唯一id用户团队排名php/json
编辑 我明白了:) 这当然不是优化,但它可以工作:唯一id用户团队排名php/json,php,json,Php,Json,编辑 我明白了:) 这当然不是优化,但它可以工作: $db = json_decode( file_get_contents( $location ), true ); $users = $db['users']; $scoreTeam1 = 0; $scoreTeam2 = 0; $scoreTeam3 = 0; $scoreTeam4 = 0; foreach ($users as $user) { $team = $user['team']; $userScore = $
$db = json_decode( file_get_contents( $location ), true );
$users = $db['users'];
$scoreTeam1 = 0;
$scoreTeam2 = 0;
$scoreTeam3 = 0;
$scoreTeam4 = 0;
foreach ($users as $user) {
$team = $user['team'];
$userScore = $user['userScore'];
switch ($team) {
case 1:
$scoreTeam1 = $scoreTeam1 + $userScore;
break;
case 2:
$scoreTeam2 = $scoreTeam2 + $userScore;
break;
case 3:
$scoreTeam3 = $scoreTeam3 + $userScore;
break;
case 4:
$scoreTeam4 = $scoreTeam4 + $userScore;
break;
}
}
初始问题
我想从以下json文件中创建一个php团队排名:
"users": {
"uniqueID": {
"team": 1,
"userScore": 2500,
}, "uniqueID": {
"team": 2,
"userScore": 1235,
}, "uniqueID": {
"team": 3,
"userScore": 6582,
}, "uniqueID": {
"team": 4,
"userScore": 1200,
}, "uniqueID": {
"team": 1,
"userScore": 9875,
}, "uniqueID": {
"team": 2,
"userScore": 500,
}, "uniqueID": {
"team": 3,
"userScore": 12,
}, "uniqueID": {
"team": 4,
"userScore": 695,
}, "uniqueID": {
"team": 1,
"userScore": 332,
}, etc.
}
我有4个团队,我想我需要创建4个变量:
$scoreTeam1
$scoreTeam2
$scoreTeam3
$scoreTeam4
致以最诚挚的问候。您可以通过以下功能实现您想要的:
$result = array_reduce($db['users'], function($carry, $team){
if (isset($carry[$team['team']]) ) {
$carry[$team['team']] = $carry[$team['team']] + $team['userScore'];
} else {
$carry[$team['team']] = $team['userScore'];
}
return $carry;
});