用于对字符串中的字符进行计数/格式化的xquery递归
我试图在xquery 1.0中找到以下问题的解决方案。我只需要单独使用递归函数。输入的字符始终仅在A-Z之间用于对字符串中的字符进行计数/格式化的xquery递归,xquery,Xquery,我试图在xquery 1.0中找到以下问题的解决方案。我只需要单独使用递归函数。输入的字符始终仅在A-Z之间 1. AAABBCCD --> A1A2A3-B1B2-C1C2-D1 2. ABBAB --> A1-B1B2-A2-B3 谢谢。到目前为止，我所做的尝试没有任何结果。首先，您可能希望将字符串拆分为单个字符，这可以通过多种方式完成（例如，使用fn:substring（…）），我将使用fn:string-to-codepoints（$str）： 示例：local:t
1. AAABBCCD --> A1A2A3-B1B2-C1C2-D1
2. ABBAB --> A1-B1B2-A2-B3
谢谢。到目前为止，我所做的尝试没有任何结果。首先，您可能希望将字符串拆分为单个字符，这可以通过多种方式完成（例如，使用
fn:substring（…）fn:string-to-codepoints（$str）local:to chars（'ABC'）（'A'，'B'，'C'）- 一个接一个地检查每个字符
- 获取并更新其在地图中的计数
- 决定是否在输出字符串中的破折号之前添加破折号
- 然后再发生
declare function local:format($chars, $counts, $last, $out) {
if(empty($chars)) then $out
else (
let $c := head($chars), (: current char :)
$new-chars := tail($chars), (: rest of the chars :)
$count := ($counts($c), 1)[1], (: old count if present, `1` otherwise :)
$cc := $c || $count, (: char with count :)
$new-out := if($c eq $last) then $out || $cc (: updated output string :)
else $out || '-' || $cc,
$new-counts := map:put($counts, $c, $count + 1) (: updated map :)
return local:format($new-chars, $new-counts, $c, $new-out)
)
};
declare function local:format-string($str) {
let $chars := local:to-chars($str)
return local:format($chars, map{}, head($chars), '')
};
local:format string（'HELLO'）'H1-E1-L1L2-O1'declare function local:format($chars, $counts, $last, $out) {
if(empty($chars)) then $out
else (
let $c := $chars[1],
$new-chars := subsequence($chars, 2),
$pos := string-to-codepoints($c) - 64,
$count := $counts[$pos],
$cc := $c || $count,
$new-out := if($c eq $last) then concat($out, $cc)
else concat($out, '-', $cc),
$new-counts :=
(
subsequence($counts, 1, $pos - 1),
$count + 1,
subsequence($counts, $pos + 1)
)
return local:format($new-chars, $new-counts, $c, $new-out)
)
};
declare function local:format-string($str) {
let $chars := local:to-chars($str),
$counts := for $i in 1 to 26 return 1
return local:format($chars, $counts, head($chars), '')
};
1. AAABBCCD --> A1A2A3-B1B2-C1C2-D1
2. ABBAB --> A1-B1B2-A2-B3
string-join(
(for $s in .,
$indS in 1 to string-length($s),
$cp in string-to-codepoints($s)[$indS]
return
(
'-'[$indS gt 1 and $cp ne string-to-codepoints($s)[$indS -1]]
,
concat(codepoints-to-string($cp),
for $i in 1 to count(index-of(string-to-codepoints($s), $cp))
return
$i[index-of(string-to-codepoints($s), $cp)[$i] eq $indS]
)
)
),
''
)
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="t">
<xsl:sequence select=
"string-join(
(for $s in .,
$indS in 1 to string-length($s),
$cp in string-to-codepoints($s)[$indS]
return
(
'-'[$indS gt 1 and $cp ne string-to-codepoints($s)[$indS -1]]
,
concat(codepoints-to-string($cp),
for $i in 1 to count(index-of(string-to-codepoints($s), $cp))
return
$i[index-of(string-to-codepoints($s), $cp)[$i] eq $indS]
)
)
),
''
)
"/>
</xsl:template>
</xsl:stylesheet>
<z>
<t>AAABBCCD</t>
<t>ABBAB</t>
<t>aBcBaBcc</t>
</z>
A1A2A3-B1B2-C1C2-D1
A1-B1B2-A2-B3
a1-B1-c1-B2-a2-B3-c2c3
->>>>>还要注意，输入字符可以有任何值，而不仅仅是[A-Z]。使用简单的序列将XQuery 3.0、基于地图的解决方案很好地转换为某种XQuery 1.0方式。但是我认为，XQuery1中没有
headtail子序列concat| |