Hi there,

I am new to Php and MySql.

However, i have been learning for some time now.

I get the above error and my line 77 has the following statement:
$result=mysql_query($query,$conn);

Any ideas? If you need more of the code please let me know.

Thanks a lot,

Sofia

You have an error in your query. It isn't returning a valid resource. Print out $query, execute it in phpmyadmin/mysql. Then you ll know what's the error. You can also give die(mysql_error());
ie., $result=mysql_query($query,$conn) or die(mysql_error());

Hi, I have found that error now. i called the $conn in the included file $connection.

I feel rather daft now, but at least i have a different error to play with now :o)

Thanks for the rapid response

You are welcome!

Hi nav33n,

coul you please help me.. im trying to display quiz questions from database but i get an the same error as desribed above . please see my code

the error points to line 3 and 7.

1 <?php

2 include("contentdb.php");

3 $display = mysql_query("SELECT * FROM $table ORDER BY id",$db);

4 if (!$submit) {


echo "<form method=post action=$PHP_SELF>";
echo "<table border=0>";

7 while ($row = mysql_fetch_array($display)) {


Please please please... i desperately need your help.

Thank in advance

Select * from $table ? What does $table have ? It must be empty.

my table is defined in config. file

<?
$database = "quiz";
$hostname = "localhost";
$table = "quiz";
?>
as you see table is called quiz and it has data in it.

Shall I recall
$display = mysql_query("SELECT * FROM $table ORDER BY id",$db);

to

$display = mysql_query("SELECT * FROM $quiz ORDER BY id",$db); ???

thanks

Nope. Just print out the query. ie., instead of do,
Execute your query in phpmyadmin/mysql console and see what's the problem. Or, you can also try putting die statement after mysql_query to see the error.