我正在调用 API 来获取我的朋友和为此获得的数组,我需要“过滤”在数据库中找到的 ID 的数组。现在结果只是“空”。
这是数组(定义为变量 $friends):
[0]=>
array(2) {
["name"]=>
string(17) "One friends name"
["id"]=>
string(8) "FRIEND_ID"
}
[1]=>
array(2) {
["name"]=>
string(13) "Another friends name"
["id"]=>
string(9) "ANOTHER_FRIEND_ID"
}
[2]=>
array(2) {
["name"]=>
string(22) "Another friends name"
["id"]=>
string(9) "ANOTHER_FRIEND_ID"
}
还有 PHP 代码,我想在其中“使用数据库中的 ID 过滤数组:
$query_top_list_friends = mysql_query("SELECT * FROM ".$DBprefix."users
WHERE center_id='" . $personal['center_id'] . "' ORDER BY
workouts DESC LIMIT 10");
$i=0;
$friends = $friends['data'];
foreach($friends as $friend)
{
while ($top_list_friends = mysql_fetch_array($query_top_list_friends))
{
if($friend['id']==$top_list_friends['fid'])
{
$i++;
echo "<div class='user'>";
echo "<span class='number'>" . $i . "</span>";
echo "<span class='name'>" . $top_list_friends['name'] . "</span>";
echo "<span class='workouts'>" . $top_list_friends['workouts'] . "</span>";
echo "</div>";
}
}
}
有什么建议么?
更新
我对此进行了一些更改,但仍然没有结果:
$friends = array($friends['data']);
while ($top_list_friends = mysql_fetch_array($query_top_list_friends)) {
if($friend[$top_list_friends['fid']]) {
$i++;
echo "<div class='user'>";
echo "<span class='number'>" . $i . "</span>";
echo "<span class='name'>" . $top_list_friends['name'] . "</span>";
echo "<span class='workouts'>" . $top_list_friends['workouts'] . "</span>";
echo "</div>";
}
}