Question

我有一个具有这种结构的数组：

$queues[n] Array (
[id] => integer
[idClient] => integer
[name] => string
[people] => integer )

其中填充了：

$query = "SELECT clients.idClient AS 'idClient', queues.idQueue AS 'idQueue', queues.name AS 'name' FROM clients, queues WHERE clients.idClient = queues.client";
    $queues = null;

    $result = mysql_query($query);
    if ($result){
        while ($queue = mysql_fetch_assoc($result)){
            $queues[] = array ("idClient" => $queue['idClient'], "id" => $queue['idQueue'], "name" => $queue['name'], "people" => 0);
        }
    }

每个'n'值与来自数据库和人的队列匹配，默认情况下设置为0.

填充数组后，我再次使用每个队列查询数据库到其他表，以获取队列中的人数，并使用如下查询：

SELECT COUNT(*) FROM peoplequeued WHERE queue ='".$queue['name']."'

然后：

$result = mysql_query($query);
        if ($result){
            $num_people = mysql_fetch_row($result);
            $queue['people'] = $num_people[0];
        }

这里发生了一些奇怪的事情。如果我在foreach中回显$ queue ['people']，它会显示它得到的值，但是如果我在返回之前预览完整数组，它会回到0。

可能会发生什么？

Answer 1

我的猜测是，您使用$queues循环遍历foreach，如下所示：

foreach ( $queues as $queue ) {
    $result = mysql_query($query);
    if ($result){
        $num_people = mysql_fetch_row($result);
        $queue['people'] = $num_people[0];
    }
}

如果是，$queue和$queues[$n]之间没有“链接”，即修改$queue，则不会修改$queues。

如果是这种情况，您应该使用$queue作为参考变量，或者修改$queues[$n]并将$n作为索引。

foreach ( $queues as &$queue ) { // add & so $queue is a reference to an element in $queues
    $result = mysql_query($query);
    if ($result){
        $num_people = mysql_fetch_row($result);
        $queue['people'] = $num_people[0];
    }
}
unset($queue); // drop the reference, otherwise you might have unexpected results after modifying $queue outside the loop

......或......

foreach ( $queues as $n => $queue ) { // store index of "current" element in $n
    $result = mysql_query($query);
    if ($result){
        $num_people = mysql_fetch_row($result);
        $queues[$n]['people'] = $num_people[0]; // change $queue to $queues[$n]
    }
}

或者，我建议您考虑将所有数据都放在一个语句中。看起来像这样的东西可能适合你：

select
    baseTable.id,
    baseTable.idClient,
    baseTable.name,
    peopleCount.count as people
from
    baseTable
    left join (
        select
            count(*) as count,
            queue
        from
            peoplequeued
        group by
            queue
    ) as peopleCount on peopleCount.queue = baseTable.name

Answer 2

听起来你可能正在使用foreach循环遍历$queues数组，如下所示：

foreach($queues as $queue) {

}

生成的每个$queue变量都是一个副本，对它的更改不会保存到$queues变量中。要按照您的意图保存更改，您需要执行以下操作：

foreach($queues as $k => $queue) {
    $queue['people'] = 10;
    $queues[$k] = $queue;
    // or, more efficiently...
    $queues[$k]['people'] = 10;
}

在PHP5中你也可以参考。见http://uk.php.net/manual/en/control-structures.foreach.php

值不保存在数组中

2 个答案: