无法弄清楚如何运行mysqli_multi_query并使用上一个查询的结果

Question

我之前从未使用过mysqli_multi_query而且它让我的大脑难以置信,我在网上找到的任何例子并没有帮助我弄清楚我想要做什么.

这是我的代码:

<?php

    $link = mysqli_connect("server", "user", "pass", "db");

    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", 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";

    $i = 0;
    $agentsresult = mysqli_multi_query($link, $agentsquery);

    while ($row = mysqli_fetch_array($agentsresult)){
        $number_of_apps = getAgentAppsWeek($row['agent_name'],$weeknum,$current_year);
        $i++;
?>

            <tr class="tr<?php echo ($i & 1) ?>">
                <td style="font-weight: bold;"><?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

    }
?>

我要做的就是运行多个查询,然后使用这4个查询的最终结果并将它们放入我的表中.

上面的代码根本不起作用,我只是得到以下错误:

警告:mysqli_fetch_array()预计参数1被mysqli_result,在C中给出布尔:\ XAMPP\htdocs中\ hydroboard\hydro_reporting_2010.php上线391

任何帮助？

Answer 1

从手册:mysqli_multi_query()返回bool指示成功.

要从第一个查询中检索结果集,可以使用mysqli_use_result()或mysqli_store_result().可以使用mysqli_more_results()和mysqli_next_result()处理所有后续查询结果.

这是一个返回多个查询的最后结果的函数:

function mysqli_last_result($link) {
    while (mysqli_more_results($link)) {
        mysqli_use_result($link); 
        mysqli_next_result($link);
    }
    return mysqli_store_result($link);
}

用法:

$link = mysqli_connect();

$query  = "SELECT 1;";
$query .= "SELECT 2;";
$query .= "SELECT 3";

mysqli_multi_query($link, $query);
$result = mysqli_last_result($link);
$row = $result->fetch_row();
echo $row[0];  // prints "3"

$result->free();
mysqli_close($link);

Answer 2

好吧，经过一些摆弄、试验和错误，并参考我在谷歌搜索中遇到的另一篇文章，我已经成功解决了我的问题！

这是新代码：

<?php

    $link = mysqli_connect("server", "user", "pass", "db");

    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", 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 class="tr<?php echo ($i & 1) ?>">
                <td style="font-weight: bold;"><?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

        }
    }
?>

在为每个查询多次粘贴 mysqli_next_result 后，它神奇地工作了！耶！我明白它为什么有效，因为我告诉它跳到下一个结果 3 次，所以它跳到查询 #4 的结果，这是我想要使用的结果。

不过，对我来说似乎有点笨重，应该只有一个类似 mysqli_last_result($link) 之类的命令，或者如果你问我的话......

感谢 rik 和 f00 的帮助，我最终到达了那里:)