好吧，经过反复试验，反复尝试，并参考了我在Google搜索中遇到的另一篇文章的参考，我已经设法解决了我的问题！
这是新的代码：
<?php $link = mysqli_connect("server", "user", "pass", "db"); if (mysqli_connect_errno()) { printf("Connect failed: %sn", mysqli_connect_error()); exit(); } $agentsquery = "CREATE TEMPORARY TABLE LeaderBoard ( `agent_name` varchar(20) NOT NULL, `job_number` int(5) NOT NULL, `job_value` decimal(3,1) NOT NULL, `points_value` decimal(8,2) NOT NULL );"; $agentsquery .= "INSERT INTO LeaderBoard (`agent_name`, `job_number`, `job_value`, `points_value`) SELECt agent_name, job_number, job_value, points_value FROM jobs WHERe YEAR(booked_date) = $current_year && WEEKOFYEAR(booked_date) = $weeknum;"; $agentsquery .= "INSERT INTO LeaderBoard (`agent_name`) SELECt DISTINCT agent_name FROM apps WHERe YEAR(booked_date) = $current_year && WEEKOFYEAR(booked_date) = $weeknum;"; $agentsquery .= "SELECt agent_name, SUM(job_value), SUM(points_value) FROM leaderboard GROUP BY agent_name ORDER BY SUM(points_value) DESC"; mysqli_multi_query($link, $agentsquery) or die("MySQL Error: " . mysqli_error($link) . "<hr>nQuery: $agentsquery"); mysqli_next_result($link); mysqli_next_result($link); mysqli_next_result($link); if ($result = mysqli_store_result($link)) { $i = 0; while ($row = mysqli_fetch_array($result)){ $number_of_apps = getAgentAppsWeek($row['agent_name'],$weeknum,$current_year); $i++;?> <tr > <td ><?php echo $row['agent_name'] ?></td> <td><?php echo $row['SUM(job_value)'] ?></td> <td><?php echo $row['SUM(points_value)'] ?></td> <td><?php echo $number_of_apps; ?></td> </tr><?php } }?>在对每个查询多次粘贴mysqli_next_result之后，它神奇地工作了！好极了！我知道它为什么起作用，因为我告诉它要跳到下一个结果3次，所以它跳到查询4的结果，这是我要使用的结果。
虽然对我来说似乎有点笨拙，但应该只提供诸如mysqli_last_result（$ link）之类的命令，或者如果您问我一个问题…
感谢rik和f00的帮助，我最终到达了:)
