quick sql question

[willwill100]

is this code valid? because it is rejected with a mysql error

[code]
$sssn = "INSERT INTO '$comp' ('Position') VALUES ('$score') WHERE 'Sail Number' = '$bnum'";
[/code]

if not what is wrong?

thanx

[shocker-z]

I think your looking for

[code]
$sssn = "INSERT INTO `$comp` (`Position`) VALUES ('$score') WHERE `Sail Number` = '$bnum'";
[/code]

are you sure u have a field with a space in in your table? (Sail Number)

[keeB]

[code]
$debug = true;
$sssn = "INSERT INTO `$comp` (`Position`) VALUES ('$score') WHERE `Sail Number` = '$bnum'";
if($debug) print $sssn;

mysql_query($sssn) or die ("Unable to insert: " . mysql_error());
[/code]

Replace this code with yours.. and post the output! :)

[willwill100]

thanx for the debugging help guyys but ive worked out that the problem was much earlier in the program!

i keep getting the error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\Will\My Documents\xampp\htdocs\sailing\rescalc.php on line 34

with this code below:

[code]$comp = $_REQUEST['comp'];
$snum = $_REQUEST['sailnum'];
$time = $_REQUEST['time'];

$result = mysql_fetch_array("SELECT `First Name` , `Last Name` FROM members WHERE `Sail Number` = '$snum'");
[/code]

i would ideally like to stick the first name and last name into separate variables, how can i do this??

[shocker-z]

[code]$comp = $_REQUEST['comp'];
$snum = $_REQUEST['sailnum'];
$time = $_REQUEST['time'];

$result = mysql_fetch_array("SELECT `First Name` , `Last Name` FROM members WHERE `Sail Number` = '$snum'");[/code]

Can you run the query straight inside the fetch array??? try this

[code]$comp = $_REQUEST['comp'];
$snum = $_REQUEST['sailnum'];
$time = $_REQUEST['time'];
$query=mysql_query("SELECT `First Name` , `Last Name` FROM members WHERE `Sail Number` = '$snum'");
$result = mysql_fetch_array($query);[/code]


That work better?

[willwill100]

sorted! thanx.

is there any reason that it works like that or is it just a random feature??

[shocker-z]

well i *think* you could run...

$result = mysql_fetch_array(mysql_query("SELECT `First Name` , `Last Name` FROM members WHERE `Sail Number` = '$snum'"));

The reason it wasn't working the way you did it is because you had not rean a query you was masicaly just doin mysql_fetch_array on the text/string "SELECT `First Name` , `Last Name` FROM members WHERE `Sail Number` = '$snum'" instead of on the query results.

[willwill100]

im having a similar problem again:

this is the code concerned:
[code]
$q= "SELECT * FROM `temp` ORDER BY `score` ASC";
$result = mysql_query ($q) or die('Problem with query: ' . $q . '<br />' . mysql_error());
$count = 1;
while ($row = mysql_fetch_assoc($result)){
$nom=$row['Name'];     

$query = "UPDATE `temp` SET `Position` = '$count' WHERE `name`='$nom'";
$result = mysql_query($query);

$count = $count + 1;
}
[/code]

i get the error message:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\Will\My Documents\xampp\htdocs\sailing\handicapcalc.php on line 30

i would have thought that because i was calling the "fetch_array" from a query this time it would have worked but obviously not. Any help!?!?!

[willwill100]

no1?? please could someone give me a hand; it is much needed, lol!!

[kenrbnsn]

Is the line quoted in the error message (line 30) the line you showed in your code snippet?

Ken

[txmedic03]

Variables inside of single quotes are not processed properly as far as I am aware. "SELECT * FROM `table` WHERE `column_name` ='".$variable."'"

[kenrbnsn]

Actually, in this case:
[code]<?php $query = "UPDATE `temp` SET `Position` = '$count' WHERE `name`='$nom'"; ?>[/code]
the variable is really inside the double quotes and do get translated.
You can prove it for yourself bu putting an [code]<?php echo $query; ?>[/code] on the next line.

Ken

[txmedic03]

Learn something new everyday. :-)

[Barand]

[!--quoteo(post=355702:date=Mar 16 2006, 10:17 PM:name=kenrbnsn)--][div class=\'quotetop\']QUOTE(kenrbnsn @ Mar 16 2006, 10:17 PM) [snapback]355702[/snapback][/div][div class=\'quotemain\'][!--quotec--]
Is the line quoted in the error message (line 30) the line you showed in your code snippet?

Ken
[/quote]

As the error message concerns "mysql_fetch_array" and the snippet uses "mysql_fetch_assoc" then I think ken's question is very relevant.