That should definitely be working. Only thing I can think of is that you're
database name isn't correct. You're certain "news" is the full name of the
database.. I mean it's not "news_com" or something like that? The $link
must be valid becuase the script isn't ending with "Couldn't make
connection". So that just leaves the mysql_select_db() function. And the
only thing that will make that function return FALSE is if it can't find the
database name for that user on the specified server.
-Kevin
----- Original Message -----
From: "Matthew Bielecki" <[EMAIL PROTECTED]>
To: "PHPCoder" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Cc: "php-general" <[EMAIL PROTECTED]>
Sent: Wednesday, July 24, 2002 2:07 PM
Subject: Re: [PHP] Help with msql_fetch_array()
> 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!!
> >
>
>
>
>
>
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