PHP将错误的值写入数据库?
我在数据库中创建了下表PHP将错误的值写入数据库?,php,mysql,database,Php,Mysql,Database,我在数据库中创建了下表 $sql = "CREATE TABLE tac_flightsize ( id int NOT NULL AUTO_INCREMENT, PRIMARY KEY (id), gameid int NOT NULL, shipid int, flightsize int )"; 我检查是否有这样的DB写入: if (!$ship->superheavy){ debug::log("db flight: " + $s
$sql = "CREATE TABLE tac_flightsize
(
id int NOT NULL AUTO_INCREMENT,
PRIMARY KEY (id),
gameid int NOT NULL,
shipid int,
flightsize int
)";
我检查是否有这样的DB写入:
if (!$ship->superheavy){
debug::log("db flight: " + $ship->flightSize);
self::$dbManager->submitFlightSize($id, $gamedata->id, $ship->id, $ship->flightSize);
}
日志条目将12显示为$ship->flightSize,因此该值是正确的
以下是实际的submitFlightSize函数:
public function submitFlightSize($shipid, $gameid, $flightSize){
try{
$sql = "INSERT INTO `B5CGM`.`tac_flightsize` VALUES(null, $gameid, $shipid, $flightSize)";
$id = $this->insert($sql);
Debug::log($sql);
}catch(Exception $e) {
$this->endTransaction(true);
throw $e;
}
}
此函数的Debug:::日志显示以下SQL条目
INSERT INTO `B5CGM`.`tac_flightsize` VALUES(null, 2535, 16238, 1)
现在,最后一个参数是1,根据前面的调试,它应该是12,并且是一秒钟前的参数
有人能解释一下我可能做错了什么吗?这里您发送了4个参数
self::$dbManager->submitFlightSize($id, $gamedata->id, $ship->id, $ship->flightSize);
但函数只接受3
public function submitFlightSize($shipid, $gameid, $flightSize)
这是你发送的4个参数
self::$dbManager->submitFlightSize($id, $gamedata->id, $ship->id, $ship->flightSize);
但函数只接受3
public function submitFlightSize($shipid, $gameid, $flightSize)