position - 如何在 XQuery 中为序列中的位置赋值？

Question

我有一个长度为 $n 的序列初始化为零：

let $seq := (for $i in (1 to $n) return 0)

我可以轻松获得职位...

return $seq[5]

...但是我该如何更新呢？（以下不起作用）

let $seq[5] := $seq[5] + 1

Answer 1

如果您正在使用支持 XQuery 3 映射的 XQuery 实现（例如 Saxon、BaseX），您可以使用这些：

declare namespace map="http://www.w3.org/2005/xpath-functions/map";

(: Fill map with square numbers :)
let $map := map:new(
  for $i in (1 to 10)
  return map:entry($i, $i*$i)
)
(: Overwrite a single value :)
let $map := map:new(($map, map:entry(2, 5)))
(: Fetch this value :)
return map:get($map, 2)

But generally it is possible to solve a problem without maps and in most cases this code will probably run faster as it will get better optimized.

Answer 2

XQuery 是一种函数式语言，而不是过程式语言，因此变量是不可变的——它们一旦分配就不能更新。你需要做这样的事情并创建一个新序列：

let $seq2 := 
  for $n at $pos in $seq
  if ($pos eq 5) 
  then $n + 1
  else $n

通常，在 XQuery 中，最好设计算法以便不需要这种类型的可变变量解决方法。如果您有需要更新的数据，请考虑将其放入数据库中。

Answer 3

...but how do I update it? (the following doesn't work)

let $seq[5] := $seq[5] + 1

Using pure XPath (which is also pure XQuery :) here is probably the shortest way to specify this:

subsequence($seq, 1, 4), $seq[5] + 1, subsequence($seq, 6)

This produces a new sequence whose items are the same as the items of $seq, except that its 5th item's value is $seq[5] + 1.

As others have noted, XPath and XQuery are functional languages and among other things this means that a variable, once defined, cannot have its value modified.