这是我的代码。
// insert reward into wallet
$sql = "
INSERT INTO `wallet` (`uid`, `created_at`, `amount`, `type`, `payment_id`) VALUES (:uid, CURRENT_TIMESTAMP, :amount, 'payment', :payment_id);
";
$sth = self::link()->prepare($sql);
// primary key makes sure payment does not get double rewarded
$sth->execute(
array(
':uid' => $referer,
':amount' => $reward,
':payment_id' => $payment_data['payment_id'],
)
);
var_dump(self::link()->errorInfo());
self::log("issuing subscription",self::LOG_LEVEL_DEBUG);
// extend referers subscription
$tid = self::link()->lastInsertId();
var_dump(self::link()->errorInfo());
self::log("using $tid as id for wallet transfer",self::LOG_LEVEL_DEBUG);
我的日志说:
[2011-07-02 20:31:44] using 0 as id for wallet transfer
但是插入查询成功,数据库记录被创建并且两个 errorInfo 输出都没有错误。
最佳答案
从查询中删除分号或分号后的空格或两者:
$sql = "
INSERT INTO `wallet` (`uid`, `created_at`, `amount`, `type`, `payment_id`) VALUES (:uid, CURRENT_TIMESTAMP, :amount, 'payment', :payment_id)
";
关于php - 为什么 PDO::lastInsertId 返回 0?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/6558885/