ben

Hi,

I have a problem with this query:


$link = mysql_connect("localhost", "root", "pw");
$dbsel = mysql_select_db('mydb', $link);
$lol='bibi';

$sql = "select count(*) as totuur from studres where where dag>= curdate()
AND logon='" . $lol . "'";
$result = mysql_query($sql);
$row = mysql_fetch_field($result,0);
// J'ai essayé avec: $row = mysql_fetch_row($result);
$totu=$row[0];


The error is on line: $row = mysql_fetch_field($result,0):
Warning: mysql_fetch_field(): supplied argument is not a valid MySQL result
resource in c:\Inetpub\wwwroot\myfile.php on line 74


Thanks
Ben

kimmo laine

You have WHERE twice in "...from studres where where dag...", try what
happens if you remove the extra where. This is one of the reasons why sql
keywords should be written in CAPITAL letters. It's easier to spot the error if you write:
"SELECT COUNT(*) AS totuur FROM studres WHERE WHERE dag>= CURDATE() AND
logon='$lol'"

--
"En ole paha ihminen, mutta omenat ovat elinkeinoni." -Perttu Sirviö
antaatulla.sikanautaa@gmail.com.NOSPAM.invalid

jerry stuckle

And if you would have checked the result of the mysql_query() call, you
would have found your problem.

$result = mysql_query($sql);
if ($result) {
$row = mysql_fetch_field($result,0);
// J'ai essayé avec: $row = mysql_fetch_row($result);
$totu=$row[0];
}
else
echo "MySQL error: " . mysql_error();


--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================

kimmo laine

You have WHERE twice in "...from studres where where dag...", try what
happens if you remove the extra where. This is one of the reasons why sql
keywords should be written in CAPITAL letters. It's easier to spot the error if you write:
"SELECT COUNT(*) AS totuur FROM studres WHERE WHERE dag>= CURDATE() AND
logon='$lol'"

--
"En ole paha ihminen, mutta omenat ovat elinkeinoni." -Perttu Sirviö
antaatulla.sikanautaa@gmail.com.NOSPAM.invalid

jerry stuckle

And if you would have checked the result of the mysql_query() call, you
would have found your problem.

$result = mysql_query($sql);
if ($result) {
$row = mysql_fetch_field($result,0);
// J'ai essayé avec: $row = mysql_fetch_row($result);
$totu=$row[0];
}
else
echo "MySQL error: " . mysql_error();


--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================