我正在制作一个用户管理系统,我想做的是当用户登录时,他看不到(例如销售员工)他看不到其他部门的链接,我想做的很简单,但我不知道为什么不工作,我的代码是:
$departmentidquery = mysql_query("SELECT department_id FROM users WHERE username = '".$username."'");
$departmnetid = mysql_fetch_row ($departmentidquery);
<ul>
<li><a href="logout.php">Logout.</a></li>
<li <?php if ($departmnetid[0]!=1){?>style="display:none"<?php } ?>> <a href="admin.php">Admin Pgae</a> </li>
<li <?php if ($departmnetid[0]!=2){?>style="display:none"<?php } ?>> <a href="sales.php">sales Pgae</a> </li>
<li <?php if ($departmnetid[0]!=3){?>style="display:none"<?php } ?>> <a href="tech.php">Tech Pgae</a> </li>
</ul>
错误是:
SCREAM: Error suppression ignored for
( ! ) Notice: Undefined variable: departmnetid in C:'wamp'www'Ticket.sys'index.php on line 49
SCREAM: Error suppression ignored for
( ! ) Notice: Undefined variable: departmnetid in C:'wamp'www'Ticket.sys'index.php on line 50
SCREAM: Error suppression ignored for
( ! ) Notice: Undefined variable: departmnetid in C:'wamp'www'Ticket.sys'index.php on line 51
我确信从查询中,我已经尝试了display:none
与<div>
及其工作的技巧,是否有更好的方法来做到这一点?
你没有像你应该的那样移出一个PHP块…这段代码应该是
<?php
$departmentidquery = mysql_query("SELECT department_id FROM users WHERE username = '".$username."'");
$departmnetid = mysql_fetch_row ($departmentidquery);
?>
<ul>
<li><a href="logout.php">Logout.</a></li>
<li <?php if ($departmnetid[0]!=1){?>style="display:none"<?php } ?>><a href="admin.php">Admin Pgae</a> </li>
<li <?php if ($departmnetid[0]!=2){?>style="display:none"<?php } ?>><a href="sales.php">sales Pgae</a> </li>
<li <?php if ($departmnetid[0]!=3){?>style="display:none"<?php } ?>><a href="tech.php">Tech Pgae</a> </li>
</ul>
检查是否从mysql_fetch_row
mysql实例可能有问题(实例关闭,无效的用户名/通行证)。你试过这个吗?
if (!$departmnetid) {
echo 'Could not run query: ' . mysql_error();
exit;
}