从MySQL查询构建多维PHP数组

Question

我正在尝试从MySQL查询创建多维数组。 该阵列深3层。 每个用户都有一个membershipid ，此成员资格ID可以具有多个characterid 。 每个这些characterid可以具有多个属性。

Membershipid-> characterid（最多3个）-> light，racehash。

    $query = "SELECT p.membershipid, p.displayname, c.characterid, c.light, c.racehash
    FROM Players as p
    LEFT JOIN Characters as c
    ON c.membershipid = p.membershipid";
    $result = $mysqli->query($query) or die("SQL Error: Members konden niet worden opgehaald.");

    // Loop through the requested data, add to an array
    while ($data = mysqli_fetch_assoc($result)) {
        $categorylist[ $data['membershipid'] ] = array(
                    'displayname' => $data['displayname'],
              array('characterid' => $data['characterid'],
                    'light' => $data['light'],
                          'racehash' => $data['racehash']
            )
                    );

    } // end while statement

// A test to see how many items in array          
echo "<pre>\n";
print_r($categorylist);    
echo "</pre>\n";  

此方法有效，但每个人仅显示一个字符，而查询结果显示每个membership id每个character id membership id 。

    Array
(
    [4611686018428931875] => Array
        (
            [displayname] => White_Anomaly
            [0] => Array
                (
                    [characterid] => 2305843009264668680
                    [light] => 554
                    [racehash] => 3887404748
                )

        )

    [4611686018437972738] => Array
        (
            [displayname] => Bpunisher7
            [0] => Array
                (
                    [characterid] => 2305843009277456739
                    [light] => 392
                    [racehash] => 2803282938
                )

        )

SQL查询的结果如下：

我可能在数组定义上做错了，每次都覆盖了我的characterid，因为我只得到了最新的characterid 。

首选数组：

Array
(
    [4611686018428931875] => Array
        (
            [displayname] => White_Anomaly
            [characters] => Array
                (
                    [0] => Array
                        (
                            [characterid] => 2305843009264668678
                            [light] => 370
                            [racehash] => 3887404748
                        )
                    [1] => Array
                        (
                            [characterid] => 2305843009264668680
                            [light] => 554
                            [racehash] => 3887404748
                        )
                )
        )

    [4611686018437972738] => Array
        (
            [displayname] => Bpunisher7
            [characters] => Array
                (
                    [0] => Array
                        (
                            [characterid] => 2305843009265241161
                            [light] => 534
                            [racehash] => 2803282938
                        )
                    [1] => Array
                        (
                            [characterid] => 2305843009265241163
                            [light] => 524
                            [racehash] => 3887404748
                        )
                    [2] => Array
                        (
                            [characterid] => 2305843009277456739
                            [light] => 392
                            [racehash] => 3887404748
                        )
                )
        )
)

Answer 1

如果您允许我稍微重组结果数组，这应该是一种更好的方法：

// Loop through the requested data, add to an array
while ($data = mysqli_fetch_assoc($result)) {
    extract($data,EXTR_PREFIX_ALL,'data');
    $categorylist[$data_membershipid]['displayname']     = $data_displayname;
    $categorylist[$data_membershipid][$data_characterid] = ['light'    => $data_light,
                                                            'racehash' => $data_racehash];

} // end while statement

如您所见，我已经将“字符ID”移到了子数组的键上。 这样做是为了避免覆盖其他字符。

我还对数组使用了新的表示法。 其中的extract()只是使代码更易于阅读，但您可以摆脱它。

也许最好将字符放在自己的数组中，如下所示：

// Loop through the requested data, add to an array
while ($data = mysqli_fetch_assoc($result)) {
    extract($data,EXTR_PREFIX_ALL,'data');
    $categorylist[$data_membershipid]['displayname'] = $data_displayname;
    $categorylist[$data_membershipid]['characters'][$data_characterid] = ['light'    => $data_light,
                                                                          'racehash' => $data_racehash];

} // end while statement

Answer 2

如果$ categorylist [$ data ['membershipid']]存在，则下一个字符将覆盖它。

您必须检查$ categorylist [$ data ['membershipid']]是否存在。 在这种情况下，请使用array_push（）添加第二个字符。 否则，您的代码仍然有效。