我目前从我的关注者数据库获取accountID并将其输出到JSON。但是,我不是从数据库中获取accountID,而是如何从另一个表“Accounts”获取用户信息并将其附加到JSON?
我目前的代码是:
$accountID = NULL;
if (isset($_GET['accountID'])) {
$accountID = mysql_real_escape_string($_GET['accountID']);
}
else {
exit("No accountID set");
}
$query = mysql_query("SELECT * FROM Following WHERE `followingUserID` = '$accountID`");
$rows = array();
while($r = mysql_fetch_assoc($query)) {
$rows[] = $r;
}
header('Content-type: application/json');
exit(json_encode($rows));
答案 0 :(得分:2)
我会继续下去并假设一些事情:
你有两张桌子:
继
followId,followUserID,someOtherColumns
帐户
userID,someOtherAccountCollumns
鉴于此,修复很简单:
// Switch this:
$query = mysql_query("SELECT * FROM Following WHERE `followingUserID` = '$accountID`");
// By this:
$query = mysql_query("
SELECT f.*, a.*
FROM Following as f
JOIN Accounts as a on f.followingUserID = a.userID
WHERE `f.followingUserID` = '$accountID`
");
您需要将列名修改为正确的名称,但这可以帮助您获取下表中任何给定用户的所有帐户信息。
祝你好运!来源: