Return Results error

[Knightime]

Hello all, I am trying to build a query that a user can do by selecting something in a drop-down box. I am getting a "line 1" syntax error though on the return results code. Can someone check out the code for obvious errors please and reply with any ideas on how to write if correctly?

Here is the code for the select form with the drop-down box that seems to work fine.

<form action="driverraceresults.php" method="post">
<table cellpadding="0" cellspacing="5" border="0" width="450">
<tr><td class="scheduleheader">View a driver's Race Results</td></tr>
<tr><td> 
<?php
$sql = "SELECT Driver FROM season04 ORDER by Driver ASC";
$results = mysql_query($sql);
print("<select width=\"150\" style=\"width: 150px\" name=\"select\">\n" );
while ($row = mysql_fetch_array($results)):
$name = $row["Driver"];
print("<option value=\"$name\">$name\n" );
endwhile;
print("</select>\n" );
?> 
</td><tr>
<td align="center"><br /><input class="lefttext" type="submit" name="submit" value="Submit"></td>
</tr> 
</table>
</form>

Here is the code for the return results page as listed in the "action" for the form up above:

print"<table class=\"userstats\" border=\"1\" style=\"width:85%;\">
<tr class=\"scheduleheader\"><td>Finish</td><td>Start</td><td>Car#</td><td>Driver</td><td>Interval</td><td>Laps</td><td>Led</td>
<td>Pts</td><td>Status</td></tr>";
$query=" SELECT * FROM nc04, lv04, atl04 WHERE Driver = " .$_REQUEST['select'];
$result=mysql_query($query);
if (!$result) { 
echo("<p>Error performing query: " . mysql_error() . "</p>" ); 
exit(); 
}
if ($row = @mysql_fetch_array($result, MYSQL_ASSOC)) {
print"<tr class=\"scheduledata\"><td>$row[Finish]</td><td>$row[Start]</td><td>$row[Car_Number]</td><td>$row[Driver]</td>
<td>$row[Interval_Difference]</td><td>$row[Laps]</td><td>$row[Led]</td><td>$row[Points]</td><td>$row[Status_End]</td></tr>";
}
print"</table>";
?> 

I am new to this type of function so please be gentle, lol

Thank you, Troy

[Gybbyl]

<?php
while ($row = mysql_fetch_array($results)):
$name = $row["Driver"];
print("<option value=\"$name\">$name\n" );
endwhile;
?> 

I don't believe you can access an array index in this manner: $name = $row["Driver"]; without using mysql_fetch_assoc() instead of mysql_fetch_array. Don't quote me on that, but try using fetch_assoc instead of fetch_array

[-Ice-php]

<?php
$name = array();
$max = mysql_num_rows($results);
while ($row = mysql_fetch_array($results)):
$name[] = $row["Driver"];
endwhile;

for($i = 0; $i<$max; $i++){
echo "<option value=\"$name[$i]\">$name[$i]\n";
}

Im not quite sure but the while statement just keeps replaces each <option> statement
just try it and see what happens :S


?>

[Knightime]

Guys, thanks for the replies. The problem is not with the drop-down at all. It passes fine. I only put it up for reference. The problem is the return results coding.

somewhere in here:

$query=" SELECT * FROM nc04, lv04, atl04 WHERE Driver = " .$_REQUEST['select'];

I am getting the error.

[Knightime]

Also, if there is a Driver name in the database with an * or other html special characters, how do I escape it in the submit form?

[Knightime]

Anyone have an idea how to troubleshoot or fix the error?

[Sillysoft]

Anyone have an idea how to troubleshoot or fix the error?

If you feel this line is giving the error:

$query=" SELECT * FROM nc04, lv04, atl04 WHERE Driver = " .$_REQUEST['select'];

Then try this:

$query = "SELECT * FROM nc04,lv04,atl04 WHERE Driver = $_REQUEST[select]";

Since the query is in double quotes it will still read $_REQUEST[select]

P.S. Add single quotes around $_REQUEST[select] if it isnt a number

Silly