【发布时间】:2015-06-09 23:28:29
【问题描述】:
下面是我的 php 的简化版本,来自 wordpress 简码。我如何从所有使用 jquery 的玩家那里获得名字?
部分代码:
$profile_personal_info['id'] = $player->ID;
$profile_personal_info['Firstname'] = $fname;
$profile_personal_info['Lastname'] = $lname;
$profile_details_stats['S'] = $playerShots;
$profile_details_stats['F'] = $playerFouls;
$personalInfo = array();
foreach ( $profile_personal_info as $key => $value ) {
$personalInfo[$key] = $value;
}
$playerStatistics = array();
foreach ( $profile_details_stats as $key => $value ) {
$playerStatistics[$key] = $value;
}
$response[$player->ID] = array_merge($personalInfo, $playerStatistics);
}
$response['player-'.$player->ID] = array_merge($personalInfo, $playerStatistics);
}
这是来自
的输出jQ.getJSON( "PATH_TO_PHP_FILE", function( data ) {
console.log(data);
}
player-151: Object
personal: Object
Firstname:"Jason"
Lastname:"Smith"
为了简洁,我简化了代码。
【问题讨论】:
标签: php jquery mysql arrays json