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ช่วยตรวจสอบโค๊ตหน่อยครับผมต้องการวันที่ ที่คำนวนจำนวนวันได้มาคิดดอกเบี้ยครับ |
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คือผมใช้ในการ print($time); ออกมาคร๊าฟ แต่ไม่สามารถนำามาคำนวนดอกเบี้ยได้ ที่ต้องการคือเอาวันที่คำนวนได้มาคูณกับดอกเบี้ยครับช่วย ด่วนครับ
$sum_interest=$interest/30*คูณจำนวนวันที่คำนวนได้; เช่น 200*21/30=140 บาท นี่คือจำนวนเงินที่ต้องชำระครับ
Code (PHP)
//ตรงนี้ครั บทีต้องการหาดอกเบี้ยที่ต้องชำระ
$sum_interest=$interest/30*คูณจำนวนวันที่คำนวนได้;
<?php
$rdate = mktime(8,0,0,3,30,2011);
$ftart = mktime(8,0,0,3,30,2011);
$online=$rdate-$ftart;
$day = intval( $online / 86400 ); // จำนวนวัน
//$hours = intval( ( $online % 86400 ) / 3600 ); // จำนวน ชั่วโมง
// $mins = intval( ( ( $online % 86400 ) % 3600 ) / 60 ); // จำนวน นาที
// $secs = intval( ( ( ( $online % 86400 ) % 3600) % 60 ) ); // จำนวน วินาที
// print "$online -- $day -- $hours-- $mins-- $secs ";
?>
<?
function dateDiv($t1,$t2){ // ส่งวันที่ที่ต้องการเปรียบเทียบ ในรูปแบบ มาตรฐาน 2006-03-27 21:39:12
$t1Arr=splitTime($t1);
$t2Arr=splitTime($t2);
$Time1=mktime($t1Arr["h"], $t1Arr["m"], $t1Arr["s"], $t1Arr["M"], $t1Arr["D"], $t1Arr["Y"]);
$Time2=mktime($t2Arr["h"], $t2Arr["m"], $t2Arr["s"], $t2Arr["M"], $t2Arr["D"], $t2Arr["Y"]);
$TimeDiv=abs($Time2-$Time1);
$Time=intval($TimeDiv/86400); // จำนวนวัน
// $Time["H"]=intval(($TimeDiv%86400)/3600); // จำนวน ชั่วโมง
// $Time["M"]=intval((($TimeDiv%86400)%3600)/60); // จำนวน นาที
//$Time["S"]=intval(((($TimeDiv%86400)%3600)%60)); // จำนวน วินาที
return $Time;
}
function splitTime($time){ // เวลาในรูปแบบ มาตรฐาน 2006-03-27 21:39:12
$timeArr["Y"]= substr($time,2,2);
$timeArr["M"]= substr($time,5,2);
$timeArr["D"]= substr($time,8,2);
$timeArr["h"]= substr($time,11,2);
$timeArr["m"]= substr($time,14,2);
$timeArr["s"]= substr($time,17,2);
return $timeArr;
}
//------------------------------ ตัวอย่างการใช้งาน
$t1="$datenow";
$t2="$datetime";
//print "<br> $t1 <br> $t2 <br> ";
$time=dateDiv($t1,$t2);
print($time);
?>
Tag : PHP, JavaScript
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ประวัติการแก้ไข 2013-05-08 12:54:56
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Date :
2013-05-08 00:42:15 |
By :
liwkalg |
View :
841 |
Reply :
9 |
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PHP DateDiff / TimeDiff
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Date :
2013-05-08 06:38:37 |
By :
mr.win |
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ตอนนี้คำนวนหาส่วนต่างของวันได้ยังครับ จำนวนวันอะ
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Date :
2013-05-08 12:21:51 |
By :
Krungsri |
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จากรูปแบบ $t1 และ $t2
น่าจะใช้ Function นี้ก็น่าจะได้ผลต่างของวัน
Code (PHP)
<?php
function DateTimeDiff($strDateTime1,$strDateTime2)
{
return (strtotime($strDateTime2) - strtotime($strDateTime1))/ ( 60 * 60 ); // 1 Hour = 60*60
}
echo "Date Time Diff = ".DateTimeDiff("2008-08-01 00:00","2008-08-01 19:00");
?>
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ประวัติการแก้ไข 2013-05-08 12:53:16
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Date :
2013-05-08 12:52:46 |
By :
Krungsri |
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ลองดูครับ
Code (PHP)
$numdate = DateTimeDiff("2008-08-01 00:00","2008-08-01 19:00");
$sum_interest=$interest/30*$numdate;
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Date :
2013-05-08 15:34:59 |
By :
Krungsri |
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Load balance : Server 03
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