Reputation: 3391
I have a PHP function called GetQA which returns data from MySQL:
$result = mysqli_query($con, $sqlQuery) or die(mysqli_error()$con);
I'd like to add a dynamically generated variable result to the end. I don't want to store that variable in the database.
What I'd like to is something like this:
$result = Array(mysqli_query($con, $sqlQuery), myVariable);
Is this possible?
Upvotes: 2
Views: 67
Reputation: 11298
Assuming $myVariable
will be the same for all rows, I can think of two ways off the top of my head to do this:
$result = mysqli_query($con, $sqlQuery) or die(mysqli_error($con));
while ($row = mysqli_fetch_assoc($result)) {
$row['myVariable'] = $myVariable;
// do whatever you want to $row now, $myVariable will be at the end
}
Or:
$sqlQuery = "SELECT field1, field2, field3, '" . $myVariable . "' as myVariable FROM table";
$result = mysqli_query($con, $sqlQuery) or die(mysqli_error($con));
while ($row = mysqli_fetch_assoc($result)) {
// $myVariable will be in $row['myVariable']
}
Keep in mind that in that last example, you will need to make sure $myVariable is properly escaped.
Upvotes: 1