[英]Build a multidimensional PHP array from MySQL query
我正在嘗試從MySQL查詢創建多維數組。 該陣列深3層。 每個用戶都有一個membershipid
,此成員資格ID可以具有多個characterid
。 每個這些characterid
可以具有多個屬性。
Membershipid-> characterid(最多3個)-> 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,但每個人僅顯示一個字符,而查詢結果顯示每個membership id
每個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
)
)
我可能在數組定義上做錯了,每次都覆蓋了我的characterid,因為我只得到了最新的characterid
。
首選數組:
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
)
)
)
)
如果您允許我稍微重組結果數組,這應該是一種更好的方法:
// 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' => $data_racehash];
} // end while statement
如您所見,我已經將“字符ID”移到了子數組的鍵上。 這樣做是為了避免覆蓋其他字符。
我還對數組使用了新的表示法。 其中的extract()
只是使代碼更易於閱讀,但您可以擺脫它。
也許最好將字符放在自己的數組中,如下所示:
// Loop through the requested data, add to an array
while ($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' => $data_racehash];
} // end while statement
如果$ categorylist [$ data ['membershipid']]存在,則下一個字符將覆蓋它。
您必須檢查$ categorylist [$ data ['membershipid']]是否存在。 在這種情況下,請使用array_push()添加第二個字符。 否則,您的代碼仍然有效。
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