More PHP, ;) [Archive] - ForumsHQ

[Durin]

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in D:\Xitami\webpages\newphp2.php on line 37



this is an error i get when i run a php program. This is part of an online tutorial and i dont know if the database is wrong or what. I have tried to find some more information on it, but i suppose i cannot follow simple instructions...any help would be nice...thanks for bearing with me guys ;).

[Durin]

\n";

echo "NamePosition\n";



while ($myrow = mysql_fetch_row($result)) {

printf("%s %s%s\n",

$myrow[1], $myrow[2], $myrow[3]);

}



echo "\n";



?>











thats the code for the page...that might help ;)

[War Hawk]

lemmi give it a lookie see....





put around your damn code so the BB displays it right....

[War Hawk]

my bad... I'm a dumbass... just ignore my stupidity... lemmi look at it here

[War Hawk]

first off DON'T USE PRINT!! use ECHO... its MUCH MUCH MUCH better!!!

[War Hawk]

try using mysql_fetch_array($result) instead of mysql_fetch_row($result)

[Durin]

heh i figured that part out now...i didnt fill the database, duh ;)



but now i have a new prob: this one is another one that should all me to start using the PHP_SELF function, that allows the page to call itself





Notice: Undefined variable: id in D:\Xitami\webpages\newphp2.php on line 15



Notice: Undefined variable: PHP_SELF in D:\Xitami\webpages\newphp2.php on line 41

Bob Smith



Notice: Undefined variable: PHP_SELF in D:\Xitami\webpages\newphp2.php on line 41

John Roberts



Notice: Undefined variable: PHP_SELF in D:\Xitami\webpages\newphp2.php on line 41

Brad Johnson





whats that all about?? code for page is as follows:













", $myrow["first"]);



printf("Last name: %s\n", $myrow["last"]);



printf("Address: %s\n", $myrow["address"]);



printf("Position: %s\n", $myrow["position"]);



} else {



// show employee list



$result = mysql_query("SELECT * FROM employees",$db);



if ($myrow = mysql_fetch_array($result)) {



// display list if there are records to display



do {



printf("%s %s\n", $PHP_SELF, $myrow["id"], $myrow["first"], $myrow["last"]);



} while ($myrow = mysql_fetch_array($result));



} else {



// no records to display



echo "Sorry, no records were found!";



}



}







?>





















hehe thanks alot ;)

[War Hawk]

hrm $_server['PHP_SELF'] perhaps??

[Durin]

Notice: Undefined variable: id in D:\Xitami\webpages\newphp3.php on line 15



Notice: Undefined variable: _server in D:\Xitami\webpages\newphp3.php on line 41

Bob Smith



Notice: Undefined variable: _server in D:\Xitami\webpages\newphp3.php on line 41

John Roberts



Notice: Undefined variable: _server in D:\Xitami\webpages\newphp3.php on line 41

Brad Johnson





hehe thats the new errors, i dunno. I gotta get to bed now, so i wont be working on it much til tommorow AM, so dont worry about stressing too much. Thanks for the quick replies :D

[Durin]

whoa, fun. I changed it to have 'PHP_SELF' and now the error is:



Notice: Undefined variable: id in D:\Xitami\webpages\newphp3.php on line 15

Bob Smith

John Roberts

Brad Johnson





and the code of the page is:











", $myrow["first"]);

printf("Last name: %s\n", $myrow["last"]);

printf("Address: %s\n", $myrow["address"]);

printf("Position: %s\n", $myrow["position"]);

} else {



// show employee list



$result = mysql_query("SELECT * FROM employees",$db);



if ($myrow = mysql_fetch_array($result)) {

// display list if there are records to display

do {

printf("%s %s\n", 'PHP_SELF', $myrow["id"], $myrow["first"], $myrow["last"]);

} while ($myrow = mysql_fetch_array($result));

} else {

// no records to display

echo "Sorry, no records were found!";

}

}



?>











fun huh? Ok, NOW i am going to bed!



http://www.chadlehmkuhl.slyip.com/newphp3.php

test her out see what it does, :D