I have this code:
$db_billing=mysqli_connect(localhost,metheuser,mypass,billing);
if (mysqli_connect_error()) { die("Can't connect: " . mysqli_connect_error()); }
mysqli
//$dbname = 'billing';
$sql = "SHOW TABLES";
$result = mysql_query($sql); // line 53
Now mysql, What are you doing?
Yes. 3 lashing. Thanks. I am not likely to neglect again remembering that mysql and mysqli are different and have different syntax.
unfortunately I am still in over my head enough to have to ask..
Here is what I have now:
$db_billing=mysql_connect(localhost,metheuser,mypass,billing);
if (!$db_billing) { die('Could not connect: ' . mysql_error()); }
$sql = "SHOW TABLES";
$result = mysql_query($sql);
foreach(mysql_fetch_assoc($result) as $k => $v) { //line 62
$ssql = "DESCRIBE ".mysql_real_escape_string($v);
$rresult = mysql_query($ssql);
echo "<b>".$k."</b>:<br />\n";
echo "<pre>\n";
print_r(mysql_fetch_assoc($rresult));
echo "</pre>\n";
echo "<br />\n";
}
giving this error:
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/meee/public_html/somedir/test.php on line 62
I read about:
-mysql_fetch_assoc
-mysql_query
-SHOW TABLES
but do not see why this should be failing. Why isn't $result a ' valid MySQL result resource'?
-G