queries dont work when files are uploaded

grumpy

Hello-
My php pages all work perfectly when they are run on my local computer, but when I upload them to the hosting company server they all fail with the following errors:

"Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in (somepage)"

"Couldn't execute query."

Any ideas why? Thanks
(Its not because of global variables. Frankly, PHP is very frustrating. I don't see what's so great about it. Its much more difficult to work with than Cold Fusion)

cmburns69

I would say you need to examine the connection to MySQL.

The error message you describe says that a connection to the DB was never established in the first place. This can be done with the following code:

$conn = mysql_connect(...);
if(!$conn) {
    echo "Could not connect to the database!";
    die();
}

Without seeing your code (or knowing something about your configuration), I can't really help more.

Tom Brown
Netnexus: Where gamers play
http://www.netnexus.com

grumpy

Hi-
thanks for your help, but I think I'm connected to the database.

Here is a sample test page which is returning a different error, a parse error on line 23. It should return a list of cities. Any other ideas?

<?php

// Default page display
// Connect to the database server
$dbcnx = @mysql_connect("localhost", "username", "password");
if (!$dbcnx) {
echo( "<p>Unable to connect to the " .
"database server at this time.</p>" );
exit();
}
// Select the db_test database
if (! @mysql_select_db("db_test") ) {
echo( "<p>Unable to locate the " .
"database at this time.</p>" );
exit();
}

$sql = mysql_query("SELECT * FROM employers where active = '1' ");

// execute SQL query and get result
$result = mysql_query($sql)
while ($row = mysql_fetch_array($result) ) {

echo (" " . $row["city] . " ");
}
?>

stolzyboy

you are only connecting to mysql, not the database, you need to execute the select database, use $conn = mysql_select_db(), then do the if(!$conn) junk

cmburns69

I see in your code that your querying the database twice. The first time passing it an SQL string, the second time passing it the result set. (You're also missing a semicolon after the second query)

$sql = mysql_query("SELECT * FROM employers where active = '1' "); 

// execute SQL query and get result 
$result = mysql_query($sql) 
while ($row = mysql_fetch_array($result) ) { 

echo (" " . $row["city] . " "); 
}

should be:

$sql = "SELECT * FROM employers where active='1' ";

// execute SQL query and get result
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)) {
    echo(" ".$row["city"]." ");
}

Tom Brown
Netnexus: Where gamers play
http://www.netnexus.com