我正在尝试执行以下PHP / MySQL查询,它对前两组工作正常,但对于其他人我收到MySQL错误,这是否写得正确?
$user =& JFactory::getUser();
$N = $user->get('name');
$username = $user->get('username');
$groups = $user->get('groups');
foreach($groups as $groupName=>$groupId)
{
}
$G=$groupName;
if ($G=="Management Staff")
$result = mysql_query("SELECT * FROM hqfjt_chronoforms_data_addupdatelead");
elseif ($G=="Website Developers")
$result = mysql_query("SELECT * FROM hqfjt_chronoforms_data_addupdatelead");
else
$result = mysql_query("SELECT * FROM hqfjt_chronoforms_data_addupdatelead WHERE createdby=$N");
当我像其他人一样登录时,我得到了:
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in C:\server2go\server2go\htdocs\chandlers\components\com_jumi\views\application\view.html.php(38) : eval()'d code on line 87
Warning: mysql_free_result() expects parameter 1 to be resource, boolean given in C:\server2go\server2go\htdocs\chandlers\components\com_jumi\views\application\view.html.php(38) : eval()'d code on line 132
答案 0 :(得分:0)
$result = mysql_query("SELECT * FROM hqfjt_chronoforms_data_addupdatelead WHERE createdby=$N");
很可疑,应该是
$result = mysql_query("SELECT * FROM hqfjt_chronoforms_data_addupdatelead WHERE createdby='$N'");
确保,$ N没有任何邪恶。
答案 1 :(得分:0)
不确定这是否只是一个糟糕的复制/粘贴作业,你进入stackoverflow但看起来你的花括号都是不合适的。认为你会想要更像这样的东西:
$username = $user->get('username');
$groups = $user->get('groups');
foreach($groups as $groupName=>$groupId) {
$G=$groupName;
if ($G=="Management Staff")
$result = mysql_query("SELECT * FROM hqfjt_chronoforms_data_addupdatelead");
elseif ($G=="Website Developers")
$result = mysql_query("SELECT * FROM hqfjt_chronoforms_data_addupdatelead");
else
$result = mysql_query("SELECT * FROM hqfjt_chronoforms_data_addupdatelead WHERE createdby=$N");
}
这样你的foreach循环实际上会执行它下面的代码。您粘贴到问题中的代码有一个for循环,它在循环中不执行任何操作,因为它是这样编写的:
foreach($groups as $groupName=>$groupId) {}
然后你继续尝试使用循环外的foreach循环中的一个变量:
$G=$groupName;
if ($G=="Management Staff")
...
哪个不起作用,因为$groupName
变量只会在foreach
循环的范围内设置。