You have an error in your SQL syntax

[$php_mysql$]

whenever i try to insert data like

 

$page = $_GET['page'];
if(!$page >= 1){
$page = 1;
}
$records_per_page = 5;
if($page == 1)
$first_record = 0;
else{
$first_record = (($page - 1) * $records_per_page );
}

 

in my form get the error message

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'page']; if(!$page >= 1){ $page = 1; } $records_per_page = 5; if($page == 1)' at line 11

 

but else my form works fine even if i insert html tags in the textarea field, what could be causing this?

[TeNDoLLA]

The error is in your SQL and the code you provided has nothing to do with that error (or it could have, if you use the variables in this code but still can't say shit with that code why your SQL query fails).

[$php_mysql$]

very strange, now that i used mysql_real_escape_string in the "..desctiption." it do not show error

[TeNDoLLA]

very strange, now that i used mysql_real_escape_string in the "..desctiption." it do not show error

 

So you had some characters in the description like apostrophes that needs to be escaped so they don't break the query structure.

[$php_mysql$]

i was trying to insert $page = $_GET['page']; into description, i thought escape string in description is not necessary

[TeNDoLLA]

i was trying to insert $page = $_GET['page']; into description, i thought escape string in description is not necessary

 

It is needed for every single bit of data input that you can't control, this including all POST and GET data.

[$php_mysql$]

thanks got to keep that in mind