Reputation: 448
I am using mysqli_real_escape_string to clean my user input before inserting it into my database. I have used it before without trouble, but for some reason this time it is not recognizing my link identifier.
//Connect to mysql server
$link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);
if (!$link) {
die('Failed to connect to server: ' . mysqli_error());
}
//Function to sanitize the values received from the form (prevents SQL injection)
function clean($str) {
$str = trim($str);
if (get_magic_quotes_gpc()) {
$str = stripslashes($str);
}
$rtstr = mysqli_real_escape_string($link, $str);
return $rtstr;
}
For some reason when I try to input information through this file, it gives me the error "Undefined variable: link" and then "mysqli_real_escape_string() expects parameter 1 to be mysqli" for every time it encounters this function.
I am very confused because everything seems to be correct, but I can't find a way around this error. Is there something I'm doing wrong here? Or is it something outside of this code causing the issue?
Upvotes: 0
Views: 184
Reputation: 8618
Just declare global $link within your function.
function clean($str) {
/* Other code */
global $link; // Add this
$rtstr = mysqli_real_escape_string($link, $str);
return $rtstr;
}
Hope this helps.
Peace! xD
Upvotes: 0
Reputation: 8537
You call $link
in your function but it is not defined in your function.
You have to pass it as a parameter or define it in the function.
Then, you have to call your function.
//Connect to mysql server
$link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);
if (!$link) {
die('Failed to connect to server: ' . mysqli_error());
}
//Function to sanitize the values received from the form (prevents SQL injection)
function clean($str,$link) {
$str = trim($str);
if (get_magic_quotes_gpc()) {
$str = stripslashes($str);
}
$rtstr = mysqli_real_escape_string($link, $str);
return $rtstr;
}
clean('test',$link);
Upvotes: 1