please help with error...

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December 5th, 2002, 12:31 AM

papasmurf

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please help with error...

i just did this basic mysql and php tutorial/article: http://www.devarticles.com/art/1/23/3

and got this error. I have no idea about it other than maybe a dir basic i missed....and help would be appreciated.
Here is the error I got when trying to view the file:

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /home/essenmin/public_html/ope.php on line 4

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/essenmin/public_html/ope.php on line 5

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/essenmin/public_html/ope.php on line 6
Name:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/essenmin/public_html/ope.php on line 7
Address:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/essenmin/public_html/ope.php on line 8
Home Number:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/essenmin/public_html/ope.php on line 9
Work Number:

December 5th, 2002, 01:03 AM

AmericanD

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Lets see your code

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December 5th, 2002, 02:02 AM

Ben Rowe

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Just making a educated guess here, but i think your providing the wrong database details

December 5th, 2002, 10:31 AM

papasmurf

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i tried two things:
where it says "essenmin_myDB" i also tried just myDb
and
in the printf's i tried "$result,0,... as well as $result,1
both alterations give same results
<code>
<html>
<body>
<?PHP
$db = @mysql_connect("localhost", "root");
mysql_select_db("essenmin_myDb",$db);
$result = mysql_query("SELECT * FROM address",$db);
printf("Name: %s \n", mysql_result($result,1,"name"));
printf("Address: %s \n", mysql_result($result,1,"address"));
printf("Home Number: %s \n", mysql_result($result,1,"hnum"));
printf("Work Number: %s \n", mysql_result($result,1,"wnum"));
?>
</body>
</html>
</code>

December 5th, 2002, 11:08 AM

papasmurf

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Join Date: Dec 2002

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Figured it out..."root" was wrong...needed my login info.
thanks tho, hope it didn't cause any headaches

December 7th, 2002, 11:19 PM

papasmurf

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Join Date: Dec 2002

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please help with error...(new error)

another error i can't figure out :P help paweese...
and i dont know if it is actually line 17 do to wrapping and such on my post, but it refers to:
$result = mysql_query("SELECT * FROM address",$db);
if ($myrow = mysql_fetch_array($result))
{ ...

[error/output]
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/essenmin/public_html/matt/ope.php on line 17
Sorry, no records were found!
[/end output]

[code]
<html>
<body>
<?PHP
$db = mysql_connect("localhost", "*****", "******");
mysql_select_db("essemin_myDb",$db);
// display individual record
if ($id)
{
$result = mysql_query("SELECT * FROM address WHERE id=$id",$db);
$myrow = mysql_fetch_array($result);
printf("Name: %s\n ", $myrow["name"]);
printf("Address: %s\n ", $myrow["address"]);
printf("Home Number: %s\n ", $myrow["hnum"]);
printf("Work Number: %s\n ", $myrow["wnum"]);
} else
{
// show employee list
$result = mysql_query("SELECT * FROM address",$db);
if ($myrow = mysql_fetch_array($result))
{
// display list if there are records to display
do
{
printf("<a href=\"%s?id=%s\">%s</a> \n", $PHP_SELF, $myrow["id"], $myrow["name"]);
} while ($myrow = mysql_fetch_array($result));
} else
{
// no records to display
echo "Sorry, no records were found!";
}
}
?>
</body>
</html>

Last edited by papasmurf : December 9th, 2002 at 12:20 AM.

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