问题描述
我对FLWOR循环有一个非常奇怪的问题,该问题以一种方式起作用,但没有另一种方式.目标是采用任意长度的字符串,并将其分解为每个只能容纳80个字符的XML节点.因此,第一次重击,效果很好:
I have a very strange problem with a FLWOR loop, that works one way but not another. The goal is to take a string of any length, and break it into XML nodes that can only hold 80 chars each. So, first whack, this works great:
for $noteLine in $noteLineArr
where $noteLine != ''
return
if (fn:string-length(fn:normalize-space($noteLine)) < 80) then (
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 1, 80)}</Descr>
</NTE>
) else if (fn:string-length(fn:normalize-space($noteLine)) > 80 and fn:string-length(fn:normalize-space($noteLine)) <= 160) then (
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 1, 80)}</Descr>
</NTE>,
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 81, 80)}</Descr>
</NTE>
) else if (fn:string-length(fn:normalize-space($noteLine)) > 160 and fn:string-length(fn:normalize-space($noteLine)) <= 240) then (
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 1, 80)}</Descr>
</NTE>,
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 81, 80)}</Descr>
</NTE>,
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 161, 80)}</Descr>
</NTE>
) else()
所以,我觉得这不是很优雅.我尝试通过将第一个元素移出if来清理它,因为它应该总是被使用,对吗?这样更少的代码?所以这是我尝试过的:
So, I get it in my head that this isn't very elegant. I try to clean it up by moving the first element out of the if, since it should always get used, right? Less code that way? So here is what I tried:
for $noteLine in $noteLineArr
where $noteLine != ''
return
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 1, 80)}</Descr>
</NTE>,
if (fn:string-length(fn:normalize-space($noteLine)) > 80 and fn:string-length(fn:normalize-space($noteLine)) <= 160) then (
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 81, 80)}</Descr>
</NTE>
) else if (fn:string-length(fn:normalize-space($noteLine)) > 160 and fn:string-length(fn:normalize-space($noteLine)) <= 240) then (
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 81, 80)}</Descr>
</NTE>,
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 161, 80)}</Descr>
</NTE>
) else()
解析器现在告诉我"noteLine的未定义变量"指向第一行"if"行.我在这里想念什么? (请注意,是的,我的确有更多方法来解决此问题,但这是第一步,而当它不幸失败时,我感到非常恐慌).
And the parser is now telling me "undefined variable at noteLine" pointing to the first "if" line. What am I missing here? (note that yes, I do have other ideas of how to clean this up even more, but this was the first simple step and when it failed miserably, I panicked).
解决方案(感谢Jens Erat)返回块需要用括号括起来,以强制其在flwor循环的每次迭代中进行完全评估.因此:
Solution (thanks to Jens Erat)The return block needs to be wrapped in parentheses in order to force it to evaluate fully on each iteration of the flwor loop. Thusly:
for $noteLine in $noteLineArr
where $noteLine != ''
return
(
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 1, 80)}</Descr>
</NTE>,
if (fn:string-length(fn:normalize-space($noteLine)) > 80 and fn:string-length(fn:normalize-space($noteLine)) <= 160) then (
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 81, 80)}</Descr>
</NTE>
) else if (fn:string-length(fn:normalize-space($noteLine)) > 160 and fn:string-length(fn:normalize-space($noteLine)) <= 240) then (
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 81, 80)}</Descr>
</NTE>,
<NTE>
<NoteRefCd>WHI</NoteRefCd>
<Descr>{fn:substring(fn:normalize-space($noteLine), 161, 80)}</Descr>
</NTE>
) else()
)
推荐答案
这是一个较小的语法问题.您必须在return
子句中创建的元素周围加上括号.为了进一步解释该问题,我提供了一个类似但简化的示例:
This is a minor syntax issue. You have to put parenthesis around the elements created in the return
clause. To further explain the issue, I'm providing a similar, but simplified example:
for $i in 1 to 3
return
<foo />,
<bar />
您希望获得以下结果(两个元素重复了三遍):
You'd expect to get following result (the two elements repeated three times):
<foo/>
<bar/>
<foo/>
<bar/>
<foo/>
<bar/>
但是,相反,您得到
<foo/>
<foo/>
<foo/>
<bar/>
这是因为查询实际上被评估为
This is because the query is actually evaluated as
(
for $i in 1 to 3
return
<foo />
),
<bar />
在您的代码中,不会得到意外的输出,而是一条错误消息.这是因为在第一个逗号,
之后未定义$noteLine
.对于以下查询,您将收到类似的错误消息:
In your code, you don't get unexpected output, but an error message. This is because $noteLine
is not defined after the first comma ,
. You would get a similar error message for following query:
for $i in 1 to 3
return
<foo>{ $i }</foo>,
<bar>{ $i }</foo>
在这里,$i
没有绑定到<bar/>
元素,因为查询被评估为
Here, $i
is not bound for the <bar/>
element, as the query is evaluated as
(
for $i in 1 to 3
return
<foo>{ $i }</foo>
),
<bar>{ $i }</foo> (: obivously, $i is not defined here :)
这篇关于错误"noteLine上的未定义变量";返回多个节点时,在FLWOR表达式中的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!