Hello
Can anyone tell me if the below php/mysql statement is wrong? When I run it on my server it gives me the error mysql_fetch_array(): supplied argument is not a valid MySQL result resource.
I have googled this problem to find out that i can put the '@' sign in front of mysql_fetch_array. This blocks the error out but still does not print out the variable I want it to! When I googled it I found so many errors the same. But none help me out!
The sql code works fine in phpmyadmin.
And I have checked the case on the text over and over.
My eyes are hurting from looking at the screen! Had to change my monitor background from blue to black!
Any ideas? I really am stuck with this!
$query = ("SELECT *
FROM cringeitwebsitequote
ORDER BY `__id` DESC
LIMIT 1 ");
while ($row = mysql_fetch_array($query)) {
$hosting = ( $row[hosting]); {
$pages = ( $row[pages]); {
$forms = ( $row[forms]); {
$cart = ( $row[cart]); {
$products = ( $row[products]); {
$logo = ( $row[logo]); {
$flash = ( $row[flash]); {
$dhtml = ( $row[dhtml]); {
$statistics = ( $row[statistics]); {
}}}}}}}}}}
print("$hosting");
This should just print the work 'No'
Thanks
Ben
Can anyone tell me if the below php/mysql statement is wrong? When I run it on my server it gives me the error mysql_fetch_array(): supplied argument is not a valid MySQL result resource.
I have googled this problem to find out that i can put the '@' sign in front of mysql_fetch_array. This blocks the error out but still does not print out the variable I want it to! When I googled it I found so many errors the same. But none help me out!
The sql code works fine in phpmyadmin.
And I have checked the case on the text over and over.
My eyes are hurting from looking at the screen! Had to change my monitor background from blue to black!
Any ideas? I really am stuck with this!
$query = ("SELECT *
FROM cringeitwebsitequote
ORDER BY `__id` DESC
LIMIT 1 ");
while ($row = mysql_fetch_array($query)) {
$hosting = ( $row[hosting]); {
$pages = ( $row[pages]); {
$forms = ( $row[forms]); {
$cart = ( $row[cart]); {
$products = ( $row[products]); {
$logo = ( $row[logo]); {
$flash = ( $row[flash]); {
$dhtml = ( $row[dhtml]); {
$statistics = ( $row[statistics]); {
}}}}}}}}}}
print("$hosting");
This should just print the work 'No'
Thanks
Ben