Warning: mysql_fetch_assoc():

[sylesia]

Ok, first off, thanks for all the help you guys have been giving me, I truely appreciate it.  Now I have a question I really cannot see what went wrong with it...

[code]
if ($var <10)
{
$request = "SELECT $try FROM bluecard WHERE UserName = '$userName' AND CardNumber = $cardNumber";
$eval = mysql_query($request, $conn);
while($row = mysql_fetch_assoc($eval)){
$$try = $row[$try]."\n";
}
}

else{
//echo $var . $try . "          ";
$request = "SELECT $try FROM dimensions WHERE UserName = '$userName' AND CardNumber = $cardNumber";
$eval = mysql_query($request, $conn);
// echo $request;
while($row = mysql_fetch_assoc($eval)){
$$try = $row[$try]."\n";
echo $$try;
}
}
[/code]

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in http://localhost/srh/bluecardreview.html on line 141


Thing is, it seems to just hate the dimensions table, and I cannot figure out why.  The if works fine, and when I output the parts in the else, everything is what I expect and want, yet that error shows.  No idea whats going on or why...
Oh, and to note, the variable $try is simply a string where the value is the new variable.  IE... $try = 'Loyalty', than $$try is actually $Loyalty. Saves me alot of redundant coding.

[fert]

change
[code]
$eval = mysql_query($request, $conn);
[/code]
to
[code]
$eval = mysql_query($request, $conn) or die(mysql_error());
[/code]

[sylesia]

You gotta be kidding me...
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'AS FROM dimensions WHERE UserName = '1' AND CardNumber = 3' at line 1

Is there anyway around that?  I need the AS title if at all possible since its the accepted value

[fert]

change this
[code]
$request = "SELECT $try FROM dimensions WHERE UserName = '$userName' AND CardNumber = $cardNumber";
[/code]
to this
[code]
$request = "SELECT $try FROM dimensions WHERE UserName = '$userName' AND CardNumber = '$cardNumber'";
[/code]

and this
[code]
$request = "SELECT $try FROM bluecard WHERE UserName = '$userName' AND CardNumber = $cardNumber";
[/code]
to this
[code]
$request = "SELECT $try FROM bluecard WHERE UserName = '$userName' AND CardNumber = '$cardNumber'";
[/code]

[sylesia]

ok, actually all I had to do was put ' ' around $try and it worked.  But now another issue, seems it only will use the else statement twice.
[code]
<?
$conn = mysql_connect("localhost", "zzz", "xxx");
mysql_select_db("evaluation", $conn);

$var = 0;

while ($var < 40)
{
switch($var){
case 0:
$try = 'UserName';
break;
case 1:
$try = 'CardNumber';
break;
case 2:
$try = 'Date';
break;
case 3:
$try = 'UnitPosition';
break;
case 4:
$try = 'Spot';
break;
case 5:
$try = 'Unit';
break;
case 6:
$try = 'SummarySeen';
break;
case 7:
$try = 'Sustain';
break;
case 8:
$try = 'Improve';
break;
case 9:
$try = 'EvaluateeName';
break;
case 10:
$try = 'UserName';
break;
case 11:
$try = 'CardNumber';
break;
case 12:
$try = 'Loyalty';
break;
case 13:
$try = 'Duty';
break;
case 14:
$try = 'Respect';
break;
case 15:
$try = 'SelflessService';
break;
case 16:
$try = 'Honor';
break;
case 17:
$try = 'Integrity';
break;
case 18:
$try = 'PersonalCourage';
break;
case 19:
$try = 'ME';
break;
case 20:
$try = 'PH';
break;
case 21:
$try = 'EM';
break;
case 22:
$try = 'CO';
break;
case 23:
$try = 'IP';
break;
case 24:
$try = 'TE';
break;
case 25:
$try = 'TA';
break;
case 26:
$try = 'CM';
break;
case 27:
$try = 'DM';
break;
case 28:
$try = 'MO';
break;
case 29:
$try = 'PL';
break;
case 30:
$try = 'EX';
break;
case 31:
$try = 'AS';
break;
case 32:
$try = 'DE';
break;
case 33:
$try = 'BD';
break;
case 34:
$try = 'LN';
break;
case 35:
$try = 'Overall';
break;
}

if ($var < 10)
{
$request = "SELECT '$try' FROM bluecard WHERE UserName = '$userName' AND CardNumber = $cardNumber";
$eval = mysql_query($request, $conn) or die(mysql_error());
while($row = mysql_fetch_assoc($eval)){
$$try = $row[$try]."\n";
}
}

if ($var > 9)
{
echo $$try;
$request = "SELECT '$try' FROM dimensions WHERE UserName = '$userName' AND CardNumber = $cardNumber";
$eval = mysql_query($request, $conn) or die(mysql_error());
while($row = mysql_fetch_assoc($eval)){
$$try = $row[$try]."\n";
}
echo $$try;
}
$var = $var + 1;
}
print $loyalty;
?>
[/code]

It prints out $$try properly the first two times, than it never does again, and it does keep going thru the loop.  The problem arrises in the If If tree at the bottom, I just wanted to show all my code.  I have tried If-Else tree and even the If-Else If, but none of them work

[sylesia]

And a note onto that.  It does actually run thru the second if, but for some reason the $row is apparently always false after the second time, and I am not sure why.

[trq]

You should actually have backticks ` around your $try variable as it may contain the sql reserved word [i]date[/i]. You should also take notice of what fert said and place quotes around $cardNumber.