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    <title><![CDATA[Questions & Answers: Calculating duration with strange date / time format ]]></title>
    <link><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format]]></link>
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    <lastBuildDate>2013-05-23T23:53:29-07:00</lastBuildDate>
             

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        <title><![CDATA[Reponse To Answer]]></title>
        <link><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751858]]></link>
        <description><![CDATA[Why don't you just wrap the entire function in a left({function},len(B1)-4)?]]></description>
        <guid><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751858]]></guid>
        <dc:creator><![CDATA[phil@...]]></dc:creator>
        <pubDate>Wed, 13 Mar 2013 11:48:17 -0700</pubDate>
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    <item>
        <title><![CDATA[Reponse To Answer]]></title>
        <link><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751737]]></link>
        <description><![CDATA[Hi,Thanks. I'm working in Excel 2013.. I have got a little further now.. I have successfully changed [01:02:2013][09:49:04:651] into 01/02/2013 09:49:04:651 by using the formula: =SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(B1,&quot;[&quot;,&quot;&quot;),&quot;]&quot;,&quot;&quot;,2),&quot;]&quot;,&quot; &quot;,1),&quot;:&quot;,&quot;/&quot;,2),&quot;:&quot;,&quot;/&quot;,1) BUT, i still need to strip off the last 4 digits/char (:651) which I think can be done by using the 'LEFT' option, but I cant figure out how to integrate it into the SUBSTITUTE formula above. Any ideas? Thanks]]></description>
        <guid><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751737]]></guid>
        <dc:creator><![CDATA[chrispx001]]></dc:creator>
        <pubDate>Wed, 13 Mar 2013 02:38:34 -0700</pubDate>
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        <title><![CDATA[Reponse To Answer]]></title>
        <link><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751728]]></link>
        <description><![CDATA[HI,1. Yes, excel 2013 (I should have put that before!, sorry)2. Original file format is .csv]]></description>
        <guid><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751728]]></guid>
        <dc:creator><![CDATA[chrispx001]]></dc:creator>
        <pubDate>Wed, 13 Mar 2013 02:35:08 -0700</pubDate>
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    <item>
        <title><![CDATA[Just forcing a display format]]></title>
        <link><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751709]]></link>
        <description><![CDATA[Won't change the underlying data type. It sounds like the data type for those cells have changed from date/time to text. Can't be more specific w/o details. ( spreadsheet version, input data source, input/output example)ADDENDUM:This is some VBA code to check cell contents. Might help differentiate between datetime and text...http://www.j-walk.com/ss/excel/tips/tip62.htm]]></description>
        <guid><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751709]]></guid>
        <dc:creator><![CDATA[Charles Bundy]]></dc:creator>
        <pubDate>Tue, 12 Mar 2013 18:59:01 -0700</pubDate>
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        <title><![CDATA[I probably can't help, but two questions come to mind]]></title>
        <link><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751591]]></link>
        <description><![CDATA[1. Are you working in Excel, Works, or something else? if the last, what?2. How did the date format come to change?]]></description>
        <guid><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751591]]></guid>
        <dc:creator><![CDATA[john.a.wills@...]]></dc:creator>
        <pubDate>Tue, 12 Mar 2013 08:33:27 -0700</pubDate>
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    <item>
        <title><![CDATA[Calculating duration with strange date / time format]]></title>
        <link><![CDATA[http://www.techrepublic.com/forum/questions/101-402445/calculating-duration-with-strange-date-time-format?#msg-3751551]]></link>
        <description><![CDATA[I had a simple formula for working out a duration based on 2 columns of data. Each column had a date and a time in it in this format in it: &quot;dd/mm/yyy hh:mm&quot;The formula was this: &quot;(F2-E2)*24*60*60)&quot;Unfortunately now the date and time format has changed to be like this: &quot;[01:02:2013][04:06:05:062]&quot; so basicly there are now &quot;[]&quot; around the date and time and the time now includes seconds and milliseconds as well.The formula now longer works and I have tried to create a custom number format in the hope it will recognise the new format, but I cant get it work (either to accept the new format or read the date/time correctly)This is what i tried &quot;[&quot;dd:mm:yyyy&quot;]&quot;&quot;[&quot;hh:mm:ss:000&quot;]&quot;Does any one have any ideas how this can be achieved please? (The milliseconds are that important)Thank you in advance for any help!]]></description>
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        <dc:creator><![CDATA[chrispx001]]></dc:creator>
        <pubDate>Tue, 12 Mar 2013 06:44:32 -0700</pubDate>
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