Question

$tSQL = "insert into events(title,start,end,allday,url,customerid) VALUES(\"" . $_POST['title'] . "\", FROM_UNIXTIME($epochstart), FROM_UNIXTIME($epochend), \"$allday\", \"$url\", \"$customerid\")";
$mysqli->multi_query($tSQL);
$lasterror = $mysqli->error;



$tSQL = "update events set url = \"details.php?\"" . mysql_insert_id() . " where idevents = \"$eventid\"";
$row = $mysqli->multi_query($tSQL);
$lasterror = $mysqli->error;

echo print_r($tSQL);

我的插入语句肯定会插入记录，但mysql_insert_id（）会一直返回0.这不应该是这种方式，因为在该事件表中有一个自动递增的主键，并且运行正常。关于如何获取最后一个插入ID的任何建议？

Answer 1

您的查询是通过mysqli执行的，因此mysql函数不会保留插入的ID。相反，请使用the mysqli version：

$id = $mysqli->insert_id;

Answer 2

因为您使用的是mysqli而不是mysql，

如果使用程序样式

，只需将mysql_insert_id()替换为mysqli_insert_id()

如果使用面向对象的样式

，则将其替换为$mysqli->insert_id

Answer 3

由于您使用的是mysqli扩展程序，请更改

$tSQL = "update events set url = \"details.php?\"" . mysql_insert_id() . " where idevents = \"$eventid\"";

到

$tSQL = "update events set url = \"details.php?\"" .$mysqli->insert_id. " where idevents = \"$eventid\"";

Answer 4

因为您使用的是mysqli mysql的改进版本。

使用mysqli->insert_id代替mysql->insert_id()

$tSQL = "insert into events(title,start,end,allday,url,customerid) VALUES(\"" . $_POST['title'] . "\", FROM_UNIXTIME($epochstart), FROM_UNIXTIME($epochend), \"$allday\", \"$url\", \"$customerid\")";
    $mysqli->multi_query($tSQL);
    $lasterror = $mysqli->error;

    $lastInsId=$mysqli->insert_id();

    $tSQL = "update events set url = \"details.php?\"" . $lastInsId . " where idevents = \"$eventid\"";
    $row = $mysqli->multi_query($tSQL);
    $lasterror = $mysqli->error;

    echo print_r($tSQL);

mysql_insert_id（）返回0，即使最后一次插入成功

4 个答案: