Select date in PHP

paratracker

I have tables with DATE or DATETIME fields where queries like:

Select dateField FROM table where <whatever>

return results, but when I fetch an association from the result set, the dateField in the row always produces 1969-12-31.

I know that the dateFields are correctly populating the tables because I can run queries from the MySQL command line and see that the date fields appear as expected, but when I run identical queries within a PHP script, the dates come back 1969-12-31. I have been getting around this situation until now by only using queries with dates in the where clause (where I have no problems), but I've finally run into a situation where I need to retrieve a value from a DATETIME field.

Here's a code snippet:

$query = "SELECT accessDate FROM requests WHERE orderid = '$orderid'";
$result = mysql_query( $query );
$numRes = mysql_num_rows( $result );
if ( $numRes > 0 )
{ $row = mysql_fetch_assoc( $result );
$accessDate = date( "Y-m-d", $row['accessDate'] );
echo "accessDate = $accessDate<br>";
}

Am I missing something?

Thanks, Mike
http://pnmx.com/

bradgrafelman

Please don't hijack other users' threads. I've moved your post into a new thread of your own.
When posting PHP code, please use the board's [PHP][/PHP] bbcode tags - they make the code much, much easier to read (as well as aiding the diagnosis of syntax errors). You should edit your post now and add these bbcode tags to help others in reading/analyzing your code.
If you'll look up the [man]date/man function in the PHP manual, you'll see that it expected to be given a Unix timetamp as its second parameter. A Unix timestamp is an integer (it's the number of seconds since the Unix Epoch).

Since you have a DATE/DATETIME field in SQL, you're obviously not passing the date in the format that PHP's date() function can recognize. But then again, why use PHP? Try using the DATE_FORMAT() function in MySQL instead.