Php Mysql数据库查询错误
问题描述:
基本上,从我制作的表单我发送customer_name
,并将此代码与此代码一起拖出并与时间一起保存到数据库中。此代码一直给我这个错误:Php Mysql数据库查询错误
Fatal error: Using $this when not in object context in /home/projectu/public_html/sub/save.php on line 19
注:19条线是这一个:
$db->query($queryone);
这里是我的代码:
全Save.php
include('db-config.php');
$customer_name = $_POST['customername'];
/* Kaspersky */
$kaspersky_date = strtotime("+11 months").'|'.strtotime("+12 months");
$kaspersky = explode("|", $kaspersky_date);
$kaspersky_temp = "$customer_name got new kaspersky.";
/* PC Picked-UP */
$pickedups_date = strtotime("+1 months");
$pickedups_temp = "$customer_name picked up his computer.";
if(isset($_POST['kaspersky'])) {
$queryone = "INSERT INTO sublist (scheduled_date, customer_name, kaspersky_status, kaspersky_template)
VALUES ($kaspersky[0], $customer_name, YES, $kaspersky_temp)";
$this->query($queryone);
$querytwo = "INSERT INTO sublist (scheduled_date, customer_name, kaspersky_status, kaspersky_template)
VALUES ($kaspersky[1], $customer_name, YES, $kaspersky_temp)";
$this->query($querytwo);
}
if(isset($_POST['pickeduppc'])) {
$query = "INSERT INTO sublist (scheduled_date, customer_name, pcpickup_status, pcpickup_template)
VALUES ($pickedups_date, $customer_name, YES, $pickedups_temp)";
}
答
您不能在非对象或静态方法中使用$this
。你应该创建一个新的对象。
$db = new mysqli(......);
$db->query("SELECT ... FROM...");
+0
感谢您。我想我的查询也是错误的.. – Extelliqent
'$ this'只能在对象方法中使用。如果在对象之外使用它将会失败。 – datasage
那么$ db的值是多少?以及该对象的查询方法是什么? –
我想要做的就是运行插入查询。我怎么做?连接数据库我使用mysqli – Extelliqent