使用UPDATE数据库查询

Question

晚上所有，我有一个语法sql错误的问题

You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'where username = danny (name, url, banner, description, sponsor, date, password)' at line 1

这是我的代码

$query = "UPDATE websites where username = $login_session (name, url, banner, description, sponsor, date, password) VALUES ('$n', '$b', '$d', '0', now(), SHA('$p'))";

Answer 1

您的MySQL查询错误，如错误所示，请查看手册。

在UPDATE中你不能像table(field,field1) values('value','value1')一样使用INSERT，你也使用field='value', field1='value1'，WHERE应该在最后，正确的顺序是查询+其中+订单+限制。 MySQL并不灵活。

Answer 2

那是因为您的UPDATE语句语法错误。检查MySQL documentation for proper UPDATE syntax。我认为你打算做INSERT而不是

INSERT INTO websites (name, url, banner, description, sponsor, date, password) 
VALUES ('$login_session', '$n', '$b', '$d', '0', now(), SHA('$p'))

修改

我认为这就是你的目标

UPDATE websites SET name = '$n', 
url = '$b', 
banner = '$d', 
description = '0', 
sponsor = 'some_value_here', 
date = now(), 
password = SHA('$p')
where username = '$login_session';