Question

该任务的想法是允许用户添加和提取＆＃34; money＆＃34;来往他们的帐户。问题是我可以加钱，但我不能撤回

$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\n";

mysql_close($conn);

例如，让我们说$fund = 5和$info['userFunds'] = 20，然后变量$new_fund应为15。但相反，它等于-5。如果有人可以提供帮助，我将不胜感激。

Answer 1

首先，您使用了与db连接相关的代码：

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

然后在mysql_select_db('details_db');

之后显示并删除mysql_行

$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\n";

mysql_close($conn);

注意：请停止使用mysql_*个功能。在PHP 7中删除了mysql_* extensions。请使用PDO和MySQLi。

没有得到正确的价值作为回报

1 个答案: