Question

我正在将我的现有mysql转换为mysqli

但我无法正确使用这段代码

mysql_query("INSERT INTO `sendmsg`(`sendname`, `recievename`, `subject`, `body` , `mdate`, `mtime`) VALUES ('$sendname','$recievename','$subject','$body','$msgdate','$msgtime')");
$new_id = mysql_insert_id();
mysql_query("INSERT INTO `recievemsg`(`msgid`, `sendname`, `recievename`, `subject`, `body`, `mdate`, `mtime`, `status`) VALUES ($new_id,'$sendname','$recievename','$subject','$body','$msgdate','$msgtime','UNREAD')");

我尝试的内容如下，但没有工作

$mysqli->query("INSERT INTO `sendmsg`(`sendname`, `recievename`, `subject`, `body` , `mdate`, `mtime`) VALUES ('$sendname','$recievename','$subject','$body','$msgdate','$msgtime')");
$new_id = mysqli_insert_id();
$mysqli->query("INSERT INTO `recievemsg`(`msgid`, `sendname`, `recievename`, `subject`, `body`, `mdate`, `mtime`, `status`) VALUES ($new_id,'$sendname','$recievename','$subject','$body','$msgdate','$msgtime','UNREAD')");

问题在于$new_id = mysqli_insert_id();语句bcoz第一个查询正在执行

请帮助

Answer 1

对于mysqli面向对象的样式来获取最后插入的id使用此

$mysqli->insert_id ;

http://www.php.net/manual/en/mysqli.insert-id.php

所以你的查询将是

$mysqli->query("INSERT INTO `sendmsg`(`sendname`, `recievename`, `subject`, `body` , `mdate`, `mtime`) VALUES ('$sendname','$recievename','$subject','$body','$msgdate','$msgtime')");

$new_id = $mysqli->insert_id;

$mysqli->query("INSERT INTO `recievemsg`(`msgid`, `sendname`, `recievename`, `subject`, `body`, `mdate`, `mtime`, `status`) VALUES ($new_id,'$sendname','$recievename','$subject','$body','$msgdate','$msgtime','UNREAD')");

从mysql到mysqli的转换不起作用

1 个答案: