【发布时间】:2017-04-30 16:01:38
【问题描述】:
您好,我正在尝试以 JSON 格式合并 MySQL 结果的输出,但我很困惑如何做到这一点,所以我需要您的帮助,请告诉我如何完成这项工作,谢谢。
SQL:
$result = $db->sql_query("SELECT a.*,i.member_id,i.members_seo_name
FROM ".TBL_IPB_USER." i
LEFT JOIN ".TBL_IPB_LA." a
ON a.member_id=i.member_id
WHERE i.".$column." = '".$val."' AND a.game = '".$game."'");
while( $dbarray = $db->sql_fetchrow($result) ){
$arr[] = $dbarray;
}
return ($arr);
我的查询的 JSON 格式的正常结果和输出是:
{
"status": 200,
"result": [
{
"member_id": "1",
"member_name": "maxdom",
"ip_address": "177.68.246.162",
"session_onlineplay": "1",
"sid": "IR63374a32d1424b9288c5f2a5ce161d",
"xuid": "0110000100000001",
"serialnumber": "9923806a06b7f700a6ef607099cb71c6",
"game": "PlusMW3",
"members_seo_name": "maxdom"
},
{
"member_id": "1",
"member_name": "maxdom",
"ip_address": "81.254.186.210",
"session_onlineplay": "1",
"sid": "IR3cd62da2f143e7b5c8f652d32ed314",
"xuid": "0110000100000001",
"serialnumber": "978e2b2668ec26e77c40c760f89c7b31",
"game": "PlusMW3",
"members_seo_name": "maxdom"
}
],
"handle": "checkUSER"
}
但我想像这样合并输出和结果:
{
"status": 200,
"result": [
{
"member_id": "1",
"member_name": "maxdom",
"ip_address": [
"177.68.246.162",
"81.254.186.210"
],
"session_onlineplay": "1",
"sid": [
"IR63374a32d1424b9288c5f2a5ce161d",
"IR3cd62da2f143e7b5c8f652d32ed314"
],
"xuid": "0110000100000001",
"serialnumber": [
"9923806a06b7f700a6ef607099cb71c6",
"978e2b2668ec26e77c40c760f89c7b31"
],
"game": "PlusMW3",
"members_seo_name": "maxdom"
}
],
"handle": "checkUSER"
}
【问题讨论】:
-
合并背后的逻辑是什么?所有项目都应合并为一个项目,还是某些选项可能导致合并为多个项目?
-
请注意,由于 WHERE 子句中的
a.game = ...,您的 LEFT JOIN 将被转换为 INNER JOIN。