Re: [PHP-DB] Help w/ displaying return vals

[Russ Michell]

Ryan:

The followoing error message: 
>Warning: Supplied argument is not a valid MySQL result resource

means that a php function hasn't been supplied witht the correct number 
of arguments.  Try the following:

$link = mysql_connect("host","yourusername","yourpassword") or 
die("Died at connection.");
table_name = "yourtable";
mysql_select_db("dbname");
$sql = "SELECT * FROM yourtable";
$result = mysql_query ($result, connection);

$fields = mysql_num_fields($result);
$rows = mysql_num_rows($result);

$table = mysql_field_table($result, $i);

echo "Your '".$table."' table has ".$fields." fields and ".$rows." 
records "

- - EOP - -

//All I've done is to give mysql_query() two arguments
//Remover gaps between mysql functions anf their arguments (Which they 
//shouldn't have)

And given mysql_connect all 3 arguments it should have to make a 
successful connection to your database server. (Not sure wether you 
missed these out simply for clarity though)

Hope that all helps and makes sense!!

Cheers
Russ

#---#

 "Believe nothing - consider everything"

  Russ Michell
  Anglia Polytechnic University Webteam
  http://gertrude.sipu.anglia.ac.uk/webteam
  [EMAIL PROTECTED]
  +44 (0)1223 363271 ext 2331
  
  www.theruss.com

#---#


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Re: [PHP-DB] Help w/ displaying return vals

[JJeffman]

Use   or die( mysql_error())  on every mysql function
or command you're performing, this will tell you what is going wrong .

HTH

Jayme.

-Mensagem Original-
De: <[EMAIL PROTECTED]>
Para: <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Enviada em: quarta-feira, 14 de março de 2001 19:47
Assunto: [PHP-DB] Help w/ displaying return vals


> Here's my php program, I think $result never gets assigned a value,
staying
> null or whatever.
>
> $link = mysql_connect("host");
>
> mysql_select_db("dbname");
>
> $result = mysql_query ("SELECT * FROM table");
>
> $fields = mysql_num_fields ($result);
> $rows = mysql_num_rows ($result);
>
> $table = mysql_field_table ($result, $i);
>
> echo "Your '".$table."' table has ".$fields." fields and ".$rows." records
> "
>
> - - EOP - -
>
> The problem comes on the lines that reference the variable $result.
>
> I keep getting this in the browser... leading me to believe that $result
is
> null:
>
> Warning: Supplied argument is not a valid MySQL result resource
>
>
> help!
>
> Thankx0r
>
> Ryan
>
>
>
>
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>


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