problem with select count(*)

[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]

"Ben" <sdvs@qcsq> wrote in message
news:M8********************@scarlet.biz...

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 . "'";

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
an*******************@gmail.com.NOSPAM.invalid

[Jerry Stuckle]

Kimmo Laine wrote:

"Ben" <sdvs@qcsq> wrote in message
news:M8********************@scarlet.biz...

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 . "'";

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'"

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();
--
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Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
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