Hello there , new on mySQL and I am having errors already...

[jimmyslam]

Ok i am having this error.

Fatal error: Call to a member function on a non-object in /datos/home/d177397/public_html/public_html/html/admin/view.php on line 14

And my code is

Vars!
[code]<?

define ( "DB_SERVER", "x.x.x.x" );
define ( "DB", "d177397_news" );
define ( "DB_LOGIN", "user..." );
define ( "DB_PASSWORD", "pass..." );

?>[/code]



Connect!
[code]<?

$db = mysql_connect(DB_SERVER, DB_LOGIN, DB_PASSWORD) or die("Could not connect: ".mysql_error());
mysql_select_db(DB) or die("Could not select database: ".mysql_error());

?>[/code]


[code]<?
include_once ("vars.php");
include_once ("connect.php");


//ver infos de las tablas
//$sql->QueryRow("select * from news where 'ID'=0");
$sql->QueryRow("select * from news where 'ID'=1");


$row = $sql->data;

echo ("$row[ID]");
echo ("$row[date]");
echo ("$row[description]");
echo ("$row[links]");

mysql_close();
?>[/code]

What is going on?

I think the queryrow is wrongly used. I dont see what is the problem really...

Anyone help? Thanks

[Wildbug]

Which line is line 14?  This one:  "$row = $sql->data;"?

In this line -- [b]$sql->QueryRow("select * from news where 'ID'=1");[/b] -- I think ID shouldn't be in single quotes.  You don't really need anything around it, but you can use backticks if you really want to quote it.

[king arthur]

Where are you setting up the $sql object, as it looks like that is where the error is coming from.

[jimmyslam]

Well the error is in this line:

[code]$sql->QueryRow("select * from news where 'ID'=1");[/code]

I think  i tried doing
[code]$sql->QueryRow("select * from news where ID=1");[/code]
and didnt work either.

[Wildbug]

I agree w/ kingarthur; there's probably an error with the object creation.

But the 'ID'=1 was comparing a [i]string[/i] to a number, not a column to a number.  It would always evaluate to FALSE and was, no doubt, not what you had intended.

[jimmyslam]

I just followed other steps and is working now..

Thanks anyway! ;)