Topic: XML file to 'nested' list? $Pages that link to <a href="https://ozoneasylum.com/backlink?for=11123" title="Pages that link to Topic: XML file to &#039;nested&#039; list?" rel="nofollow" >Topic: XML file to 'nested' list?\$

Author	Thread
Kaniz Bipolar (III) Inmate From: Insane since: Jun 2003	posted 01-02-2004 06:13 I have an xml file that is something like code: <menu> <menuitem> <name>Products</name> <link>#</link> <menuitem> <name>Printing</name> <link>#</link> </menuitem> <menuitem> <name>Whatever</name> <link>#</link> <menuitem> <name>Whatever-Sub</name> <link>#</link> </menuitem> </menuitem> </menuitem> </menuitem> How could I break this down into a menu structure like -Products ---Printing ---Whatever ------Whatever-Sub ---Something ------Something1 ---------Something2 ---------Whateverblah What I have so far, is something along the lines of code: <xsl:for-each select="menu/menuitem"> </xsl:for-each> <script> var oMenu1 = new Menu(); <xsl:for-each select="menu/menuitem"> <xsl:if test="count(menuitem) = 0"> var oMenu1A = oMenu1.CreateLink(); oMenu1A.displayHtml = "<xsl:value-of select="name" />"; oMenu1A.href = "<xsl:value-of select="link" />"; </xsl:if> <xsl:if test="count(menuitem) > 0"> var oMenu1B = oMenu1.CreateMenu(); oMenu1B.displayHtml = "<xsl:value-of select="name" />"; <xsl:for-each select="menuitem"> var oMenu1B_a = oMenu1.CreateLink(); oMenu1B_a.displayHtml = "<xsl:value-of select="name" />"; oMenu1B_a.href = "<xsl:value-of select="link" />"; oMenu1B.AddItem(oMenu1B_a); </xsl:for-each> oMenu1.AddItem(oMenu1B); </xsl:if> <xsl:if test="count(menuitem) = 0"> oMenu1.AddItem(oMenu1A); </xsl:if> </xsl:for-each> However, for each 'level' of nesting I want to be able to handle, it seems like I need to do another for-each. While my current use for this is only going to have maybe 3 levels tops, its not that bad. My worry is if it needs to get any bigger, I dont want to keep adding level after level of for-each.
Moon Shadow Paranoid (IV) Inmate From: Rouen, France Insane since: Jan 2003	posted 01-03-2004 17:06 First, I must say I don't understand much of your XSL code, so forgive me if I say something idiotic. I was wondering, wouldn't it be easier to create nested lists like this ? code: <ul> <xsl:for-each select="menu/menuitem"> <ul> <li><xsl:value-of select="name" /></li> <li><xsl:value-of select="link" /></li> </xsl:for-each> </ul> .... </ul> </ul> (write he number of </ul> you need)
Kaniz Bipolar (III) Inmate From: Insane since: Jun 2003	posted 01-04-2004 22:37 I tried simply doing the for-each menuitem/menu thing, but it'd only do it for 1 layer. If any menu option had sub-options, it wasnt working.
MajorFracas Nervous Wreck (II) Inmate From: Insane since: Jul 2003	posted 01-15-2004 18:07 Could you post a sample source file and the complete XSLT file? I can see some places where there might be problems but wouldn't know for sure unless I could test it...

Topic: XML file to 'nested' list?\$

Author

Thread

Kaniz
Bipolar (III) Inmate

From:
Insane since: Jun 2003

posted 01-02-2004 06:13

I have an xml file that is something like

code:

<menu>

	<menuitem>

		<name>Products</name>

		<link>#</link>

			<menuitem>

				<name>Printing</name>

				<link>#</link>

			</menuitem>

			<menuitem>

				<name>Whatever</name>

				<link>#</link>

					<menuitem>

						<name>Whatever-Sub</name>

						<link>#</link>

					</menuitem>



			</menuitem>



	</menuitem>

</menuitem>

How could I break this down into a menu structure like

-Products
---Printing
---Whatever
------Whatever-Sub
---Something
------Something1
---------Something2
---------Whateverblah

What I have so far, is something along the lines of

code:

<xsl:for-each select="menu/menuitem">



	  </xsl:for-each>

		<script>

	var oMenu1 = new Menu();

      <xsl:for-each select="menu/menuitem">

		<xsl:if test="count(menuitem) = 0">

	var oMenu1A = oMenu1.CreateLink();

	oMenu1A.displayHtml = "<xsl:value-of select="name" />";

	oMenu1A.href = "<xsl:value-of select="link" />";

			</xsl:if>

	<xsl:if test="count(menuitem) > 0">



		

	var oMenu1B = oMenu1.CreateMenu();

	oMenu1B.displayHtml = "<xsl:value-of select="name" />";

	        <xsl:for-each select="menuitem">

			var oMenu1B_a = oMenu1.CreateLink();

			oMenu1B_a.displayHtml = "<xsl:value-of select="name" />";

			oMenu1B_a.href = "<xsl:value-of select="link" />";

			oMenu1B.AddItem(oMenu1B_a);



	  		</xsl:for-each>

	oMenu1.AddItem(oMenu1B);

		</xsl:if>

				<xsl:if test="count(menuitem) = 0">

	oMenu1.AddItem(oMenu1A);

	</xsl:if>

	    </xsl:for-each>

However, for each 'level' of nesting I want to be able to handle, it seems like I need to do another for-each. While my current use for this is only going to have maybe 3 levels tops, its not that bad. My worry is if it needs to get any bigger, I dont want to keep adding level after level of for-each.

Moon Shadow
Paranoid (IV) Inmate

From: Rouen, France
Insane since: Jan 2003

posted 01-03-2004 17:06

First, I must say I don't understand much of your XSL code, so forgive me if I say something idiotic. I was wondering, wouldn't it be easier to create nested lists like this ?

code:

<ul>



<xsl:for-each select="menu/menuitem">



<ul>

	<li><xsl:value-of select="name" /></li>

	<li><xsl:value-of select="link" /></li>	



</xsl:for-each>



</ul>          

....                

</ul>           

</ul> 



(write he number of </ul> you need)

Kaniz
Bipolar (III) Inmate

From:
Insane since: Jun 2003

posted 01-04-2004 22:37

I tried simply doing the for-each menuitem/menu thing, but it'd only do it for 1 layer. If any menu option had sub-options, it wasnt working.

MajorFracas
Nervous Wreck (II) Inmate

From:
Insane since: Jul 2003

posted 01-15-2004 18:07

Could you post a sample source file and the complete XSLT file? I can see some places where there might be problems but wouldn't know for sure unless I could test it...

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Topic: XML file to 'nested' list? $Pages that link to <a href="https://ozoneasylum.com/backlink?for=11123" title="Pages that link to Topic: XML file to 'nested' list?" rel="nofollow" >Topic: XML file to 'nested' list?\$