[SOLVED] Can someone tell me whats wrong here

[affordit]

<?php

$name="Patrick";

include("k_falls_dbinfo2.inc.php");

mysql_connect(mysql,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

 

$query = "INSERT INTO test (name, time_submitted) VALUES ('$name', now())";

mysql_query($query);

mysql_free_result( $result );

mysql_close();

echo "All Done.";

$st = strtotime("+1 week");

$dt = new DateTime("@$st");

echo ('$dt')

?>

[p2grace]

First of all you should always list what the problem is so we know where to look.

 

Are the variables $username and $password set?

 

also try updating your code to this:

 

<?php
$name="Patrick";
include("k_falls_dbinfo2.inc.php");
mysql_connect(mysql,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO `test` (`name`, `time_submitted`) VALUES ('$name', NOW())";
mysql_query($query);
mysql_close();
echo "All Done.";
$st = strtotime("+1 week");
$dt = new DateTime("@$st");
echo "$dt";
?>

 

Note, the query will not work if time_submitted is not set to timestamp, datetime, or date in mysql.

 

The biggest error I saw was the echo statement was like this echo ('$dt') which the variable is never being shown because you were using single quotes, and you were missing the semi-colon at the end.  The semi-colon at the end wouldn't break it unless there's more code after the echo statement.

[affordit]

It still does not echo $dt

[p2grace]

Ah right, try this:

 

$st = date("m/d/Y",strtotime("+1 week"));
echo $st;

[affordit]

SWEEEET Thanks p2grace that got the job done.

You have a great day

[p2grace]

No problem, glad I could help

 

Would you mind hitting the Topic Solved button, Thank you

[KrisNz]

$dt is an object that cant be converted to a string. you probably want something like

echo $dt->format("Y-m-d H:i:s");