Re: [PHP] Help with msql_fetch_array()

[Matthew Bielecki]

Well I think you were correct about not connecting to the db, but I don't 
understand why.  I wrote another little script just to test the 
connection.

<?php

$link = mysql_connect("servername","username","password")
      or die("Couldn't make connection.");
$diditwork = mysql_select_db("news", $link);

if ($diditwork <> FALSE)
   {
      echo "got the db..yeahhhh!";
      echo $link;
   }
   else
   {
      echo "didnt get db...booooo";
      echo $link;
   }
?>

This returns "didnt get db...booooooResource id #1

I don't understand how I can get a resource ID but then not be able to use 
the "news" database.  Like I described earlier, this happens with my other 
db as well.  I can connect to the db through the console or any other 
client I have on the physical server and do whatever I want with the db's, 
I'm just having problems with php.

Thanks again for your help!!









PHPCoder <[EMAIL PROTECTED]>
07/24/02 01:50 PM

 
        To:     Matthew Bielecki <[EMAIL PROTECTED]>
        cc:     php-general <[EMAIL PROTECTED]>
        Subject:        Re: [PHP] Help with msql_fetch_array()


I can almost guarantee that it's not the second line that is "failing", 
the problem here is that $result is not containing naything, and that is 
normally due to the fact that you are not connecting to the db, or the 
table "tablename" is not there.

I use the following format as my "standard" MySQL connect and query 
snippet:

$link = @mysql_connect("localhost",$username,$password) or die ('Could 
not connect!'); //@ suppresses the default error message generated by 
this function and the "or die()" bit kills the script right then and 
there should it not be able to connect.
        mysql_select_db("YOUR_DB_NAME",$link);
        $sql = "select * from your_table_name";
        if ( $result = mysql_query($sql)) {  // checks to see if $result 
contains anything before it even tries to fetch an associative array 
from it.
             $row = mysql_fetch_assoc($result);
        } else {
        echo "Empty result set!";

Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9 
out of 10 times, you don't need the array element id's that is returned 
by mysql_fetch_array.

Matthew Bielecki wrote:

>I have a couple of scripts that fail with the error of:
>Warning: mysql_fetch_array(): supplied argument is not a valid MySQL 
result 
>resource in...
>
>I'm new to both SQL and PHP and I'm wondering if I have some setting 
>turned off or what.
>
>Here's the piece of code that is failing (the second line fails):
>
>$result = mysql_db_query($dbname, "SELECT * FROM tablename ORDER BY id");
>        $row = mysql_fetch_array($result);
>
>
>Thanks for your help in advance!!
>