xml 理解 xsd:choice 和 minOccurs
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/5214555/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Understanding xsd:choice and minOccurs
提问by makes
I am having trouble understanding the behavior of the following XML schema:
我无法理解以下 XML 模式的行为：
<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<xsd:element name="rootnode">
<xsd:complexType>
<xsd:sequence>
<xsd:choice minOccurs="1" maxOccurs="2">
<xsd:element name="e1" minOccurs="1" maxOccurs="2"/>
<xsd:element name="e2" minOccurs="0" maxOccurs="1"/>
</xsd:choice>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:schema>
I expected at least one instance of either element <e1>or <e2>be required as a child of <rootnode>. Despite my expectations, an empty <rootnode>will validate against this schema:
我期望至少有一个元素的实例<e1>或被<e2>要求作为<rootnode>. 尽管有我的期望，空<rootnode>将根据此模式进行验证：
> xmllint --schema test.xsd empty.xml
<?xml version="1.0" encoding="UTF-8"?>
<rootnode>
</rootnode>
empty.xml validates
If I change the minOccursattribute of element e2to something other than "0", I get the behavior I originally expected.
如果我将minOccurselement的属性更改为e2以外的其他内容"0"，则会得到我最初预期的行为。
It seems as though the mere absenceof element
<e2>counts as an occurrence of thexsd:choicein my example.If this is the case, then how come this infinite number of occurrences does not violate the
maxOccurslimit in myxsd:choice?
在我的示例中，似乎仅缺少元素就被
<e2>视为出现了xsd:choice。如果是这种情况，那么这种无限次出现怎么不违反
maxOccursmy 中的限制xsd:choice？
回答by Michael Kay
I tell you you can go to the shops at least once and at most twice, and each time you have a choice of what to buy: you can buy apples (either one apple or two apples), or you can buy oranges (either no oranges or one orange).
我告诉你，你可以去商店至少一次，最多两次，每次你都可以选择买什么：你可以买苹果（一个苹果或两个苹果），或者你可以买橙子（要么不买，要么不买）。橙子或一个橙子）。
It's entirely possible that you will choose to go to the shops twice and on each occasion to buy no oranges. So you come back with nothing.
你完全有可能选择去商店两次，每次都不买橙子。所以你什么都没有回来。
回答by David W
Here are the allowable combinations
以下是允许的组合
Two choices:
e1 (1 - 2) + e1 (1 - 2) = e1 x (2 - 4), or
e1 (1 - 2) + e2 (0 - 1), or
e2 (0 - 1) + e1 (1 - 2), or
e2 (0 - 1) + e2 (0 - 1) = e2 (0 - 2)
One choice (but no new outcomes):
e1 (1-2), or
e2 (0-1)
e1e1, e1e1e1, e1e1e1e1
e1, e1e2, e1e1e2
e2e1, e2e1e1
empty, e2, e2e2
Note that choice[min=2 max=2] would have produced the same set of valid combinations.
请注意， choice[min=2 max=2] 会产生相同的一组有效组合。
