REMOVE leading zero and alpha characters in string in XSL...

san_yacha · July 28th, 2009, 12:51 AM

hi

i have a code in C language

Basically i want to remove leading zero and alphabets from string

example

000C77499 should become 7749
000wert499 should become 499
00erty100 should become 100

I want to do it in XSL . PLEASE HELP

In the below CODE above objective is already achieved in C language

CString Buff = "000C77499";
Buff.MakeUpper();
int len = Buff.GetLength();
int index = 0;
for(int len1 = Buff.GetLength(); len1>0;len1--)
if(0 != isdigit(Buff.GetAt(len1-1)))
{
index++;
continue;
}
else
{
break;
}
Buff = Buff.Mid(len-index,len);

PLEASE HELP

joefawcett · July 28th, 2009, 03:16 AM

You posted in the XSLT forum, do you want an XSLT solution or something in C?

mrame · July 28th, 2009, 03:19 AM

With XSLT 2.0, try the below:

Sample input xml:

<root>
<test>000c46456546</test>
<test>00ax46456546</test>
<test>0cc46456546</test>
</root>

xslt code for the above:

Code:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:output method="xml" indent="yes"></xsl:output>

<xsl:template match="test">
<xsl:value-of select=" replace(.,'([0-9]+)([a-z]+)([0-9]+)','$3')"/>
</xsl:template>
</xsl:stylesheet>

Output:

46456546
46456546
46456546

mhkay · July 28th, 2009, 04:06 AM

You need to make it clear whether you need an XSLT 1.0 or 2.0 solution.

In 2.0 you can do it using regular expressions:

replace($in, '.*?([\d]+$)', '$1')

In 1.0 you need a recursive template along the lines:

Code:

<xsl:template name="strip">
  <xsl:param name="in"/>
  <xsl:when test="translate($in, '0123456789', '')">
    <xsl:call-template name="strip">
       <xsl:with-param name="in" select="substring($in, 2)"/>
    </xsl:call-template>
  </xsl:when>
  <xsl:otherwise>
    <xsl:value-of select="$in"/>
  </xsl:otherwise>
</xsl:template>

(The translate() call removes all digits and returns true if the result is non-zero-length, that is, if the string contained a non-digit).