Mark Sargent wrote:
*Warning*: mysql_fetch_row(): supplied argument is not a valid MySQL
result resource in */var/www/html/phpmysqltable.php* on line *36*
Below are snippets of my code,
Line 22: $result = mysql_query("SELECT product_name,
product_model_number, product_serial_number FROM Products",$
Mark Sargent wrote:
Hi All,
get the following error when calling data from mysql,
*Warning*: mysql_fetch_row(): supplied argument is not a valid MySQL
result resource in */var/www/html/phpmysqltable.php* on line *36
This means that $result is not a valid result resource. You can
var_dump($result)
I rarely use the @ sign. It is what I call 'balistic programming'. You
know, a shell, once it leaves the cannon, is not controlled in any way, you
just hope it will get where you want to go.
Instead you should check for errors and recover from them as gracefully as
possible..
You will see in
Hegedűs Balázs wrote:
Hi all,
ever tried to test if your connection to the database succeeded? maybe
that's why it shouted that "...not a valid result resource..." msg.
Balazs Hegedus
Mark Sargent wrote:
Drewcore wrote:
just put a @ symbol before the function calll, eg
while ($myrow = @mysql_fetc
Hi again,
just another tip...it's from the php man:
*mysql_query()* will also fail and return *FALSE* if the user does
not have permission to access the table(s) referenced by the query.
Balazs Hegedus
Mark Sargent wrote:
Drewcore wrote:
just put a @ symbol before the function calll, eg
while ($m
Hi all,
ever tried to test if your connection to the database succeeded? maybe
that's why it shouted that "...not a valid result resource..." msg.
Balazs Hegedus
Mark Sargent wrote:
Drewcore wrote:
just put a @ symbol before the function calll, eg
while ($myrow = @mysql_fetch_row($result)) {
etc
Drewcore wrote:
just put a @ symbol before the function calll, eg
while ($myrow = @mysql_fetch_row($result)) {
etc
On 4/25/05, Mark Sargent <[EMAIL PROTECTED]> wrote:
Hi All,
get the following error when calling data from mysql,
*Warning*: mysql_fetch_row(): supplied argument is not a valid MyS
just put a @ symbol before the function calll, eg
while ($myrow = @mysql_fetch_row($result)) {
etc
On 4/25/05, Mark Sargent <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> get the following error when calling data from mysql,
>
> *Warning*: mysql_fetch_row(): supplied argument is not a valid MySQL
Hi All,
get the following error when calling data from mysql,
*Warning*: mysql_fetch_row(): supplied argument is not a valid MySQL
result resource in */var/www/html/phpmysqltable.php* on line *36
*
Below are snippets of my code,
Line 22: $result = mysql_query("SELECT product_name,
product_model_
Hi Philip and Thomas and All.
Thanks very much for yours helps.
I am Brazillian and I speak in portuguese, for this, I am sorry very much
for my bad english.
After a long time, my code works very good! But, I dont understand what was
the problem!
My code was:
$result = sql_query("select catego
Hi Janet,
I dont have know how in php.
I inserted the function mysql_error() in my code and it dont return any
message!
Thanks
Marcelo
> This means that when you did:
>
> $result = mysql_query($sql);
>
> $result did not end up with a valid resource in it. Then, when you do the
fetch row comman
??
Thanks
Marcelo
- Original Message -
From: "Thomas Seifert" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, June 08, 2003 6:50 PM
Subject: Re: [PHP] Warning: mysql_fetch_row(): supplied argument is not a
valid MySQL result resource in f:\.\none.php on line 286
What this message means?
What it is the error?
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result
resource in f:\.\none.php on line 42
This is the code.
Thanks very much.
Marcelo
";
$content .=" Últimos 10 Concursos";
$content .= "";
\\ line 42
while (list($catego
What this message means?
What it is the error?
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result
resource in f:\.\none.php on line 286
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