如何在xquery赋值中使用if else [英] how to use if else in xquery assignment
问题描述
我正在尝试使用if条件为xquery中的变量赋值。我不知道该怎么做。
I am trying to use a if condition to assign a value to a variable in an xquery. I am not sure how to do this.
这就是我的尝试:
declare namespace libx='http://libx.org/xml/libx2';
declare namespace atom='http://www.w3.org/2005/Atom';
declare variable $entry_type as xs:string external;
let $libx_node :=
if ($entry_type = 'package' or 'libapp') then
{element {fn:concat("libx:", $entry_type)} {()} }
else if ($entry_type = 'module') then
'<libx:module>
<libx:body>{$module_body}</libx:body>
</libx:module>'
此代码抛出[XPST0003] Incomplete'如果'表达错误。有人可以帮我解决这个问题吗?
This code throws an [XPST0003] Incomplete 'if' expression error. Can someone help me fix this?
此外,有人可以提供一些很好的教程来学习xqueries。
Also, can someone suggest some good tutorials to learn xqueries.
谢谢,
Sony
Thanks, Sony
推荐答案
那是因为在XQuery的条件表达式规范 else-expression 始终是必需的:
That's because in XQuery's conditional expression specification else-expression is always required:
[45] IfExpr ::= "if" "(" Expr ")" "then" ExprSingle "else" ExprSingle
所以你必须写第二个 else
子句(例如,它可能返回空序列):
So you have to write second else
clause (for example, it may return empty sequence):
declare namespace libx='http://libx.org/xml/libx2';
declare namespace atom='http://www.w3.org/2005/Atom';
declare variable $entry_type as xs:string external;
let $libx_node :=
if ($entry_type = ('package','libapp')) then
element {fn:concat("libx:", $entry_type)} {()}
else if ($entry_type = 'module') then
<libx:module>
<libx:body>{$module_body}</libx:body>
</libx:module>
else ()
... (your code here) ...
一些明显的错误也得到修复:
Some obvious bugs are also fixed:
- 不需要{}围绕计算元素构造函数;
- 很可能,你想要
if($ entry_type =('package','libapp'))
- don't need {} around computed element constructor;
- most likely, you want
if($entry_type = ('package', 'libapp'))
关于XQuery教程。 W3CSchools的XQuery教程是一个非常好的起点。
Concerning XQuery tutorials. The W3CSchools's XQuery Tutorial is a pretty good starting point.
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