Yeah I have try that... and when I : echo $field[1]
appear error msg : "Undefined offset: 1 "
- Original Message -
From: "Sam Chill" <[EMAIL PROTECTED]>
To: "PHP DB Help" <[EMAIL PROTECTED]>
Sent: Saturday, June 26, 2004 11:21 AM
Subject
Instead of
return $field[];
try
return $field;
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Hi All,
how to return array var from function ??
I have try that but just can't return it..
here's my code :(function to return any fields name from table)
function fields($table,$db_name)
{ $link=$this->database(); // this open the database connection
$fields = mysql_list_fields($db_n
save you some. ;)
Thanks for the help.
> -Original Message-
> From: Dave Smith [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, September 26, 2002 4:31 PM
> To: Hutchins, Richard
> Cc: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Return Array from Function
>
>
> T
Try single-quotes instead of double (' instead of "). Could be an
interpolation issue.
--Dave
On Thu, 26 Sep 2002, Hutchins, Richard wrote:
> Can somebody tell me what I might be doing wrong in the code below? I'm
> trying to use a function to perform an often-used db query and return the
> re
You can't start variables with a number.
---John Holmes...
- Original Message -
From: "Hutchins, Richard" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, September 26, 2002 4:18 PM
Subject: [PHP-DB] Return Array from Function
> Can somebody
Sorry, line 51 is the one that reads:
$1contentID=$row["1contentID"];
> -Original Message-
> From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, September 26, 2002 4:19 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Return Array
Can somebody tell me what I might be doing wrong in the code below? I'm
trying to use a function to perform an often-used db query and return the
resulting resource to the calling script. I keep getting this error though:
Parse error: parse error, unexpected T_LNUMBER, expecting T_VARIABLE or '$'
