is really just off the top of my head..... give it a try!
我手头没有任何东西可以测试这个，所以这真的只是我的头顶......试一试！
EDIT: tested it by now - it works, the only minor price to pay is that your XML will have to look something like this:
编辑：现在已经测试过了 - 它可以工作，唯一要付出的代价是你的 XML 必须看起来像这样：
<....>
<email>......</email>
<nationalities>
<nationality>ABC</nationality>
<nationality>CDE</nationality>
</nationalities>
<name>.....</name>
</.....>
So you get an extra node that will contain the arbitrary long list of <nationality>items.
所以你会得到一个额外的节点，它将包含任意长的<nationality>项目列表。
Marc
马克
回答by fangster
Just come across the a similar problem (I wanted to have any number of each element in any order) and solved it with a sequence of choices. Using the example above:
刚遇到一个类似的问题（我想按任何顺序排列任意数量的每个元素）并通过一系列选择解决它。使用上面的例子：
<?xml version='1.0' encoding='UTF-8'?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name='user'>
<xs:complexType>
<xs:sequence minOccurs="0" maxOccurs="unbounded">
<xs:choice>
<xs:element name="name" type="xs:string" />
<xs:element name="email" type="xs:string" />
<xs:element name="nationality" />
</xs:choice>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>
This allows you to have any number of name, email and nationality in any order.
这允许您以任何顺序拥有任意数量的姓名、电子邮件和国籍。
回答by Jeremy Stein
I think what you're looking for would go against the intent of XML. It would seems strange to have a valid XML fragment like this:
我认为您正在寻找的内容与 XML 的意图背道而驰。有这样一个有效的 XML 片段似乎很奇怪：
<user>
<nationality/>
<name/>
<nationality/>
<email/>
<nationality/>
</user>
It sounds like you're asking for something like what marc_s proposed:
听起来您要求的内容类似于 marc_s 提出的内容：
<user>
<name/>
<email/>
<nationality/>
<nationality/>
<nationality/>
<user>
which needs to get pushed into:
需要推入：
<user>
<name/>
<email/>
<nationalities>
<nationality/>
<nationality/>
<nationality/>
</nationalities>
<user>
回答by ugo
For XSD 1.0 the suggestion from leppie works.
对于 XSD 1.0，leppie 的建议有效。
The XSD
XSD
<?xml version='1.0' encoding='UTF-8'?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name='user'>
<xs:complexType>
<xs:sequence>
<xs:element name="name" type="xs:string" />
<xs:element name="email" type="xs:string" />
<xs:choice minOccurs='0' maxOccurs='unbounded'>
<xs:element name="nationality" />
</xs:choice>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>
A sample XML document that validates against the schema
根据模式进行验证的示例 XML 文档
<user>
<name>Name</name>
<email>[email protected]</email>
<nationality>Italian</nationality>
<nationality>Japanese</nationality>
<nationality>Alien</nationality>
</user>
And validation e.g. using xmllint
和验证，例如使用xmllint
xmllint --noout --schema test.xsd test.xml
test.xml validate
回答by AllenG
Or, since "USER" will be set up with multiple child elements, why not set it up as a complex type? Something like this should work.
或者，既然“USER”将设置多个子元素，为什么不将其设置为复杂类型？像这样的事情应该有效。
<xs:complexType>
<xs:sequence>
<xs:element name="Name" type="xs:string" />
<xs:element name="Password" type="xs:string" />
<xs:element minOccurs="1" maxOccurs="unbounded" name="Nationality" type="xs:string" />
</xs:sequence>
</xs:complexType>
回答by leppie
xs:choicewont work? If not, just wrap that in a sequence or vice versa.
xs:choice行不通？如果没有，只需将其按顺序包装，反之亦然。
