Question

嗨，我想知道是否有人可以提供帮助，我对MySQL非常陌生并努力让它做我想做的事情，我已经阅读了几个不同的帖子和网站，没有运气。简短而简单，我的查询有问题。

    $a = $mysqli->real_escape_string($_POST['cardInputOne']);
            $b = $mysqli->real_escape_string($_POST['cardInputTwo']);
            $c = $mysqli->real_escape_string($_POST['cardInputThree']);
            $d = $mysqli->real_escape_string($_POST['cardInputFour']);

            $incomingCardNumber = "{$a}{$b}{$c}{$d}";
            $safeIncomingCard = $mysqli->real_escape_string($incomingCardNumber);
            $fundingUser = $mysqli->real_escape_string($email);

            $fundingQuery = "SELECT f.cardValue f.cardNumber FROM funding f 
WHERE f.cardNumber = '{$safeIncomingCard}' 
AND s.email, s.accountBalance FROM shopUser s 
WHERE s.email = '{$fundingUser}'" . 
"UPDATE s.accountBalance = s.accountBalance + f.cardValue";

Answer 1

如果您想根据另一个表中的数据更新表，那么您需要使用它的更新：

update shopUser, funding
set shopUser.accountBalance = shopUser.accountBalance + funding.cardValue
where shopUser.email='...' and funding.cardNumber='...'

php变量将取代...

Answer 2

听起来像开启者想要用另一个数据更新表。一般来说，这可能是：

UPDATE
  table1 AS target,
  (SELECT column1, column2 FROM table2) AS source
SET
  target.column3 = source.column1
WHERE
  target.column4 = source.column2

当然，您必须选择正确的源/目标列。

MySQL从一个表转到另一个表

2 个答案: