Error in xslt formation

Question

User-254743730 posted

Hi,

Iam using xml to XSLT Operation. I did code . but i got only partial output... i did't get where i want  to corrct my code...

 my excepted output is un ordered list like this

List

• Menu1
• Menu2

Item

•   lst1
• lst2
• lst3

 like that.. but i got out put like this

List 

menu1

List

Menu2

for contious all item will be displayed.

i ll show my xslt file

?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl">
    <xsl:output method="xml" indent="yes"/>
    <xsl:strip-space  elements ="*"/>

    <xsl:template match="Menus">
        <div>
            <xsl:if test ="Menu" ></xsl:if>
            <ul >
                <xsl:apply-templates/>
            </ul>
        </div>
    </xsl:template>

    <xsl:template match="Menu">
        <li>
            <a>
                <xsl:apply-templates select="Menu"/>
                  <xsl:attribute name="href">
                    <xsl:value-of select="LinkAddress"/>
                </xsl:attribute>
                <xsl:value-of select="DisplayTitle" />
            </a>
            <xsl:for-each select="SubMenu">
                <ul>
                    <li>
                        <a>
                            <xsl:apply-templates select="SubMenu"/>
                            <xsl:attribute name="href">
                                <xsl:value-of select="LAddress"/>
                            </xsl:attribute>
                            <xsl:value-of select="Title" />
                        </a>
                    </li>
                </ul>
            </xsl:for-each>
        </li>
    </xsl:template>
</xsl:stylesheet>

can any one tell me what is wrong..

Answer 1

User1777983149 posted

Required xsl file and xml file,  then use  transforms merg the you will get proper result as per required,  manage xsl file

Answer 2

User-254743730 posted

Hi,

I  found solution.

This is the solution link

http://forums.asp.net/t/1640334.aspx

Answer 3

User-573138384 posted

For the output you mentioned, you need to call the Menu template twice. First to pull only menus and second to pull submenus. You can use mode for this purpose.

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl">
  <xsl:output method="xml" indent="yes"/>
  <xsl:strip-space  elements ="*"/>

  <xsl:template match="Menus">
    <div>
      List<br />
      <ul >
        <xsl:apply-templates mode="Menu"/>
      </ul>
      Item<br />
      <ul >
        <xsl:apply-templates mode="SubMenu"/>
      </ul>
    </div>
  </xsl:template>

  <xsl:template match="Menu" mode="Menu">
    <li>
      <a>
        <xsl:apply-templates select="Menu"/>
        <xsl:attribute name="href">
          <xsl:value-of select="LinkAddress"/>
        </xsl:attribute>
        <xsl:value-of select="DisplayTitle" />
      </a>     
    </li>
  </xsl:template>

  <xsl:template match="Menu" mode="SubMenu">
    <xsl:for-each select="SubMenu">
        <li>
          <a>
            <xsl:apply-templates select="SubMenu"/>
            <xsl:attribute name="href">
              <xsl:value-of select="LAddress"/>
            </xsl:attribute>
            <xsl:value-of select="Title" />
          </a>
        </li>
    </xsl:for-each>
  </xsl:template>
</xsl:stylesheet>

Answer 4

User2056829991 posted

Hi

Can you put your XML data to me? May I can help you.

Best Regards