Php 如何使用表别名从mysqli_fetch_assoc获取结果

我对PHP非常陌生，如果我的select语句有表联接，并且这些联接是用表别名处理的，我不知道如何从

mysqli\u fetch\u assoc（）

函数中获取结果

代码如下所示：

$sql = 'SELECT T1.update_text, T1.created_at, T2.username, T3.group_name   FROM updates AS T1
        INNER JOIN users AS T2 ON T2.user_id = T1.user_id_fk
        INNER JOIN groups AS T3 ON T3.group_id = T1.group_id_fk
        WHERE T1.user_id_fk = "1"';

以及之后的PHP代码：

if (mysqli_num_rows($result) > 0) {
   while($row = mysqli_fetch_assoc($result)) {
   printf ("%s (%s)\n",$row["username"],$row["T1.created_at"]);
   }

---

对于“用户名”行，我没有收到错误，但没有显示任何内容。对于第二个，我得到一个索引未定义的错误。（这只是一段代码）

如下更改SQL查询：

$sql = 'SELECT T1.update_text as update_text, T1.created_at as created_at, T2.username as username, T3.group_name as group_name   FROM updates AS T1
        INNER JOIN users AS T2 ON T2.user_id = T1.user_id_fk
        INNER JOIN groups AS T3 ON T3.group_id = T1.group_id_fk
        WHERE T1.user_id_fk = "1"';

 $result = $mysqli->query($sql);
    if (mysqli_num_rows($result) > 0) {
       while($row = $results->fetch_object()) 
       {
           echo $row->username.$row->created_at."<br/>";
       }
    }

然后更改PHP代码，如下所示：

$sql = 'SELECT T1.update_text as update_text, T1.created_at as created_at, T2.username as username, T3.group_name as group_name   FROM updates AS T1
        INNER JOIN users AS T2 ON T2.user_id = T1.user_id_fk
        INNER JOIN groups AS T3 ON T3.group_id = T1.group_id_fk
        WHERE T1.user_id_fk = "1"';

 $result = $mysqli->query($sql);
    if (mysqli_num_rows($result) > 0) {
       while($row = $results->fetch_object()) 
       {
           echo $row->username.$row->created_at."<br/>";
       }
    }

$result=$mysqli->query（$sql）；
如果（mysqli_num_行（$result）>0）{
而（$row=$results->fetch_object（））
{
echo$row->username.$row->created_at.“
”；
}
}

你的mysqli_查询在哪里？正如我所说，这只是一个片段。如果我删除别名，代码“起作用”，但不显示任何内容…$result=mysqli\u query（$conn，$sql）是的，我的代码中有此语句…但尚未将其发布到此处。您的数据库变量中的用户名为空，请转储您的行以查看内容