xslt设计以获得所需的结果 [英] xslt design to get the desired result
问题描述
我的xml如下
< root >
< Keyword_Details >
< link > www.google.com < / link >
< casename > EX PARTE JIM < / casename >
< docket > 否。 114. < / docket >
< date > 2016年1月22日。< / date >
< / Keyword_Details >
< Keyword_Details >
< link > www.test.com < / link >
< casename > EX PARTE LOWE < / casename >
< docket > 否。 2140. < / docket >
< date > 2016年1月22日。< / date >
< / Keyword_Details >
< / root >
我想要的结果如下所示是链接到链接标记之间的网址的超链接在这种情况下,它是www.google.com
EX PARTE JIM(提交日期:2016年1月22日。)
案卷号114.
我想要的结果如下所示是炒作rlinks链接到链接标签之间的网址在这种情况下是www.test.com
EX PARTE LOWE(提交日期:1月22日) ,2016年。)
案卷号2140。
我希望结果为超链接,点击结果时应该重定向到< link>标签之间给出的页面(网址)。
我希望现在能够清楚。
更新鲜
提前谢谢
这是XML数据:
< root >
< keyword_details >
< span class =code-keyword>< link > www.google。 com < / link >
< casename > EX PARTE JIM < / casename >
< docket > 否。 114. < / docket >
< date > 2016年1月22日。< / date >
< / keyword_details >
< keyword_details >
< link > www.test.com < / link >
< casename > EX PARTE LOWE < / casename >
< docket > 否。 2140. < / docket >
< date > 2016年1月22日。< / date >
< / keyword_details >
< / root >
Nest是使用以下代码设计xslt文件:
< html xsl:version = 1.0 xmlns:xsl = < span class =code-keyword> http://www.w3.org/1999/XSL/Transform xmlns = http://www.w3.org/1999/xhtml >
< body style = font-family:Arial; font-size:12pt; >
< xsl:for-each 选择 = root / keyword_details >
< span < span class =code-keyword>>
< span >
< a href = http:// {link} target = _ blank > < xsl:value-of select = link / > < / a >
< / span >
< br / >
< span >
< span style = font-weight:bold >
< xsl:value-of 选择 = casename / >
(提交日期:< xsl:value-of 选择 = date / > )
< / span >
< / span >
< div < span class =code-attribute> style = font-weight:bold >
Docket < xsl:value-of 选择 < span class =code-keyword> = docket / >
< / div >
< / span >
< br / >
< / xsl:for-each >
< span class =code-keyword>< / body >
< / html >
一旦你设计了xslt,请按照以下链接实现显示数据:
使用XSL转换和ASP.NET XML控件将XML数据呈现为HTML [< a href =http://www.codeproject.com/Articles/469723/Rendering-XML-Data-as-HTML-using-XSL-Transformatiotarget =_ blanktitle =New Window> ^
my xml is as below
<root>
<Keyword_Details>
<link>www.google.com</link>
<casename>EX PARTE JIM </casename>
<docket>No. 114.</docket>
<date>January 22, 2016.</date>
</Keyword_Details>
<Keyword_Details>
<link>www.test.com</link>
<casename>EX PARTE LOWE</casename>
<docket>No. 2140.</docket>
<date>January 22, 2016.</date>
</Keyword_Details>
</root>
I want the result as below which are hyperlinks linked to the urls between link tags in this case it is www.google.com
EX PARTE JIM (Date Submitted: January 22, 2016.)
docket No. 114.
I want the result as below which are hyperlinks linked to the urls between link tags in this case it is www.test.com
EX PARTE LOWE (Date Submitted: January 22, 2016.)
docket No. 2140.
I want the result as hyperlinks, when clicked on result it should redirect to the page(url) given between <link>tags.
I hope am clear now.
Am a fresher
Thanks in advance
Here is XML data:
<root> <keyword_details> <link>www.google.com</link> <casename>EX PARTE JIM </casename> <docket>No. 114.</docket> <date>January 22, 2016.</date> </keyword_details> <keyword_details> <link>www.test.com</link> <casename>EX PARTE LOWE</casename> <docket>No. 2140.</docket> <date>January 22, 2016.</date> </keyword_details> </root>
Nest is to design a xslt file using below code:
<html xsl:version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns="http://www.w3.org/1999/xhtml"> <body style="font-family:Arial;font-size:12pt;"> <xsl:for-each select="root/keyword_details"> <span> <span> <a href="http://{link}" target="_blank"><xsl:value-of select="link"/></a> </span> <br/> <span> <span style="font-weight:bold"> <xsl:value-of select="casename" /> (Date Submitted: <xsl:value-of select="date" />) </span> </span> <div style="font-weight:bold"> Docket <xsl:value-of select="docket" /> </div> </span> <br/> </xsl:for-each> </body> </html>
Once you designed xslt, follow below link to implement to display data:
Rendering XML Data as HTML using XSL Transformation and ASP.NET XML Control[^]
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