Xquery返回多个值,而不是一个值
考虑xml文件:Xquery返回多个值,而不是一个值,xquery,Xquery,考虑xml文件: <warehouse> - may be some tag on upper level of dep present, may be not <dep no ="2"> <item> <itemName> Item 1 </itemName> <supplier> Supp1 </supplier> <supplier> S
<warehouse>
- may be some tag on upper level of dep present, may be not
<dep no ="2">
<item>
<itemName> Item 1 </itemName>
<supplier> Supp1 </supplier>
<supplier> Supp2 </supplier>
</item>
</dep>
<dep no ="1">
<item>
<itemName> Item 2 </itemName>
<supplier> Supp1 </supplier>
<supplier> Supp3 </supplier>
</item>
</dep>
.......
</warehouse>
这是我的代码:
<ul>
{
for $sup in distinct-values(doc("warehouse.xml")//supplier)
for $item in doc("warehouse.xml")//item[supplier = $sup]
return <li>{data($sup)} <ul> {
let $n:= $item/itemName
let $y:= doc("warehouse.xml")//dep[//itemName=$n]/@no
for $n1 in $n return <li>{data($n1), data($y)} </li>} </ul> </li>
}
</ul>
我明白了
你能告诉我可能出了什么问题吗?非常感谢。作为XQuery的新手,我可以写以下内容 输入:
<warehouse>
<something/>
<dep no ="2">
<item>
<itemName>Item1</itemName>
<supplier>Supp1</supplier>
<supplier>Supp2</supplier>
</item>
</dep>
<anything/>
<dep no ="1">
<item>
<itemName>Item2</itemName>
<supplier>Supp1</supplier>
<supplier>Supp3</supplier>
</item>
</dep>
<whatever/>
</warehouse>
$doc//item[supplier/text() = $supplier]/itemName/text()
这意味着您获取输入中供应商名称等于给定供应商名称的每个项目,然后获得这些(过滤)项目的名称就这样。希望我有点道理 非常感谢你的答复。我用的是氧气,这里似乎没有定义“组别”。()有什么建议吗?@PiotrL。哦,是的,这在XQuery 3.0中是一个很好的额外定义。您的XQuery处理器似乎还不支持它。我已经更新了我的答案,所以它不使用此功能。
Supp1
Item1, 1,2, ...
Supp1
Item2, 1,2, ...
<warehouse>
<something/>
<dep no ="2">
<item>
<itemName>Item1</itemName>
<supplier>Supp1</supplier>
<supplier>Supp2</supplier>
</item>
</dep>
<anything/>
<dep no ="1">
<item>
<itemName>Item2</itemName>
<supplier>Supp1</supplier>
<supplier>Supp3</supplier>
</item>
</dep>
<whatever/>
</warehouse>
<ul>{
(: the ../text() can be dropped from everywhere :)
let $doc := doc("input.xml")
for $supplier in distinct-values($doc//supplier/text())
let $items := $doc//item[supplier/text() = $supplier]/itemName/text()
return <li>{$supplier}: {string-join($items, ", ")}</li>
}</ul>
<ul>
<li>Supp1: Item1, Item2</li>
<li>Supp2: Item1</li>
<li>Supp3: Item2</li>
</ul>
distinct-values($doc//supplier/text())
$doc//item[supplier/text() = $supplier]/itemName/text()