Confuddled

Haktar

can someone please tell me why this isnt working

$totalrows = mysql_num_rows(mysql_query("SELECT * FROM messages WHERE to='$myid'"));

im getting this error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/thephoen/public_html/inbox.php on line 168

rikmoncur

I was getting this too.
I was missing the line:

mysql_select_db($database_yadda, $yadda);

when I added that it worked ok. dunno if it'll help you?

Haktar

thanks for the reply but i have this at the top of the page

mysql_connect("localhost", "*", "") or die("There was an error connecting to the mysql server.");
mysql_select_db("**");

ClarkF1

Have you tried separating it out.....it's easier to find at which stage the problem is.

$query =  "SELECT * FROM messages WHERE to='$myid'";
$result = mysql_query($query);
$totalrows = mysql_num_rows($result);

You could even try:

$query =  "SELECT * FROM messages WHERE to='".$myid."'";

Haktar

🙁 still didnt work could it be because i have

mysql_connect("localhost", "*****", "******") or die("There was an error connecting to the mysql server."); 
mysql_select_db("*****"); 

$username = $_SESSION['MM_Username'];

$result = mysql_query("SELECT * FROM users WHERE username='$username'"); 

$r = mysql_fetch_assoc($result);

$rank=$r['rank']; 
$turns=$r['turns']; 
$bankcredits=$r['bankcredits'];
$messages=$r['messages'];
$credits=$r['credits'];
$myid=$r['id'];

at the top of the page

Houdini

Where on that page did you start the session for $SESSION['MM_Username'] to work?
session_start() must be called somewhere for you to be able to use the $SESSION variables.

Haktar

at the very top i have

<?php
//initialize the session
if (!isset($_SESSION)) {
session_start();
}

ahundiak

$sql = "SELECT * FROM users WHERE username='$username'";
echo $sql; /* Make sure you have the expected statement */

$result = mysql_query($sql) or die(mysql_error()); /* See what mysql thinks about it */

Haktar

thanks 😃, this is what i got on the page

SELECT * FROM users WHERE username='haktar'
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/thephoen/public_html/inbox.php on line 182

leatherback

Haktar,

what is the code you are using? According to your error, you have moved away from trying to get a number of rows, and are now getting an array?

Anyway, the error is telling you that the variable you used in mysql_fetch_array is
- Not the result of a mysql_query()
or
- The query-string in the query was not valid.

Haktar

hi 🙂, just now this is what i have

$limit = 20;//Limit of results here
$page = $_GET['page'];//Gets the page
echo $myid;

if(empty($page))//If the page is empty
{
$page = '1';//sets the page to 1
};
$start = ($page-1)*$limit;//set the start page
$start = round($start,0);//rounds it
$result=mysql_query("SELECT * FROM messages WHERE to='$myid' ORDER BY timesent ASC LIMIT $start, $limit

while ($r = mysql_fetch_array($result)) 
{ 

$sentusername=mysql_query("SELECT username FROM users WHERE id='$r[id]'");

echo "<table>"; 
echo "<tr> 
<td width=195 height=9><a href=profile.php?id=$r[id]>$sentusername</a></td>
<td width=205>$r[subject]</td>
<td width=160>$r[timesent]</td>
<td width=83></td>
<td width=92></td>
<td width=98></td>
<td></td></tr>"; 
}; 
echo "</table>";

leatherback

First of all: Get an editor with syntax highlighting. I am sure you have spotted an error in your code now:

$result=mysql_query("SELECT * FROM messages WHERE to='$myid' ORDER BY timesent ASC LIMIT $start, $limit

is not ended properly

$result=mysql_query("SELECT * FROM messages WHERE to='$myid' ORDER BY timesent ASC LIMIT $start, $limit");

works better

leatherback

Secondly:

$sentusername

is not a variable you can just use in the text.

You have to use a construct such as mysql_fetch_array($sentusername) to get the actual variables. (See www.bio-vision.nl/coding/PhP.php if you need more explanation)

Haktar

ill check that out now thanks 🙂, i had actually ended that line properly but i copied it in here in two sections and mised the end 🙁 sorry

Haktar

leatherback wrote:
Secondly:

$sentusername

is not a variable you can just use in the text.

You have to use a construct such as mysql_fetch_array($sentusername) to get the actual variables. (See www.bio-vision.nl/coding/PhP.php if you need more explanation)

sorry but im not following this part, cant i do it like that if it should only return one value

ahundiak

Haktar wrote:
thanks 😃, this is what i got on the page

SELECT * FROM users WHERE username='haktar'
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/thephoen/public_html/inbox.php on line 182

Please read what I posted more carefully. The above message did not come from:

$result = mysql_query($sql) or die(mysql_error());

The problem seems to lie in your query. Queries will not automatically echo out an error message (which in my opinion is one of the greatest mistakes in the mysql implementation). . Instead, you only get an error when you try to use the results and hence the 'supplied argument' errors.

You need to explicitly check to see if the query failed or not. That is what the 'or die' does.

Haktar

Hi, just tried this

$limit = 20;//Limit of results here
$page = $_GET['page'];//Gets the page
echo $myid;

if(empty($page))//If the page is empty
{
$page = '1';//sets the page to 1
};
$start = ($page-1)*$limit;//set the start page
$start = round($start,0);//rounds it
$sql = "SELECT * FROM users WHERE username='$username'"; 
echo $sql; /* Make sure you have the expected statement */ 

$result = mysql_query($sql) or die(mysql_error()); /* See what mysql thinks about it */ 

while ($r = mysql_fetch_array($result)) 
{ 

$sentusername=mysql_query("SELECT username FROM users WHERE id='$r[id]'");

echo "<table>"; 
echo "<tr> 
<td width=195 height=9><a href=profile.php?id=$r[id]>$sentusername</a></td>
<td width=205>$r[subject]</td>
<td width=160>$r[timesent]</td>
<td width=83></td>
<td width=92></td>
<td width=98></td>
<td></td></tr>"; 
};

and i didnt get any errors, but it didnt work i got this

7SELECT * FROM users WHERE username='haktar'

Resource id #5

Resource id #5 ( this is where $sentusername should be)

Haktar

thanks very much everybody, got it to work 😃