Marklogic-如何在Xquery中分配动态变量
我尝试了下面提到的XQueryMarklogic-如何在Xquery中分配动态变量,xquery,marklogic,marklogic-8,Xquery,Marklogic,Marklogic 8,我尝试了下面提到的XQuery declare variable $path as xs:string :="D:\Mongo\"; let $uri :="/MJ/1932/Vol1/Part1/387.xml" let $x := fn:normalize-space(fn:replace($uri,"/"," ")) for $i in fn:tokenize($x, " ") let $j := fn:concat($path,$i) retu
declare variable $path as xs:string :="D:\Mongo\";
let $uri :="/MJ/1932/Vol1/Part1/387.xml"
let $x := fn:normalize-space(fn:replace($uri,"/"," "))
for $i in fn:tokenize($x, " ")
let $j := fn:concat($path,$i)
return($j)
实际产出
D:\Mongo\MJ
D:\Mongo\1932
D:\Mongo\Vol1
D:\Mongo\Part1
D:\Mongo\387.xml
D:\Mongo\MJ
D:\Mongo\MJ\1932
D:\Mongo\MJ\1932\Vol1
D:\Mongo\MJ\1932\Vol1\Part1
D:\Mongo\MJ\1932\Vol1\Part1\387.xml
预期产出
D:\Mongo\MJ
D:\Mongo\1932
D:\Mongo\Vol1
D:\Mongo\Part1
D:\Mongo\387.xml
D:\Mongo\MJ
D:\Mongo\MJ\1932
D:\Mongo\MJ\1932\Vol1
D:\Mongo\MJ\1932\Vol1\Part1
D:\Mongo\MJ\1932\Vol1\Part1\387.xml
请告诉我如何更改动态变量值。XQuery是一种函数式编程语言,这意味着变量是不可变的。不能简单地递增或追加到已定义的变量。通常,使用递归函数来构造结果 这个例子(还有一些更简洁的例子,我想把各个部分分开,简单易懂)递归地创建路径,每次执行时都附加另一个级别。
$path
前缀单独附加,以避免混淆不同的任务
declare variable $path as xs:string :="D:\Mongo\";
declare variable $uri as xs:string := "/MJ/1932/Vol1/Part1/387.xml";
declare function local:add-path($parts as xs:string*) as xs:string* {
let $head := $parts[1]
let $tail := $parts[position() > 1]
return
if ($head)
then (
$head,
for $path in local:add-path($tail)
return string-join(($head, $path), "\")
)
else ()
};
for $uri in local:add-path(fn:tokenize(fn:normalize-space(fn:replace($uri,"/"," ")), " "))
return concat($path, $uri)
在这种特定情况下,另一种方法是在位置计数器上循环,并将零件连接到此位置:
declare variable $path as xs:string :="D:\Mongo\";
declare variable $uri as xs:string := "/MJ/1932/Vol1/Part1/387.xml";
let $parts := fn:tokenize(fn:normalize-space(fn:replace($uri,"/"," ")), " ")
for $i in (1 to count($parts))
return concat($path, string-join($parts[position() <= $i], '\'))
将变量$path声明为xs:string:=“D:\Mongo\”;
将变量$uri声明为xs:string:=“/MJ/1932/Vol1/Part1/387.xml”;
让$parts:=fn:tokenize(fn:normalize space(fn:replace($uri,“/”,“”)),“”)
对于$i in(1计算($parts))
返回concat($path,string join($parts[position()XQuery是一种函数式编程语言,它意味着变量是不可变的。您不能简单地递增或追加到定义的变量。通常,使用递归函数来构造结果
这个例子(还有一些更简洁的例子,我想让各个部分分开并易于理解)递归地创建路径,每次执行时都附加另一个级别。$path
前缀是单独附加的,以避免混淆不同的任务
declare variable $path as xs:string :="D:\Mongo\";
declare variable $uri as xs:string := "/MJ/1932/Vol1/Part1/387.xml";
declare function local:add-path($parts as xs:string*) as xs:string* {
let $head := $parts[1]
let $tail := $parts[position() > 1]
return
if ($head)
then (
$head,
for $path in local:add-path($tail)
return string-join(($head, $path), "\")
)
else ()
};
for $uri in local:add-path(fn:tokenize(fn:normalize-space(fn:replace($uri,"/"," ")), " "))
return concat($path, $uri)
在这种特定情况下,另一种方法是在位置计数器上循环,并将零件连接到此位置:
declare variable $path as xs:string :="D:\Mongo\";
declare variable $uri as xs:string := "/MJ/1932/Vol1/Part1/387.xml";
let $parts := fn:tokenize(fn:normalize-space(fn:replace($uri,"/"," ")), " ")
for $i in (1 to count($parts))
return concat($path, string-join($parts[position() <= $i], '\'))
将变量$path声明为xs:string:=“D:\Mongo\”;
将变量$uri声明为xs:string:=“/MJ/1932/Vol1/Part1/387.xml”;
让$parts:=fn:tokenize(fn:normalize space(fn:replace($uri,“/”,“”)),“”)
对于$i in(1计算($parts))
返回concat($path,string join($parts[position())