PHP / MySQL error

[lambada]

Little error in my PHP code I can't for the life of me figure out.
Anyone out there know how to solve it??

Heres the error:

Catchable fatal error: Object of class mysqli could not be converted to string in /home/lambada/public_html/secretconfessions/dev/includes/entry/view.php on line 5

Here's the code

$db = mysqli_connect("localhost", "USER", "PASS", "DB") or die("Database Unavailable");
$query = "SELECT id, text, datetime, ipaddress, user FROM entries WHERE moderated = 1 ORDER BY id DESC";
$selectentries = mysqli_query("$db", "$query"); 

Any ideas?

Thanks

lambada (BTW: I removed the db details. I'm not THAT idiotic :P

EDIT: Apparently I am - I forgot to say which was line 5. *looks sheepish* Line 5 is the

$selectentries = mysqli_query("$db", "$query"); 

line

[Brandon]

Which line is line 5?

[Bryon]

I don't have time to test this out for myself.. But.. I believe I know what's going on.

$selectentries = mysqli_query("$db", "$query"); 

$db contains a 'mysqli-object.' In your mysqli_query(), you're putting that variable's value between double-quotes, which would make it look like this when PHP attempts to use it:

 $selectentries = mysqli_query("[MySQLi Object]", "$query"); 

It's attempting to convert the mysqli object to a string, which.. Cannot be done.

Try to take out the quotes around $db and see if you still get the error.

[Micro]

Yeh, just remove the quotes from the $db and $query, they are variables so they can just be there on their own.

[lambada]

Ok, I did what you guys asked and it didn't print out anything, then I modified my code to use a while loop instead of foreach at Brandons suggestion, and it still didn't work. The final piece of the puzzle was that I was using mysqli_fetch_row which returns numerically indexed arrays, when I was referring in my code to string indexed arrays so I changed it to mysqli_fetch_assoc and it works like a charm.

Thanks for all your help guys.

[Zenax]

well i always thought a query would be done like this, but then again im a noob to PHP so im not entirely sure!

include("conn.inc.php");

$connection = mysql_connect($hostname,$username,$password)
    or die("Sorry Cannot Connect");
    
$dbselect = mysql_select_db("$dbname",$connection)
    or die("Sorry Cannot Connect to the db");

$style = mysql_query("SELECT stylesheet FROM style");        // Queries the database for the stylesheet table 
while ($stylerow = mysql_fetch_array($style,MYSQL_ASSOC))     // Chooses the row

print $stylerow{'stylesheet'};                                // Prints out the stylesheet to the page 

Please correct me if I am wrong, as I am still learning PHP so the more I learn the better I get at helping people with their problems!

[t2t2t]

well i always thought a query would be done like this, but then again im a noob to PHP so im not entirely sure!

$username = "root";
$password = "1234";
$hostname = "localhost";

$db_conn = mysql_connect($username, $password, $hostname)
       or die("Cannot connect!");

$query = "SELECT id, text, datetime, ipaddress, user FROM entries WHERE moderated = 1 ORDER BY id DESC";

$sql = mysql_select_db($query, $db_conn)
      or die("Cannot Connect!"); 

Please correct me if I am wrong, as I am still learning PHP so the more I learn the better I get at helping people with their problems!

You gotta work on it...

<?
mysql_connect('HOST','USERNAME','PASSWORD') or die("Cannot connect!");
mysql_select_db('table') or die("Cannot Connect!");
$query = "SELECT id, text, datetime, ipaddress, user FROM entries WHERE moderated = 1 ORDER BY id DESC";
$answer = mysql_query($query);