Reputation: 17
After getting the user-info from my sql database I would like to check if some of the fields are empty and continue the script based on that. A simplified piece of code would look like this:
$userData = mysql_query("SELECT * FROM user WHERE user='".$user."' LIMIT 1");
if(mysql_num_rows($data) == 1){
$u_info = mysql_fetch_assoc($data);
if(empty($u_info['u_mobile'])){
echo 2;
exit();
} else {
echo 1;
exit();
}
} else {
echo 3;
exit();
}
The problem is the empty statement checking the recieved field. I've tried using empty(), isset(), not_null() and array_key_exists() with no luck and can't get around to what I'm doing wrong.
I also tried if($u_info['u_mobile']) == '' || $u_info['u_mobile']) == NULL)
but that doesnæt work either.
Why is this, or how can I go about getting this information?
I need to collect the user-information and send them to fill out the information I don't have...
Upvotes: 0
Views: 195
Reputation: 5
$userData = mysql_query("SELECT * FROM user WHERE user='".$user."' LIMIT 1");
if(mysql_num_rows($userData ) == 1){
$u_info = mysql_fetch_assoc($userData );
if(empty($u_info['u_mobile'])){
echo 2;
exit();
} else {
echo 1;
exit();
}
} else {
echo 3;
exit();
}
Please Run code..I think it will be compile better it was minor mistake
Upvotes: 0
Reputation: 1427
You're setting the query result to $userData
but then you're using mysql_fetch_assoc($data);
-- doh. You need to pass the variable that you set the query result to:
$u_info = mysql_fetch_assoc($userData);
It's OK, it is still 10AM EST so this can happen in the morning =) I suggest that you turn on PHP error reporting. PHP would have alerted you that the array values were trying to access do not exist, saving you a lot of wasted frustration.
Upvotes: 2