问题描述
我在 PHP 中运行以下查询:
I have the following query running in PHP:
$ticketTotal = mysql_query("SELECT SUM(`tickets_issued`) FROM `tb_att_registered_attendants` WHERE `confirmation_code`!='000000'");
但是当我返回 $ticketTotal 时，我得到 Resource id #33 并且当我转储变量时，我得到 resource(33) 类型(mysql结果).当我在 phpMyAdmin 中运行完全相同的查询时，我得到了正确的结果.我似乎在谷歌上找不到太多东西.怎么回事?
But when I return $ticketTotal, I get Resource id #33 and when I dump the variable, I get resource(33) of type (mysql result). When I run the exact same query in phpMyAdmin, I get the correct result. I can't seem to find much on google. What is going on?
提前感谢您的帮助.
推荐答案
$ticketTotal 不保存您的查询结果.您仍然必须实际获取它们.
$ticketTotal doesn't hold your query results. You still have to actually fetch them.
while ($row = mysql_fetch_assoc($ticketTotal))
{
print_r($row);
}
请不要在新代码中使用 mysql_* 函数.它们不再被维护并被正式弃用.看到 红框?改为了解 准备好的陈述，并使用 PDO 或 MySQLi- 这篇文章将帮助你决定哪一个.如果您选择 PDO，这里有一个很好的教程.
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.
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