PHP函数没有选择数据并显示数据库中的mysql_num_rows

时间:2014-01-01 16:08:43

标签: php

我正在使用这个PHP函数:

if(!function_exists("SelectQuery")) {
    function SelectQuery ($sql) {
        global $conn;
        $SelectQuery = mysql_query($sql,$conn);
        $SelectQuery_Results=array();
        while($SelectQuery_Row = mysql_fetch_array($SelectQuery)) {
            $SelectQuery_Results[] = $SelectQuery_Row;
        }
        return $SelectQuery_Results;
    }
}

然后尝试在此处调用它:

$sql="SELECT * from tickets where ticketnumber = '".$_GET["seq"]."' ";
$ticket = SelectQuery($sql);
foreach($ticket as $ticket) {
    echo mysql_num_rows($ticket);
}

但它出现了这个错误:

Warning: mysql_num_rows() expects parameter 1 to be resource, array given in /home/integra/public_html/admin/helpdesk/reviewtickets.php on line 101

我做错了什么?

如果我回音$ticket["sequence"];有效,只有mysql_num_rows($ticket);无效

2 个答案:

答案 0 :(得分:0)

更改你的foreach循环:

foreach($ticket as $t) {
    echo mysql_num_rows($t);
}

答案 1 :(得分:0)

您应该将mysql_query()结果传递给mysql_num_rows()。

$SelectQuery = mysql_query($sql,$conn);
echo mysql_num_rows($SelectQuery);