无法计算如何运行mysqli_multi_query并使用上一个查询的结果

Question

我以前从未使用过mysqli_multi_query，它在我的大脑中摇摆不定，我在网上找到的任何例子都无法帮助我准确地知道我想要做什么。

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

    }
?>

我所要做的就是运行一个多个查询，然后使用这4个查询的最终结果并将它们放入我的表中。

上面的代码实际上根本不起作用，我只得到以下错误：

警告: C:\xampp\htdocs\hydroboard\hydro_reporting_2010.php ()要求参数1为mysqli_result，在第391行中给出的布尔值为mysqli_result

有什么帮助吗？

Answer 1

好吧，经过一些摆弄，尝试和错误，并参考了另一个帖子，我在谷歌搜索中遇到的，我设法解决了我的问题！

以下是新代码：

<?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

        }
    }
?>

在对每个查询多次插入mysqli_next_result之后，它神奇地工作了！耶！我理解它为什么工作，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。

虽然对我来说有点笨重，但是应该有一个类似mysqli_last_result($link)之类的命令，或者如果你问我.

谢谢你的帮助和f00，我终于到了那里:)

Answer 2

从手册：mysqli_multi_query()返回一个指示成功的bool。

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

下面是一个函数，它返回多个查询的最后一个结果：

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 3

我将简化您要做的事情，方法是创建一个存储过程，它将生成领导/代理统计数据，然后从php (单次调用)调用它，如下所示：

这里的完整脚本：http://pastie.org/1451802

或者，您可能可以将单个查询合并为单个select/group by语句。

见此处：http://pastie.org/1451842

select
 leaders.agent_name, 
 sum(leaders.job_value) as sum_job_value, 
 sum(leaders.points_value) as sum_points_value 
from
(
select 
 agent_name, 
 job_number, 
 job_value, 
 points_value 
from 
 jobs 
where 
 year(booked_date) = 2011 and weekofyear(booked_date) = 2
union all
select distinct
 agent_name,
 0,0,0
from
 apps
where 
 year(booked_date) = 2011 and weekofyear(booked_date) = 2
) leaders
group by
 agent_name 
order by sum_points_value desc;

存储过程

drop procedure if exists list_leaders;

delimiter #
create procedure list_leaders
(
in p_year smallint unsigned,
in p_week tinyint unsigned
)
begin

  create temporary table tmp_leaders(
    agent_name varchar(20) not null,
    job_number int unsigned not null default 0, -- note the default values
    job_value decimal(3,1) not null default 0,
    points_value decimal(8,2) not null default 0
  )engine=memory;

  insert into tmp_leaders (agent_name, job_number, job_value, points_value) 
    select agent_name, job_number, job_value, points_value from jobs 
    where year(booked_date) = p_year and weekofyear(booked_date) = p_week;

  insert into tmp_leaders (agent_name) -- requires default values otherwise you will get nulls
    select distinct agent_name from apps
    where year(booked_date) = p_year and weekofyear(booked_date) = p_week;

  select 
    agent_name, 
    sum(job_value) as sum_job_value, 
    sum(points_value) as sum_points_value 
   from
    tmp_leaders
   group by
    agent_name order by sum_points_value desc;

  drop temporary table if exists tmp_leaders;

end#

delimiter ;

call list_leaders(year(curdate()), weekofyear(curdate()));

PHP脚本

<?php

ob_start(); 

try
{
    $db = new mysqli("localhost", "foo_dbo", "pass", "foo_db", 3306);

    if ($db->connect_errno) 
        throw new exception(sprintf("Could not connect: %s", $db->connect_error));

    $sqlCmd = "call list_leaders(2011, 2)";
    $result = $db->query($sqlCmd);

    if(!$result) throw new exception(sprintf("Invalid query : %s", $sqlCmd));

    if($db->affected_rows <= 0){
        echo "no leaders found !";
    }
    else{
        $leaders = $result->fetch_all(MYSQLI_ASSOC);
        foreach($leaders as $ldr){
            // do stuff
            echo $ldr["agent_name"], "<br/>";
        }
    }
}
catch(exception $ex)
{
    ob_clean(); 
    echo sprintf("zomg borked - %s", $ex->getMessage());
}

if(!$db->connect_errno) $db->close();
ob_end_flush();
?>

现在比较简单--希望它能帮上忙:)