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

[i4VisualMedia]

Hi guys,
I am fairly new to PHP and mySQL and am still learning the ropes.

I am getting the following error:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/i4visua/public_html/cmsadmin/editmembers.php on line 84

the purpose of the script is to show the just the names of all the companies in the database using a loop.

My code is as follows:

PHP Code:

82    $query = "SELECT * FROM 'members'";
83    $result = mysql_query($query);
84    $num_results = mysql_num_rows($result);
85    
86    for ($i=0; $i <$num_results; $i++)
87    {
88    $row = mysql_fetch_array($result);
89    echo '<br /> ';
90    echo stripslashed($row['company']);
91    } 


Any help would be much appriciated

Regards
Pete

[MRMAN]

instead of using the mysql_num_rows you can do it like this

PHP Code:

$result = mysql_query($query);

while($data = mysql_fetch_array($result))
{
print $data["company"];
} 


but i think the problem is the quotes round the table name,

to see if there is a error do this

PHP Code:

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


[i4VisualMedia]

thankyou very much.
I have one more question, I am trying to pass a varible (namely the ID of the database entry) between two pages using the following:

PHP Code:

$output = "<a href='memberedit?id=".$row['id']."'>Edit</a>";
echo $output; 


When the link is clicked, I am being taken to the address:
www.mydomain.com/memberedit?id=
note the missing ID.
Can anybody help?
incase you need it, my query is as follows:

PHP Code:

$query = "SELECT ID,company FROM `members`"; 


Cheers guys
Pete

[MRMAN]

SELECT ID

$row['id']

id is not the same as ID

it should be $row['ID']

[Tyree]

Man, that seems familiar! Oh yeah, it's because the same exact thing is in your other thread.

Hope you got it straight, bro!