XML-XSLT : How to compare two dates which are in S

问题:

I know this question might get repeated and also I have went through similar articles and question but I have not found the exact solution.

Now the question I am using XSLT or XPATH to transform the xml. Here in XML two string variables are there. one is OldDate and second is CurrentDate.

Ex : $oldDate = '29.05.2015 15:25:06'
     $currentDate ='27.07.2015 14:28:02'.

Now I want to compare those two dates.

If  $oldDate > $currentDate  then 'OK' else 'Not Ok'.

As I am new to use XSLT and XPATH I did not get that how to proceed from the given answers in other articles.

It will be great-full to provide a perfect solution for this.

Thank you.

回答1:

XSLT (2.0) recognizes dates in YYYY-MM-DD format only, and date-times in YYYY-MM-DDThh:mm:ss format only. In order to compare the strings as dates, (or, in this case, date-times), you must first convert them to valid date-times. Since you need to do this more than once, it would be convenient to construct a function for this purpose:

XSLT 2.0

<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:my="http://www.example.com/my"
exclude-result-prefixes="xs my">
<xsl:output method="xml" version="1.0" encoding="utf-8" indent="yes"/>
<xsl:strip-space elements="*"/>

<xsl:variable name="oldDate" select="'29.05.2015 15:25:06'" />
<xsl:variable name="currentDate" select="'27.07.2015 14:28:02'" />

<xsl:function name="my:string-to-datetime">
    <xsl:param name="string"/> 
    <xsl:variable name="parts" select="tokenize($string,'\.|\s')"/>
    <xsl:sequence select="xs:dateTime(concat($parts[3], '-', $parts[2], '-', $parts[1], 'T', $parts[4]))" />
</xsl:function>

<xsl:template match="/">
    <result>
        <xsl:value-of select="if (my:string-to-datetime($oldDate) gt my:string-to-datetime($currentDate)) then 'OK' else 'Not Ok'" />
    </result>
</xsl:template>

</xsl:stylesheet>

Result

<?xml version="1.0" encoding="utf-8"?>
<result>OK</result>

Note that this assumes your strings come in a DD.MM.YYYY hh:mm:ss format - i.e. that the days are padded to two digits - otherwise there's more work to be done.

回答2:

Given this input XML:

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <oldDate>29.05.2015 15:25:06</oldDate>
    <currentDate>27.07.2015 14:28:02</currentDate>
</root>

and this stylesheet:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="2.0">

    <xsl:output omit-xml-declaration="yes"/>

    <xsl:variable name="raw_oldDate" select="substring-before(root/oldDate, ' ')"/>
    <xsl:variable name="raw_currentDate" select="substring-before(root/currentDate, ' ')"/>


    <xsl:template match="/">
        <xsl:choose>
            <xsl:when test="number(concat(substring($raw_oldDate, 7, 4), 
                substring($raw_oldDate, 4, 2),
                substring($raw_oldDate, 1, 2),
                translate(substring-after(root/oldDate, ' '), ':', '')))
                &lt;
                number(concat(substring($raw_currentDate, 7, 4), 
                substring($raw_currentDate, 4, 2),
                substring($raw_currentDate, 1, 2),
                translate(substring-after(root/currentDate, ' '), ':', '')))">
                <xsl:text>OK</xsl:text>
            </xsl:when>
            <xsl:otherwise>
                <xsl:text>not OK</xsl:text>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>

</xsl:stylesheet>

it produces:

OK