problem with date()

[ejwf]

Hello
I want to set a variable using date() to use in a later query but it keeps falling over and I get a sql error message!
the code I have got is;

//$today = date('Y-m-d'); // this is the actual code I plan to use but it is commented out because I am testing the code using the line below

$today = date('2004-12-07'); // this line is for testing purposes only

// PART 1 - Send "we are open" email

$body1 = "Hello,
Please note that we are open.
We kindly request you log on.
regards
'$Name'
";

$query = "SELECT Ref FROM tblDate WHERE OpenEarly = " . $today;
$result = mysql_query($query)
	or die(mysql_error());
while ($row = mysql_fetch_array($result));
  $Ref[] = $row["Ref"];

  foreach ($Ref as $value)
  {

$sql = "SELECT Name FROM tblConces WHERE ConcesId == '$Ref'";
$result1 = mysql_query($sql)
	or die(mysql_error());
$row1 = mysql_fetch_array($result1);
  $Name = $row1["Name"];

$sql = "SELECT Email FROM tblUser WHERE Level >= 3 AND Ref == '$Ref'";
$result2 = mysql_query($sql)
	or die(mysql_error());
while($row2 =  mysql_fetch_array($result2))
  $Email[] = $row2["Email"];

foreach($Email as $value)
{
mail ("$Email", "subject", "$body1");
}

}

but I keep getting the error message You have an error in your SQL syntax near '== 2004-12-07' at line 1 which is the line $query = "SELECT Ref FROM tblDate WHERE OpenEarly = " . $today;
Note - I get the same error message if I use the actual date() code $today = date('Y-m-d'); and change the date on the database when I try to make the code work

any ideas where I am going wrong?

[devinemke]

what exact field type is OpenEarly? i assume it is a standard DATE field. the [man]date[/man] functions purpose is to format a unix timestamp into a readable date. if you simply what to format today's date then do:

$today = date('Y-m-d');

then your query would be:

$query = "SELECT Ref FROM tblDate WHERE OpenEarly = '$today'";

[ejwf]

thanks for your reply

the field OpenEarly is a standard mysql date field formatted to YYY-MM-DD

I have changed the code to

// have set date to
$today = date(Y-m-d);

$query = "SELECT Ref FROM tblDate WHERE OpenEarly = '$today'";
 

but I still get the error message
You have an error in your SQL syntax near '== 'Array'' at line 1

but if I just leave the date variable as

$today = date();

I get the error message
Warning: Wrong parameter count for date() in /home/---------/-/-/-/xxx.php on line 9
You have an error in your SQL syntax near '== 'Array'' at line 1

any ideas?
thanks

[devinemke]

your forgot the quotes:

$today = date('Y-m-d');

[ejwf]

thanks for that - have put in the single quotes but still get the same error message
You have an error in your SQL syntax near '== 'Array'' at line 1

[devinemke]

what do you see when you do this:

$today = date('Y-m-d');
print_r($today);

[ejwf]

I get

2004-12-13

[devinemke]

Originally posted by ejwf
I get

2004-12-13

that's rather odd considering today is 2004-12-23

[ejwf]

sorry typo....
yes I get the correct date.....

[devinemke]

what do you see when you do:

$today = date('Y-m-d');
$query = "SELECT Ref FROM tblDate WHERE OpenEarly = '$today'";
$result = mysql_query($query);
if ($result) {exit('OK');} else {exit(mysql_error());}

[ejwf]

i get

OK

[devinemke]

then you query works fine, your problem lies elsewhere

[ejwf]

any ideas where to look first?

TIA