Question

Sql响应如下所示：名称|位置|日期|地址。行可以包含相同的名称，位置和日期，但每次都会更改地址。所以，我需要把这个响应放到这样的数组中：

while($r = mysqli_fetch_assoc($q_res)){
    $i = 0;

    if(!in_array($r['watcher_name'], $daily_quests)){          
        $daily_quests[$i]['name'] = $r['name'];
        $daily_quests[$i]['date'] = $r['date'];
        $daily_quests[$i]['locality'] = $r['locality'];
    }
    else{
        $daily_quests['address'] = $intrnal_addr[$i => $r['address']];
    }

}

我已尝试使用此代码，但确实无法工作=（

=INDEX(A:A;ROUND(ROW(A1)/2);1)

Answer 1

试试这个：

//Initialisation
$name = '';
$daily_quests = array();
$i = -1;

while($r = mysqli_fetch_assoc($q_res)){

    //If the row is different than precedent
    if (strcmp($r['name'], $name) != 0)
    {
        $i++;
        $name = $r['name'];
        $daily_quests[$i] = array();
        $daily_quests[$i]['name'] = $r['name'];
        $daily_quests[$i]['date'] = $r['date'];
        $daily_quests[$i]['locality'] = $r['locality'];
        $daily_quests[$i]['address'] = array(); 
    }
    // For each row, we must add address
    $daily_quests[$i]['address'][] = $r['address']
}

但是，我们假设名称是唯一的。与id进行比较更安全。

Answer 2

您可以通过对MySQL查询进行分组来解决此问题：

select
Name, Date, Locality, group_concat(Address SEPARATOR '|') as addresses
from __table__
group by Name, Date, Locality;

然后在php：

while($r = mysqli_fetch_assoc($q_res)){
    $r['Address'] = explode('|', $r['addresses']);
    unset($r['addresses']);
    $daily_quests[] = $r;
}

解析对数组的sql响应

2 个答案: