XSLT find and replace carriage returns -

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I need to change the carriage return from input XML.

The input is as follows:

I am trying to delete the carousis return using the following function: 

  & lt; Xsl: template name = "string-replace-all" & gt; & Lt; Xsl: param name = "text" /> & Lt; Xsl: Ultimate name = "Replaced" /> & Lt; Xsl: Ultimate name = "by" /> & Lt; XSL: Select & gt; & Lt; Xsl: when test = "in ($ text, instead of $)" & gt; & Lt; Xsl: Select the value = "substring-before ($ text, $ replace)" /> & Lt; Xsl: Select Value = "by $" /> & Lt; Xsl: call-template name = "string-replace-all" & gt; & Lt; With xsl: choose the ultimate name = "text" = "substring-after ($ text, $ location)" /> & Lt; With xsl: select param with name = "replace" = "$ replace" /> & Lt; Xsl: select-by-name = "by" = "$ by" /> & Lt; / XSL: Call-templates & gt; & Lt; / XSL: When & gt; & Lt; XSL: otherwise & gt; & Lt; Xsl: Select Value = "$ text" /> & Lt; / XSL: otherwise & gt; & Lt; / XSL: Select & gt; & Lt; / XSL: Templates & gt;   
 is calling it as:  
  & lt; Xsl: when test = "label = 'notes / comments'" & gt; & Lt; Xsl: element name = "comments" & gt; & Lt; Xsl: call-template name = "string-replace-all" & gt; & Lt; With xsl: select param name = "text" = "value" /> & Lt; Xsl: with-param name = "change" & gt; & Amp; #xA; & Lt; / Xsl: with-param & gt; & Lt; Xsl: -name = "'." "/" Gt; & lt; / XSL: call-templates & gt; & lt; / XSL: element & gt; & lt; / XSL: When & gt;   
 But there is no success yet, I'm hoping this character ( & amp; # xA; ) is passing through the wrong But I can not work on it.  
  UPDATE   
 As it turns into this case, more white space than just carriage returns Removing characters is acceptable and general-location () my needs   
 
 
 Have you checked that you have  & amp; #xD;  besides  & amp; #xA; ?

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