我的SQL是:
INSERT INTO `p_newsbase` (`id`,`cid`,`title`,`author`,`date_time`) VALUES (NULL,'2','啊','啊','1233')
在Console下面运行时完全没有问题的,但是复制到PHP代码中,会给我报参数错误呢:<?php
$mysqli = new mysqli('localhost','root','root','news') or die(mysqli_error());
$mysqli->query("INSERT INTO `p_newsbase` (`id`,`cid`,`title`,`author`,`date_time`) VALUES (NULL,'2','啊','啊','1233')") or die(mysqli_error());
?>Warning: mysqli_error() expects exactly 1 parameter, 0 given in C:\Documents and Settings\Administrator\Zend\workspaces\DefaultWorkspace7\userSYS\test.php on line 3
INSERT INTO `p_newsbase` (`id`,`cid`,`title`,`author`,`date_time`) VALUES (NULL,'2','啊','啊','1233')
在Console下面运行时完全没有问题的,但是复制到PHP代码中,会给我报参数错误呢:<?php
$mysqli = new mysqli('localhost','root','root','news') or die(mysqli_error());
$mysqli->query("INSERT INTO `p_newsbase` (`id`,`cid`,`title`,`author`,`date_time`) VALUES (NULL,'2','啊','啊','1233')") or die(mysqli_error());
?>Warning: mysqli_error() expects exactly 1 parameter, 0 given in C:\Documents and Settings\Administrator\Zend\workspaces\DefaultWorkspace7\userSYS\test.php on line 3
$mysqli->connect_error && die('Connect Error: ' . $mysqli->connect_error);不能那样new的