没有得到正确的价值作为回报

Question

The idea of the task is to allow the user to add and withdraw "money" to and from their account. The problem is I can add money, but I can't withdraw it

$funds = $_POST['funds'];
$withdraw_or_add = $_POST['list'];

if($withdraw_or_add == "add")  
{
  $sql = "UPDATE users SET userFunds = '".$funds."' WHERE userId = 1";
}   
else  
{
  $info = mysql_query("SELECT * FROM users WHERE userId = '1'");
  $info = mysql_fetch_assoc($info);
  $new_fund = $info['userFunds'] - $funds;
  $sql = "UPDATE users SET userFunds = '".$new_fund."' WHERE userId = 1";  
}


mysql_select_db('details_db');
$retval = mysql_query( $sql, $conn );

if(! $retval ) {
    die('Could not update data: ' . mysql_error());
}

echo "Updated data successfully
";

mysql_close($conn);

So for example, let's say $fund = 5 and $info['userFunds'] = 20 then the variable $new_fund should be 15. But instead it equals -5. If anyone can help it would be much appreciated.

Answer 1

Firstly page of top you put used db connection related code :

$conn = mysql_connect('localhost', 'user', 'pass');
mysql_select_db('details_db');

and then bellow and removed mysql_select_db('details_db'); line after mysql_

$funds = $_POST['funds'];
$withdraw_or_add = $_POST['list'];

if($withdraw_or_add == "add")  
{
  $sql = "UPDATE users SET userFunds = '".$funds."' WHERE userId = 1";
}   
else  
{
  $info = mysql_query("SELECT * FROM users WHERE userId = '1'");
  $info = mysql_fetch_assoc($info);
  $new_fund = $info['userFunds'] - $funds;
  $sql = "UPDATE users SET userFunds = '".$new_fund."' WHERE userId = 1";  
}


//mysql_select_db('details_db');
$retval = mysql_query( $sql, $conn );

if(! $retval ) {
    die('Could not update data: ' . mysql_error());
}

echo "Updated data successfully
";

mysql_close($conn);

Note: Please stop using mysql_* functions. mysql_* extensions have been removed in PHP 7. Please used PDO and MySQLi.