Trying to fill a PHP variable with a mysql query. Then updating another table with the data from that query

Question

First post, here it goes.

So this is the code that I have so far:

include('Connection/connect-test.php');
$selected1 = $_POST['selected'];
$sqlget = "SELECT paymentid FROM highschoolpayment WHERE hsgameid = '$selected1'";
$sqldata = mysqli_query($dbcon, $sqlget); 
$sqlupdate = "UPDATE highschool SET paymentid = '$sqldata' WHERE hsgameid = '$selected1'";
mysqli_query($dbcon, $sqlupdate);

What I'm trying to do is grab the 'paymentid' from the 'highschoolpayment' table and store that value into the $sqldata variable (line 4). Then I want to update a value in the 'highschool' table using the value that I got from line 4 as well as a value that was pulled from a POST submission (line 6). I know for a fact that the first 3 lines execute as they should. It is after those lines when things become iffy. I don't see the form (reappear) like I normally would when everything else is working. To me, this indicates that the PHP has successfully run. I go to the 'highschool' table but I don't see the value (paymentid) that I am expecting to see. I personally can't think of a single reason why this wouldn't work, but, I am not that experienced in PHP or MySQL so I am open to any help that I can get.

I hope this makes sense without seeing the structure of the tables but if I need to post those, let me know. I've spent a couple hours trying to troubleshoot this problem but with no forward progress.

Thanks!

Answer 1

try like this,

include('Connection/connect-test.php');
$selected1 = $_POST['selected'];
$sqlupdate = "UPDATE highschool SET paymentid = (select paymentid FROM highschoolpayment WHERE hsgameid = '$selected1') where hsgameid = '$selected1'";   
mysqli_query($dbcon, $sqlupdate);

Answer 2

Assuming this query returns only one row:

$sqldata = mysqli_query($dbcon, $sqlget);
$row = mysqli_fetch_array($sqldata);
$paymentid = $row['paymentid']; // then use $paymentid in the next query
$sqlupdate = "UPDATE highschool SET paymentid = '$paymentid' 
              WHERE hsgameid = '$selected1'";
if(mysqli_query($dbcon, $sqlupdate)){
   echo 'Update successfull';
} else {
   echo 'Update query is wrong. The query generated was <br />'.$sqlupdate;
}

Answer 3

you need to do fetch_assoc(), and while you are at it you should parameterize your query to make it more secure, good practice for the future. here is what your code should look like

$selected1 = $_POST['selected'];
$connect = mysqli_connect("localhost","user","pass","database");//i connect this way to my database

//the first statement that will get your paymentid
$stmt = $connect->prepare("SELECT paymentid FROM highschoolpayment WHERE hsgameid = ?")
mysqli_stmt_bind_param($stmt, 's', $selected1);//'s' is for string, 'i' for int, google rest
$stmt->execute();   
$result = $stmt->get_result();

while($row = $result->fetch_assoc()){//it fetches each id 

    //the second statement that will use the payment id and update the database
    $stmt2 = $connect->prepare("UPDATE highschool SET paymentid = ? WHERE hsgameid = ? ;")
    mysqli_stmt_bind_param($stmt2, 'ss',$row['paymentid'], $selected1 );//'s' is for string, 'i' for int, google rest
    $stmt2->execute();
    $stmt2->close();

}

$stmt->close();

I just threw this quickly together, so if anyone sees something wrong don't hesitate to edit it or mark it down if completely wrong, Would rather that.