我下面有一个Input json
[
{
"fileName": "abc",
"value": [
{
"SSC": {
"Payload": [
{
"Ledger": [
{
"Line": [
{
"AccountingCode": [
"402050046767"
],
"AccountingPeriod": [
"00720201243"
]
},
{
"AccountingCode": [
"203010334567"
],
"AccountingPeriod": [
"00720201234"
]
}
]
}
]
}
]
}
}
]
},
{
"fileName": "abc",
"value": [
{
"SSC": {
"Payload": [
{
"Ledger": [
{
"Line": [
{
"AccountingCode": [
"40205004"
],
"AccountingPeriod": [
"0072020"
]
},
{
"AccountingCode": [
"20301033"
],
"AccountingPeriod": [
"0072020"
]
}
]
}
]
}
]
}
}
]
}
]
我的预期输出是
[
{
"fileName": "abc",
"value": [
{
"SSC": {
"Payload": [
{
"Ledger": [
{
"Line": [
{
"AccountingCode": [
"402050046767"
],
"AccountingPeriod": [
"00720201243"
]
},
{
"AccountingCode": [
"203010334567"
],
"AccountingPeriod": [
"00720201234"
]
},
{
"AccountingCode": [
"40205004"
],
"AccountingPeriod": [
"0072020"
]
},
{
"AccountingCode": [
"20301033"
],
"AccountingPeriod": [
"0072020"
]
}
]
}
]
}
]
}
}
]
}
]
以下是我尝试过但未获得正确输出的xslt
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
version="3.0"
xmlns="http://www.w3.org/2005/xpath-functions"
xpath-default-namespace="http://www.w3.org/2005/xpath-functions"
expand-text="yes">
<xsl:param name="input"/>
<xsl:output method="text"/>
<xsl:template name="xsl:initial-template">
<xsl:variable name="input-as-xml" select="json-to-xml($input)"/>
<xsl:variable name="transformed-xml" as="element(array)">
<array>
<map>
<string key="fileNames">
<xsl:for-each select="$input-as-xml//string[@key='fileName']">
<xsl:variable name="file" select="../string[@key='fileName']"/>
<xsl:value-of select="$file"/>
</xsl:for-each>
</string>
<array key="value">
<map>
<array key="SSC">
<map>
<array key="Payload">
<map>
<array key="Ledger">
<xsl:for-each select="$input-as-xml//array[@key='Ledger']">
<map>
<xsl:choose>
<xsl:when test="$input-as-xml//string[@key='fileName'] = 'abc'">
<xsl:value-of select="$input-as-xml//array[@key='Lines']"/>
</xsl:when>
</xsl:choose>
</map>
</xsl:for-each>
</array>
</map>
</array>
</map>
</array>
</map>
</array>
</map>
</array>
</xsl:variable>
<xsl:value-of select="xml-to-json($transformed-xml)"/>
</xsl:template>
</xsl:stylesheet>
如果文件名为abc，我想从json中选择“行”，然后结合在“分类帐”中选择的值
任何人都可以帮助我了解如何在xslt中实现这一目标。
答案 0 :(得分：0)
您可以执行此操作而无需转换为XML并再次返回：只需将JSON作为映射和数组的结构进行操作即可。类似于
<xsl:template name="xsl:initial-template">
<xsl:variable name="in" select="json-doc('input.json')"/>
<xsl:variable name="out" as="map(*)*">
<xsl:for-each-group select="$in?*" group-by="?fileName">
<xsl:map>
<xsl:map-entry key="'fileName'" select="current-grouping-key()"/>
<xsl:map-entry key="'value'" select="
array {
map {'SSC':
map {'Payload':
array {
map {'Ledger':
array {
map{'Line': array:join(current-group()?
value?*?SSC?Payload?*?Ledger?*?Line)}
}
}
}
}"/>
</xsl:map>
</xsl:for-each-group>
</xsl:variable>
<xsl:sequence select="array{$out}"/>
</xsl:template>
我还没有对此进行测试，并且在解决问题的复杂路径中可能存在一些错误，但是我希望它能够传达出这个想法。
（请注意，如果XSLT 3.0带有xsl:array指令，这会更容易。如果您准备使用Saxon扩展，则可以使用saxon:array来弥补这一空白。）