[英]how to select all rows in sql table when a condition is meet
我有這樣的桌子。
+----+---------+-----------+
| id | user_id | friend_id |
+----+---------+-----------+
| 1 | 1 | 20 |
| 2 | 4 | 20 |
| 3 | 6 | 20 |
+----+---------+-----------+
我正在開發一個friend_list系統,但是現在被代碼困住了,問題是我想回顯所有他們的friend_id等於20的user_id,但是當我運行我的代碼時,它只會回顯第一個user_id,請問我要去嗎我的代碼有誤,應該為我提供幫助並進行修復。
if(isset($_SESSION['em'])){
$eml = $_SESSION['em'];
$sql = "select id from users where email='$eml'";
$res = mysqli_query($conn,$sql);
$row = mysqli_fetch_assoc($res);
$id = $row['id'];
$sql_friend_list = "select user_id from friend where friend_id='$id'";
$res_friend_list = mysqli_query($conn,$sql_friend_list);
$row_friend_list = mysqli_fetch_assoc($res_friend_list);
$fid = $row_friend_list['user_id'];
echo $fid;
}
因此,現在用戶已登錄,並且用戶ID等於t0 20,然后代碼檢查登錄用戶的ID是否等於20的friend_id。
使用循環遍歷關聯數組。 用以下代碼替換最后三行代碼:
while($row = mysqli_fetch_assoc($res_friend_list)){
echo $row['user_id'];
}
我希望這會有所幫助。
嘗試
if(isset($_SESSION['em'])){
$eml = $_SESSION['em'];
$sql = "select id from users where email='$eml'";
$res = mysqli_query($conn,$sql);
$row = mysqli_fetch_assoc($res);
$id = $row['id'];
$sql_friend_list = "select user_id from friend where friend_id='$id'";
$res_friend_list = mysqli_query($conn,$sql_friend_list);
while($row = mysqli_fetch_assoc($res_friend_list){
$fid = $row['user_id'];
echo $fid;
}
}
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