Re: [PHP] php- something that i don't understand

2001-07-17 Thread Sheridan Saint-Michel 

If you call it with no parameters it just uses the defaults...

host:port = 'localhost:3306'
username = name of the user that owns the server process
password = empty password

Check out
http://www.php.net/manual/en/function.mysql-connect.php

Sheridan

- Original Message -
From: Sam Masiello <[EMAIL PROTECTED]>
To: Yassel Omar Izquierdo Souchay <[EMAIL PROTECTED]>;
<[EMAIL PROTECTED]>
Sent: Tuesday, July 17, 2001 2:42 PM
Subject: RE: [PHP] php- something that i don't understand


>
> Pardon my ignorance if you can really do it this way (because I have never
> seen it done like this):
>
> In line 4, I wasn't aware that you could call mysql_connect() without any
> parameters.  Can you?
>
> HTH
>
> Sam Masiello
> Software Quality Assurance Engineer
> Synacor
> (716) 853-1362 x289
> [EMAIL PROTECTED]
>
>  -Original Message-
> From: Yassel Omar Izquierdo Souchay [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, July 17, 2001 4:38 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP] php- something that i don't understand
>
> Hey guys
> here is my code
>
> 4: mysql_connect() or die ("Problemas conectandose a la base de datos");
> 5: $query="select * from info where FirstName='$FirstName' and
> 6: LastName='$LastName' and email='$email'";
> 7: $result = mysql_db_query("users", $query);
> *8: $r=mysql_fetch_array($result); 
> 9: $count=$r["count"];
>
> and i follow redcieving this message
>
> Warning: Supplied argument is not a valid MySQL result resource in
> c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 8
>
>
> thanks for your answers
> Yassel
>
>
>
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Fw: [PHP] php- something that i don't understand

2001-07-17 Thread Jason Bell 

Oops.. forgot to CC the group..
- Original Message - 
From: "Jason Bell" <[EMAIL PROTECTED]>
To: "Yassel Omar Izquierdo Souchay" <[EMAIL PROTECTED]>
Sent: Tuesday, July 17, 2001 12:42 PM
Subject: Re: [PHP] php- something that i don't understand 


> try this:
> 
> mysql_connect("YOUR_DB_HOST","DB_USERNAME","DB_PASSWORD");
> mysql_select_db("DB");
> $query = "select * from info where FirstName='$FirstName' and
> LastName='$LastName' and email='$email'";
> $result = mysql_query($query);
> 
> 
> - Original Message -----
> From: "Yassel Omar Izquierdo Souchay" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Tuesday, July 17, 2001 1:37 PM
> Subject: [PHP] php- something that i don't understand
> 
> 
> Hey guys
> here is my code
> 
> 4: mysql_connect() or die ("Problemas conectandose a la base de datos");
> 5: $query="select * from info where FirstName='$FirstName' and
> 6: LastName='$LastName' and email='$email'";
> 7: $result = mysql_db_query("users", $query);
> *8: $r=mysql_fetch_array($result); 
> 9: $count=$r["count"];
> 
> and i follow redcieving this message
> 
> Warning: Supplied argument is not a valid MySQL result resource in
> c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 8
> 
> 
> thanks for your answers
> Yassel
> 
> 
> 


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RE: [PHP] php- something that i don't understand

2001-07-17 Thread Sam Masiello 


Pardon my ignorance if you can really do it this way (because I have never
seen it done like this):

In line 4, I wasn't aware that you could call mysql_connect() without any
parameters.  Can you?

HTH

Sam Masiello
Software Quality Assurance Engineer
Synacor
(716) 853-1362 x289
[EMAIL PROTECTED]

 -Original Message-
From:   Yassel Omar Izquierdo Souchay [mailto:[EMAIL PROTECTED]]
Sent:   Tuesday, July 17, 2001 4:38 PM
To: [EMAIL PROTECTED]
Subject:    [PHP] php- something that i don't understand

Hey guys
here is my code

4: mysql_connect() or die ("Problemas conectandose a la base de datos");
5: $query="select * from info where FirstName='$FirstName' and
6: LastName='$LastName' and email='$email'";
7: $result = mysql_db_query("users", $query);
*8: $r=mysql_fetch_array($result); 
9: $count=$r["count"];

and i follow redcieving this message

Warning: Supplied argument is not a valid MySQL result resource in
c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 8


thanks for your answers
Yassel



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Re: [PHP] php- something that i don't understand

2001-07-17 Thread Miles Thompson 

Yassel,

Do two things:

1. Echo $query, to ensure that it actually is what you think it is
2. Run the query, if you have access, at the mysql console and see what it 
returns.

Then you'll know what to fix.

Miles

At 03:37 PM 7/17/01 -0500, Yassel Omar Izquierdo Souchay wrote:
>Hey guys
>here is my code
>
>4: mysql_connect() or die ("Problemas conectandose a la base de datos");
>5: $query="select * from info where FirstName='$FirstName' and
>6: LastName='$LastName' and email='$email'";
>7: $result = mysql_db_query("users", $query);
>*8: $r=mysql_fetch_array($result); 
>9: $count=$r["count"];
>
>and i follow redcieving this message
>
>Warning: Supplied argument is not a valid MySQL result resource in 
>c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 8
>
>
>thanks for your answers
>Yassel


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Re: [PHP] php- something that i don't understand

2001-07-17 Thread Christopher Ostmo 

Yassel Omar Izquierdo Souchay pressed the little lettered thingies in this order...

> Hey guys 
> here is my code
> 
> 4: mysql_connect() or die ("Problemas conectandose a la base de datos"); 5:
> $query="select * from info where FirstName='$FirstName' and 6:
> LastName='$LastName' and email='$email'"; 7: $result =
> mysql_db_query("users", $query); *8: $r=mysql_fetch_array($result);
>  9: $count=$r["count"];
> 
> and i follow redcieving this message
> 
> Warning: Supplied argument is not a valid MySQL result resource in
> c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 8
> 
> 

Change line 7 to this:
$result = mysql_db_query("users", $query) or die(mysql_error());

What is the error after you do that?

Christopher Ostmo
a.k.a. [EMAIL PROTECTED]
AppIdeas.com
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[PHP] php- something that i don't understand

2001-07-17 Thread Yassel Omar Izquierdo Souchay 

Hey guys 
here is my code

4: mysql_connect() or die ("Problemas conectandose a la base de datos"); 
5: $query="select * from info where FirstName='$FirstName' and 
6: LastName='$LastName' and email='$email'"; 
7: $result = mysql_db_query("users", $query); 
*8: $r=mysql_fetch_array($result); 
9: $count=$r["count"];

and i follow redcieving this message

Warning: Supplied argument is not a valid MySQL result resource in 
c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 8


thanks for your answers
Yassel