June 19, 2013
How to check if there aren't any result after querying and execute another statement?
K20GH’s Question:
I’m trying to do something fairly simple as per below. Basically, I want to check to see if a user exists using their ‘gid’. If they don’t exist, add them to the database, if they do, echo “User Already Exists”
For whatever reason, it will always add a new user, even if their gid exists.
$checkuser = "SELECT gid FROM user WHERE gid = '$id'";
$adduser = "INSERT INTO user (gid, name, pic) VALUES ('$id','$name','$img')";
if (!mysql_num_rows($checkuser))
{
if (!mysqli_query($con,$adduser))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
}
else {
echo "User Already Exists";
}
There is many things wrong with your code. First and the most important is you have a query a statement, if you do not query, you will not receive any result-set and the resultant is new registration every time.
Function
is_resource()will help you decide whether or not the query
returns valid result.
Add here is the correct way to solve your problem.
$checkuser = "SELECT gid FROM user WHERE gid = '$id'";
$adduser = "INSERT INTO user (gid, name, pic) VALUES ('$id','$name','$img')";
$result = mysqli_query($con, $checkuser));
// ^-- You have to query first to find out if there are any users or not
if(is_resource($result) && mysqli_num_rows($result) > 0) {
// ^ This will confirm the query has returned a result and it contains rows
echo "User Exists";
} else {
// If it does not you can finally add it
$result = mysqli_query($con, $adduser) or die(mysqli_error());
echo "User added";
}
