How can the resource be invalid if I'm getting the results back? I'd
appreciate any explaination.
What results are you getting?...
$connection_id = @mysql_connect ('localhost', 'root') or die('No
connection.');
$database_id = @mysql_select_db ('cgcms', $connection_id) or die('No
In article 046401c0fee3$a6c789e0$6401a8c0@Lynchux100,
[EMAIL PROTECTED] (Richard Lynch) wrote:
$result_id = @mysql_query(DESC $table_name $field_name);
I don't think DESC $table_name $field_name is valid SQL, is it?...
DESC is DESCENDING in an ORDER BY clause...
Perhaps if you
dunno
but my it's just failing at the mysql_connect()
shouldn't you specify the password too?
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
The following code fragment works properly, in that I get the output I'm
expecting. However, if I don't disable
On Wed, 27 Jun 2001 06:31, [EMAIL PROTECTED] wrote:
The following code fragment works properly, in that I get the output
I'm expecting. However, if I don't disable error messages with @, I get
an error message from PHP saying: Warning: Supplied argument is not a
valid MySQL result resource in
- Original Message -
From: David Robley [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; PHP General [EMAIL PROTECTED]
Sent: Wednesday, June 27, 2001 8:47 AM
Subject: Re: [PHP] Strange error on mysql_fetch_array
On Wed, 27 Jun 2001 06:31, [EMAIL PROTECTED] wrote:
The following code fragment
From experience with oracle, desc only works in an sql prompt, not through
code. So use the more lengthy method
in mysql:
SHOW FIELDS FROM $table_name
in oracle:
SELECT c.column_name, c.data_type, c.data_length, c.data_precision,
c.data_scale, c.nullable FROM
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