Question

我在PHP / SQLite 3（PDO）中运行此查询

场景：新的驱动程序插入到驱动程序表中，现有的汽车立即与他链接：

DRIVERS

driver_id [PK]
DRIVER_NAME

CARS

car_id [PK]
fk_driver_id [FK]

$ qr =＆＃34; INSERT INTO drivers（driver_name）VALUES（＆＃39; {$ _ GET [＆＃39; driver_name＆＃39;]}＆＃39;）;承诺; UPDATE汽车SET fk_driver_id =（SELECT last_insert_rowid（））WHERE car_id = {$ _ GET [＆＃39; car_id＆＃39;]};＆＃34 ;; $ stmt = $ dbh-＆gt; prepare（$ qr）; $ result = $ stmt-＆gt; execute（）;

它插入了驱动程序，但没有更新cars表并且也没有产生错误。

如果我使用SQLite Spy使用相同的查询，它会起作用。

在PHP中，只有我将其分为两部分才能实现：

$qr = "INSERT INTO drivers (driver_name) VALUES ('{$_GET['driver_name']}'); COMMIT; ";
$stmt = $dbh->prepare($qr);
$result = $stmt->execute();

$qr = "UPDATE cars SET fk_driver_id=( SELECT last_insert_rowid() ) WHERE car_id={$_GET['car_id']};";
$stmt = $dbh->prepare($qr);
$result = $stmt->execute();

如果PHP代码不能在单个语句中工作，那么PHP代码有什么问题？

Answer 1

尝试这种方式：

$qr = "INSERT INTO drivers (driver_name) VALUES ('{$_GET['driver_name']}'); ";
$stmt = $dbh->prepare($qr);
$result = $stmt->execute();
$lastId = $dbh->lastInsertId();
$dbh->commit();

$qr = "UPDATE cars SET fk_driver_id=? WHERE car_id={$_GET['car_id']};";
$stmt = $dbh->prepare($qr);
$result = $stmt->execute(array($lastId));

SQLite：last_insert_rowid（）在同一个INSERT语句中

1 个答案: