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";
答案 0 :(得分:0)
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...
答案 1 :(得分:0)
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。