开发者

XSL - Removing the filename from the path string

开发者 https://www.devze.com 2023-01-19 10:17 出处:网络
I\'ve got a SharePoint problem which I need some help with. I\'m creating some custom ItemStyles to format the output of a Content Query Webpart (CQWP) but I need to insert a \"view all\" button into

I've got a SharePoint problem which I need some help with. I'm creating some custom ItemStyles to format the output of a Content Query Webpart (CQWP) but I need to insert a "view all" button into the output.

View all needs to point to: http://www.site.com/subsite/doclibrary1/Forms/AllItems.aspx

All the individual files in the document library have the link of: http://www.site.com/subsite/doclibrary1/FileName.doc

So what I need is some XSL functions to strip FileName.doc from the end of the string.

I've tried using substring-before($variable, '.') to get rid of the .doc, but I then need to find a way to use substring-after to search for the LAST forward slash in the series and truncate the orphaned filename.

Using @Mads Hansen's post, this is the code which resolved the problem:

Template in ItemStyle.xsl

<xsl:template name="ImpDocs" match="Row[@Style='ImpDocs']" mode="itemstyle">
    <xsl:variable name="SafeLinkUrl">
        <xsl:call-template name="OuterTemplate.GetSafeLink">
            <xsl:with-param name="UrlColumnName" select="'LinkUrl'"/>
        </xsl:call-template>
    </xsl:variable>
    <xsl:variable name="ViewAllLink">
        <xsl:call-template name="OuterTemplate.getCleanURL">
            <xsl:with-param name="path" select="@LinkUrl"/>
        </xsl:call-template>
    </xsl:variable>
    <div class="DocViewAll">
        <a href="{$ViewAllLink}Forms/AllItems.aspx" title="View all">View All</a>
        <!--Any other code you need for your custom ItemStyle here-->
    </div>
</xsl:template>

Template in ContentQueryMain.xsl

<xsl:template name="OuterTemplate.getClea开发者_运维知识库nURL">
    <xsl:param name="path" />
    <xsl:choose>
        <xsl:when test="contains($path,'/')">
            <xsl:value-of select="substring-before($path,'/')" />
            <xsl:text>/</xsl:text>
            <xsl:call-template name="OuterTemplate.getCleanURL">
                <xsl:with-param name="path" select="substring-after($path,'/')" />
            </xsl:call-template>
        </xsl:when>
        <xsl:otherwise />
    </xsl:choose>
</xsl:template>


Executing this stylesheet produces: http://www.site.com/subsite/doclibrary1/

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">
<xsl:template match="/">

    <xsl:call-template name="getURL">
        <xsl:with-param name="path">http://www.site.com/subsite/doclibrary1/FileName.doc</xsl:with-param>
    </xsl:call-template>
</xsl:template>

    <xsl:template name="getURL">
        <xsl:param name="path" />
        <xsl:choose>
            <xsl:when test="contains($path,'/')">
                <xsl:value-of select="substring-before($path,'/')" />
                <xsl:text>/</xsl:text>
                <xsl:call-template name="getURL">
                    <xsl:with-param name="path" select="substring-after($path,'/')" />
                </xsl:call-template>
            </xsl:when>
            <xsl:otherwise />
        </xsl:choose>
    </xsl:template>

</xsl:stylesheet>

The getURL template makes a recursive call to itself when there are "/" characters in the string. While there are still "/" characters, it spits out the values before the slash, and then invokes itself. When it reaches the last one, it stops.


The given solutions are not able to handle url's without filename and extension at the end (Path to folder)

I changed the ideas above to include this aswell...

   <xsl:template name="getPath"> 
        <xsl:param name="url" /> 
        <xsl:choose> 
        <xsl:when test="contains($url,'/')"> 
                <xsl:value-of select="substring-before($url,'/')" /> 
                <xsl:text>/</xsl:text> 
                <xsl:call-template name="getPath"> 
                    <xsl:with-param name="url" select="substring-after($url,'/')" /> 
                </xsl:call-template> 
        </xsl:when > 
        <xsl:otherwise>
            <xsl:if test="not(contains($url,'.'))"> 
            <xsl:value-of select="$url" /> 
            </xsl:if>
        </xsl:otherwise> 
    </xsl:choose>
</xsl:template> 

Btw. Why does MS still not support XSLT 2.0!, i saw people complainin bout that back in 2007 -.-'


If you are using XSLT 2.0 (or more specifically, XPath 2.0), then you should be able to use the replace function, using a regular expression to capture the substring before the last "/": http://www.w3.org/TR/xpath-functions/#func-replace

Unfortunately, "replace" did not exist in XSLT 1.0, so it depends on what XSLT processor you are using as to whether this will work for you.


This stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="url">
        <xsl:variable name="vReverseUrl">
            <xsl:call-template name="reverse"/>
        </xsl:variable>
        <xsl:call-template name="reverse">
            <xsl:with-param name="pString" 
                            select="substring-after($vReverseUrl,'/')"/>
        </xsl:call-template>
    </xsl:template>
    <xsl:template name="reverse">
        <xsl:param name="pString" select="."/>
        <xsl:if test="$pString">
            <xsl:call-template name="reverse">
                <xsl:with-param name="pString" select="substring($pString,2)"/>
            </xsl:call-template>
            <xsl:value-of select="substring($pString,1,1)"/>
        </xsl:if>
    </xsl:template>
</xsl:stylesheet>

With this input:

<url>http://www.site.com/subsite/doclibrary1/FileName.doc</url>

Output:

http://www.site.com/subsite/doclibrary1

One line XPath 2.0:

string-join(tokenize(url,'/')[position()!=last()],'/')


See my answer to this question and use the same technique (@Alejandro's answer essentially copies this).

0

精彩评论

暂无评论...
验证码 换一张
取 消