我有以下PHP代码。我希望返回插入的最后一个id:
$sql = $this->dbLink->prepare("
INSERT INTO VesselTransits (CreatedByUserID, TerminalID, VesselArrivalDate, VesselNameID)
VALUES (?,?,?,?)
");
//i=integer, s=string
$sql->bind_param('iisi', $userID, $terminalID, $arrivalDate, $vesselNameID);
$userID = $this->invoiceData['userID'];
$terminalID = $this->invoiceData['terminalID'];
$arrivalDate = date('Y-m-d', strtotime($this->invoiceData['berthDate']));
$vesselNameID = $this->invoiceData['vesselNameID'];
$sql->execute();
$vesselTransitID = $sql->insert_id;
//echo('RecordInvoice line '.__LINE__.'<pre>');print_r($vesselTransitID);echo('</pre>');exit();
return($vesselTransitID);
我的DB架构如下:
CreatedByUserID int
CreationDate datetime
ModifiedByUser int
ModifyDate datetime
TerminalID int
VesselArrivalDate datetime
VesselNameID int
VesselTransitID int [auto_increment]
我已经把厨房的水槽扔过去了,但一点也没有。无论我做什么,$VesselTransitID
总是未定义的。我已经验证了实际上有一个插入。我只剩下运行另一个select并返回最大VesselTransitID的(可怕的)选项,这并不能保证正确的结果。
code>insert_id属性是mysqli类(连接)的属性,而不是语句类的属性
也是如此
$vesselTransitID = $this->dbLink->insert_id;