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thanks musicman. Now i see an extra blank space at the end of second variable.
But I dont understand how it came? I used a trace function inside my flash file before sending the variable to php, and...

thanks mate :cap:

Why is this happening?

I have this code in my php...


$abs_usrstr=mysql_query("SELECT * FROM glist WHERE guest='$_POST[g]'");
while($row = mysql_fetch_array($abs_usrstr)){
...

:(

lol, bye back2012

Anyone there? Is it possible to do this?

Sorry. Its a private community website.

I would like to know - is it possible(using php script loaded into an iframe) to display alternative content instead of the actual HTML content? If thats...

This is getting even more stranger...



When I inserted the below code....

$copystr="GRP_".$groupFullName."_users";
if(mysql_query("INSERT INTO $copystr (usrip, user, status) VALUES...

I am afraid I cant do that because my "homepage.html" has links to all other sections in my website. The website has a seperate Welcome page called "index.html" that has only a 2-line text. I dont...

thanks for your reponse :)

Thats right this visitor doesn't seem to visit any other pages except my homepage.
Is there a way I can show an "alternative content" to this user without displaying...

Still i get the error message.

I tried to echo my SQL statement to check whats happening with my variables...



if(mysql_query("INSERT INTO $groupFullName (usrip, user, status) VALUES...

I have 2 seperate websites that deals with entirely different subjects. I maintain a database of users who visit my websites. Yesterday as i was going through the user records I noticed someone has...

I even tried removing the Quotes. Still i get the same error


INSERT INTO $groupFullName (usrip, user, status) VALUES ('$ipaddr', '$thisuser', '$isonline')

Why do i keep getting this kind of error message?



Here is my php code


$ipaddr=$_SERVER['REMOTE_ADDR'];
$mycnct=mysql_connect("abc","xyz","pass");
if($mycnct){