MySQL SUM 查询问题 php

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 They are no longer maintained and are officially deprecated.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.

这篇关于MySQL SUM 查询问题 php的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！