Date and Time difference

sharma.s.prakash · 23-08-2011, 10:33 AM

/**
* To fetch the difference between two dates
* @param date $firstDate
* @param date $secondDate
*/

PHP Code:


			
function dateDiffSQL($firstDate, $secondDate) {

    $conn=mysqli_connect($DBSERVER, $DBUSER, $DBPASSWORD, $DBNAME) or die('can not connect');

    $qryDiff = "SELECT DATEDIFF('$secondDate', '$firstDate') as diff";

    $rsDiff = mysqli_query($conn, $qryDiff);

    $rowDiff = mysqli_fetch_assoc($rsDiff);

    return $rowDiff[diff];

}

/**
* To fetch the time elapsed between the two dates
* @param datetime $startDtTime
* @param datetime $endDtTime
*/

PHP Code:


			
function timeDiffSql($startDtTime, $endDtTime) {

    $conn=mysqli_connect($DBSERVER, $DBUSER, $DBPASSWORD, $DBNAME) or die('can not connect');

    $qryTimeDiff="SELECT TIMEDIFF('$endDtTime', '$startDtTime') as hrs_taken";

    $rsTimeDiff=mysqli_query($conn,$qryTimeDiff);

    $rowTimeDiff=mysqli_fetch_assoc($rsTimeDiff);

    $hrsTaken=$rowTimeDiff[hrs_taken];

    return $hrsTaken;

}

kaman · 23-08-2011, 02:20 PM

Very useful! Thx for sharing.

sharma.s.prakash · 24-08-2011, 10:08 AM

thanx for your prompt response

amit · 07-09-2011, 02:51 PM

Hi I dont want to handle it through DB.. any other way??

AI

sharma.s.prakash · 08-09-2011, 09:22 AM

Try This

PHP Code:


			
$varDate1='1980-01-27';
$varDate2='2011-06-07';
echo funcDateDiff($varDate1,$varDate2);

function funcDateDiff($varDate1,$varDate2) {
    if($varDate1<=$varDate2) {
        $arrDate1=explode('-',$varDate1);
        $arrDate2=explode('-',$varDate2);
        $arrDateDiff=array();
        if($arrDate2[2]<$arrDate1[2]) {
            $arrDateDiff[2]=$arrDate2[2]+30-$arrDate1[2];
            $arrDate2[1]=$arrDate2[1]-1;
        } else {
            $arrDateDiff[2]=$arrDate2[2]-$arrDate1[2];
        }
        if($arrDate2[1]<$arrDate1[1]) {
            $arrDateDiff[1]=$arrDate2[1]+12-$arrDate1[1];
            $arrDate2[1]=$arrDate2[1]-1;
        } else {
            $arrDateDiff[1]=$arrDate2[1]-$arrDate1[1];
        }
        $arrDateDiff[0]=$arrDate2[0]-$arrDate1[0];
        $varDateDiff=$arrDateDiff[0].' Year '.$arrDateDiff[1].' Month '.$arrDateDiff[2].' Days ';
    } else {
        $varDateDiff='invalid date';
    }
    return $varDateDiff;
}

Output

Code:

        31 Year 4 Month 10 Days

sharma.s.prakash · 08-09-2011, 09:24 AM

Well come on septanet

sachin · 08-09-2011, 09:49 AM

Excellent Satya Prakash

and

Welcome Amit

23-08-2011, 10:33 AM	#1
sharma.s.prakash Senior Member Join Date: Aug 2011 Posts: 110	Date and Time difference /** * To fetch the difference between two dates * @param date $firstDate * @param date $secondDate / PHP Code: `function dateDiffSQL($firstDate, $secondDate) { $conn=mysqli_connect($DBSERVER, $DBUSER, $DBPASSWORD, $DBNAME) or die('can not connect'); $qryDiff = "SELECT DATEDIFF('$secondDate', '$firstDate') as diff"; $rsDiff = mysqli_query($conn, $qryDiff); $rowDiff = mysqli_fetch_assoc($rsDiff); return $rowDiff[diff]; }` /* * To fetch the time elapsed between the two dates * @param datetime $startDtTime * @param datetime $endDtTime */ PHP Code: `function timeDiffSql($startDtTime, $endDtTime) { $conn=mysqli_connect($DBSERVER, $DBUSER, $DBPASSWORD, $DBNAME) or die('can not connect'); $qryTimeDiff="SELECT TIMEDIFF('$endDtTime', '$startDtTime') as hrs_taken"; $rsTimeDiff=mysqli_query($conn,$qryTimeDiff); $rowTimeDiff=mysqli_fetch_assoc($rsTimeDiff); $hrsTaken=$rowTimeDiff[hrs_taken]; return $hrsTaken; }`

07-09-2011, 02:51 PM	#4
amit Junior Member Join Date: Sep 2011 Posts: 1	No DB Access Hi I dont want to handle it through DB.. any other way?? AI

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23-08-2011, 02:20 PM	#2
kaman Senior Member Join Date: Aug 2011 Posts: 111	Very useful! Thx for sharing.

24-08-2011, 10:08 AM	#3
sharma.s.prakash Senior Member Join Date: Aug 2011 Posts: 110	thanx for your prompt response

08-09-2011, 09:22 AM	#5
sharma.s.prakash Senior Member Join Date: Aug 2011 Posts: 110	Try This PHP Code: $varDate1='1980-01-27'; $varDate2='2011-06-07'; echo funcDateDiff($varDate1,$varDate2); function funcDateDiff($varDate1,$varDate2) { if($varDate1<=$varDate2) { $arrDate1=explode('-',$varDate1); $arrDate2=explode('-',$varDate2); $arrDateDiff=array(); if($arrDate2[2]<$arrDate1[2]) { $arrDateDiff[2]=$arrDate2[2]+30-$arrDate1[2]; $arrDate2[1]=$arrDate2[1]-1; } else { $arrDateDiff[2]=$arrDate2[2]-$arrDate1[2]; } if($arrDate2[1]<$arrDate1[1]) { $arrDateDiff[1]=$arrDate2[1]+12-$arrDate1[1]; $arrDate2[1]=$arrDate2[1]-1; } else { $arrDateDiff[1]=$arrDate2[1]-$arrDate1[1]; } $arrDateDiff[0]=$arrDate2[0]-$arrDate1[0]; $varDateDiff=$arrDateDiff[0].' Year '.$arrDateDiff[1].' Month '.$arrDateDiff[2].' Days '; } else { $varDateDiff='invalid date'; } return $varDateDiff; } Output Code: 31 Year 4 Month 10 Days

08-09-2011, 09:24 AM	#6
sharma.s.prakash Senior Member Join Date: Aug 2011 Posts: 110	Well come on septanet

08-09-2011, 09:49 AM	#7
sachin Senior Member Join Date: Aug 2011 Posts: 131	Excellent Satya Prakash and Welcome Amit