Categories:
.NET (357)
C (330)
C++ (183)
CSS (84)
DBA (2)
General (7)
HTML (4)
Java (574)
JavaScript (106)
JSP (66)
Oracle (114)
Perl (46)
Perl (1)
PHP (1)
PL/SQL (1)
RSS (51)
Software QA (13)
SQL Server (1)
Windows (1)
XHTML (173)
Other Resources:
How To Display a Past Time in Days, Hours and Minutes
How To Display a Past Time in Days, Hours and Minutes? - MySQL FAQs - Managing Tables and Running Queries with PHP Scripts
✍: FYIcenter.com
You have seen a lots of Websites are displaying past times in days, hours and minutes. If you want to do this yourself, you can use the TIMEDIFF() SQL function. Note that the TIMEDIFF() function can only handle time range within 839 hours (about 33 days). So it works only for past times within one month or so.
The following tutorial exercise shows you how to use TIMEDIFF() to present a past time in days, hours, and minutes:
<?php
include "mysql_connection.php";
$pastTime = "2006-06-29 04:09:49";
$sql = "SELECT HOUR(timeDiff) AS hours,"
. " MINUTE(timeDiff) AS minutes FROM ("
. " SELECT TIMEDIFF(NOW(), '".$pastTime."')"
. " AS timeDiff FROM DUAL) subQuery";
print("SQL = $sql\n");
$rs = mysql_query($sql, $con);
while ($row = mysql_fetch_assoc($rs)) {
print("$pastTime was ".$row['hours']." hours, "
. $row['minutes']." minutes ago.\n");
}
mysql_free_result($rs);
$sql = "SELECT (HOUR(timeDiff) DIV 24) AS days,"
. " (HOUR(timeDiff) MOD 24) AS hours,"
. " MINUTE(timeDiff) AS minutes FROM ("
. " SELECT TIMEDIFF(NOW(), '".$pastTime."')"
. " AS timeDiff FROM DUAL) subQuery";
print("SQL = $sql\n");
$rs = mysql_query($sql, $con);
while ($row = mysql_fetch_assoc($rs)) {
print("$pastTime was ".$row['days']." days, "
. $row['hours']." hours, "
. $row['minutes']." minutes ago.\n");
}
mysql_free_result($rs);
mysql_close($con);
If today is you run this script, you will get something like this:
SQL = SELECT HOUR(timeDiff) AS hours, MINUTE(timeDiff) AS minutes FROM ( SELECT TIMEDIFF(NOW(), '2006-06-29 04:09:49') AS timeDiff FROM DUAL) subQuery 2006-06-29 04:09:49 was 115 hours, 2 minutes ago. SQL = SELECT (HOUR(timeDiff) DIV 24) AS days, (HOUR(timeDiff) MOD 24) AS hours, MINUTE(timeDiff) AS minutes FROM ( SELECT TIMEDIFF(NOW(), '2006-06-29 04:09:49') AS timeDiff FROM DUAL) subQuery 2006-06-29 04:09:49 was 4 days, 19 hours, 2 minutes ago.
Warning again, this script only works if the past time is less than 33 days ago.
2007-05-11, 7370👍, 0💬
Popular Posts:
Can Two Forms Be Nested? - XHTML 1.0 Tutorials - Understanding Forms and Input Fields Can two forms ...
What is the output of printf("%d")? 1. When we write printf("%d",x); this means compiler will print ...
How to create arrays in JavaScript? We can declare an array like this var scripts = new Array(); We ...
What is the difference between const char* p and char const* p? In const char* p, the character poin...
What are the different accessibility levels defined in .NET ? Following are the five levels of acces...