我有律师,lawyerBadges,mysql数据库中的徽章表。我想将徽章中的所有徽章ID加载到数组中。(allBadges [])。我想将律师的所有徽章ID加载到另一个数组(lawyerBadges [])。然后我想检查allBadges元素是否在lawyerBades.how中。
到目前为止,这是我的代码。
$username = $_SESSION['username'];
getPoints();
function getPoints(){
global $connection;
global $username;
$pointCount_query =mysqli_query( $connection,"SELECT * FROM lawyer WHERE username='".$username."'");
$pointCount=mysqli_fetch_array($pointCount_query);
$points= (int)$pointCount['points'];
addBadges($points);
}
function addBadges($user_points){
global $connection;
global $username;
$badge_array = array();
$badgeList=mysqli_query($connection,"SELECT bId FROM badge ");
$badges=mysqli_fetch_array($badgeList);
$lawyers_badges=mysqli_query($connection,"SELECT bID FROM lawyerbadge WHERE username='".$username."'");
while($row=mysqli_fetch_assoc($lawyers_badges)){
$badge_array[]=$row;
}
//this is the problem area
foreach( $badge_array as $value){
echo $value.'<br />';
}
}
答案 0 :(得分:0)
你需要2个循环。外部循环将遍历所有徽章。内循环将遍历所有律师徽章。如果未找到外环徽章,则会回显徽章
foreach( $badgeList as $badge){
$badge_found_flag = false;
foreach( $badge_array as $lawyerBadge){
$badge_found_flag = true;
}
if($badge_found_flag == false){
echo $badge.'<br />';
}
}
<强>更新强>
更好的选择是使用LEFT JOIN
,这不需要手动循环,只需一个查询即可获得结果。
SELECT badge.bId FROM badge
LEFT JOIN lawyerbadge ON badge.bId = lawyerbadge.bID
WHERE username = '$username'"
AND lawyerbadge.bID IS NULL;
这只会返回您感兴趣的徽章。