[英]how to restructure the xml with xslt for list type of elements to an element
i have an XML generated in the below form from JSON to XML Conversion. 我有以下格式的JSON到XML转换生成的XML。
Full Original XML examples: 完整的原始XML示例:
Example1: 范例1:
<?xml version="1.0" encoding="UTF-8"?>
<linked-hash-map>
<entry>
<string>dataset</string>
<linked-hash-map>
<entry>
<string>id</string>
<string>120140</string>
</entry>
<entry>
<string>dataset_code</string>
<string>GDP</string>
</entry>
<entry>
<string>column_names</string>
<list>
<string>DATE</string>
<string>VALUE</string>
</list>
</entry>
<entry>
<string>frequency</string>
<string>quarterly</string>
</entry>
<entry>
<string>type</string>
<string>Time Series</string>
</entry>
<entry>
<string>premium</string>
<boolean>false</boolean>
</entry>
<entry>
<string>data</string>
<list>
<list>
<string>2016-07-01</string>
<double>18675.3</double>
</list>
<list>
<string>2016-04-01</string>
<double>18450.1</double>
</list>
<list>
<string>2016-01-01</string>
<double>18281.6</double>
</list>
</list>
</entry>
<entry>
<string>database_id</string>
<int>118</int>
</entry>
</linked-hash-map>
</entry>
</linked-hash-map>
Example 2: 范例2:
<?xml version="1.0" encoding="UTF-8"?>
<linked-hash-map>
<entry>
<string>dataset</string>
<linked-hash-map>
<entry>
<string>dataset_code</string>
<string>AAPL</string>
</entry>
<entry>
<string>column_names</string>
<list>
<string>DATE</string>
<string>Open</string>
<string>High</string>
<string>Low</string>
<string>Close</string>
</list>
</entry>
<entry>
<string>frequency</string>
<string>quarterly</string>
</entry>
<entry>
<string>type</string>
<string>Time Series</string>
</entry>
<entry>
<string>data</string>
<list>
<list>
<string>2016-07-01</string>
<double>116.45</double>
<double>117.1095</double>
<double>116.4</double>
<double>116.73</double>
</list>
<list>
<string>2016-04-01</string>
<double>18450.1</double>
<double>113.1095</double>
<double>112.4</double>
<double>100.73</double>
</list>
<list>
<string>2016-01-01</string>
<double>18281.6</double>
<double>157.1095</double>
<double>136.4</double>
<double>156.73</double>
</list>
</list>
</entry>
<entry>
<string>database_id</string>
<int>218</int>
</entry>
</linked-hash-map>
</entry>
</linked-hash-map>
the below part needs to be converted in both the xmls. 下面的部分需要在两个xml中都进行转换。
<entry>
<string>column_names</string>
<list>
<string>DATE</string>
<string>VALUE</string>
</list>
</entry>
<entry>
<string>data</string>
<list>
<list>
<string>2016-07-01</string>
<double>18675.3</double>
</list>
<list>
<string>2016-04-01</string>
<double>18450.1</double>
</list>
</list>
</entry>
How to convert this into the below formats? 如何将其转换为以下格式?
1. 1。
<entry>
<Date>2016-07-01</Date>
<Value>18675.3</Value>
</entry>
<entry>
<Date>2016-04-01</Date>
<Value>18450.1</Value>
</entry>
2. 2。
<entry>
<Date>2016-07-01</Date>
<Value>18675.3</Value>
<Date>2016-04-01</Date>
<Value>18450.1</Value>
</entry>
Note: All the data(Date, Value,data,entry,etc) are dynamic here. 注意:所有数据(日期,值,数据,条目等)在此处都是动态的。
Looking for a generic implementation of XSLT to bring the desired output. 寻找XSLT的通用实现以带来所需的输出。 if not possible with xslt, then would like to go for Java to convert this. 如果无法使用xslt,则希望Java进行转换。
Any help would be greatly appreciated.`` 任何帮助将不胜感激.``
The following stylesheet will work for both your examples. 以下样式表适用于您的两个示例。 I don't think it needs to be any more generic than that - and if it does, I don't see how. 我认为它不需要比这更通用了-如果确实如此,我不知道怎么做。
XSLT 1.0 XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/linked-hash-map">
<xsl:variable name="col-names" select="entry/linked-hash-map/entry[string='column_names']/list/string" />
<root>
<xsl:for-each select="entry/linked-hash-map/entry[string='data']/list/list">
<xsl:variable name="current-row" select="." />
<entry>
<xsl:for-each select="$col-names">
<xsl:variable name="i" select="position()" />
<xsl:element name="{.}">
<xsl:value-of select="$current-row/*[$i]" />
</xsl:element>
</xsl:for-each>
</entry>
</xsl:for-each>
</root>
</xsl:template>
</xsl:stylesheet>
As with your previous question , this assumes that the supplied column names are valid XML element names. 与上一个问题一样 ,这假定提供的列名称是有效的XML元素名称。
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