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Display MY SQL advanced Count in PHP

Hi I have this PHP script below which works for a selcect count(*) from x

However when I put a more advanced MYSQL syntax in there it gives the following error"Notice: Undefined index: total in /var/www/dupe.php on line 12 Total ="

Any ideas

// Check connection
if (mysqli_connect_errno())
  echo "Failed to connect to MySQL: " . mysqli_connect_error();

$query = "select count(*) from activecampaign.em_exclusion
Inner join activecampaign.em_subscriber
ON em_exclusion.email = em_subscriber.email";

$result = mysqli_query($con, $query);

echo $result;

1 Solution
Guy Hengel [angelIII / a3]Billing EngineerCommented:
$result = mysqli_query($con, $query);

this will only run the query, and $result will not be the "number" of the count(*), but a resource identifier. after that line, you need to fetch the row into either an object or an array, so after that line, you will do:
$result_data = mysqli_fetch_array($result);
echo $result_data[0]; 

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Since there is no total in your sample code I'm going to assume you are not showing us the real code and expecting us to be psychic.

$query = "select count(*) as total from activecampaign.em_exclusion
Inner join activecampaign.em_subscriber
ON em_exclusion.email = em_subscriber.email"; 

$result = mysqli_query($con,$query);

$rows = mysqli_fetch_row($result );

echo $rows[0];

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Ray PaseurCommented:
When in doubt, you can check the online man page for the return values from PHP functions.  Example: Ask yourself, "What is the expected return value from this function?"

Some learning resources to help you get started with things like this are available in this article.  Buy any of the books and give yourself a few weeks to read, study and practice.  In no time you'll be able to do this yourself!
ncomperAuthor Commented:
Amazing thank you
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