我正在编写一个网站,我需要传递一个SQL查询,以便将用户带回到我已经正常工作的页面,直到我添加它为止
它的输入被切断了...... eventStatus = (我知道它需要发布,作为调试运行)
(行内部标识来自SQL查询) (SQL通常也是传入的变量)
/*
//
//
//
*/
我是否在隐藏输入时出错?
Apache/2.4.18 (Win32) OpenSSL/1.0.2e PHP/7.0.6
Database client version: libmysql - mysqlnd 5.0.12-dev - 20150407 - $Id:241ae00989d1995ffcbbf63d579943635faf9972 $
PHP extension: mysqli Documentation
PHP version: 7.0.6
Server Windows 7 64 (Its a school project)
邮政编码php
<form action='' method='get'>
<textarea name ='comment' rows='4' cols = '50' value =''></textarea>
<button type='createcomment' name='createcomment' value='createcomment'>
Comment
<input type = 'hidden' name = 'internal_id' value ={$row["internal_id"]}</>
<input type = 'hidden' name = 'sql' value ='"SELECT * FROM `create_event` WHERE `eventStatus` = 'Happening' and 'approved' = 'Approved'"'</>
</button>
</form>
答案 0 :(得分:0)
通过不再传递sql而不是传递set命令的设定数字来修复它