php 不会回显内部连接查询的结果

Question

我在网站上四处寻找类似问题的解决方案，但似乎没有解决我遇到的问题。我有两张 table ； “好友列表”和“用户”。我正在尝试使用“好友列表”中的“FriendID”从“用户”表中检索信息。一切正常，直到 while($row = mysqli_fetch_array($result)){} 循环然后没有其他打印。

我的代码如下:

$query = "SELECT friendlist.FriendID, users.Name, users.Surname, users.Picture 
                FROM friendlist            
                INNER JOIN users
                ON friendlist.FriendID = users.Id
                WHERE friendlist.UserId ='".$id."'";
$result = mysqli_query($con, $query);

if(!$result){
    echo "<br/><h4>You currently do not have any friends. Please click the Find Friends button to find a friend</h4>";
}else{
    echo "<center><br/>Here is a list of all your friends:<br/>";
    echo "<table>";
    while($row = mysqli_fetch_array($result)){
        echo "<tr>";
        echo "<td>Pro Pic: <img style='width:200px; height:200px' alt='No Profile Picture' src='uploads/" .$row['Picture']. "' /></td>";
        echo "<td>Name :" .$row['Name']. "</td>";
        echo "<td>Surname :" .$row['Surname']. "</td>";
        echo "<td><form method='post' action='viewFriend.php'>";
        echo     "<input type='hidden' name='friendId' value='".$row['FriendID']."'/>";
        echo     "<input type='submit' name='View' value='View Profile'/>";
        echo "</form></td>";
        echo "</tr>";
    }
    echo "</table></center>";
}

浏览器上没有任何显示。仅显示 4 级标题文本:“这是您所有 friend 的列表”。但在那之后是一片空白。
我检查了 mySql 上的 sql 查询，它工作得很好。我不知道出了什么问题。任何帮助都感激不尽。谢谢

Answer 1

您的代码是正确的，但您只错过了 $id 变量的声明。像下面这样使用。

//initialize  the $id as.

$id = $_REQUEST['id'];

    $query = "SELECT friendlist.FriendID, users.Name, users.Surname, users.Picture 
                    FROM friendlist            
                    INNER JOIN users
                    ON friendlist.FriendID = users.Id
                    WHERE friendlist.UserId ='".$id."'";
    $result = mysqli_query($con, $query);

    if(!$result){
        echo "<br/><h4>You currently do not have any friends. Please click the Find Friends button to find a friend</h4>";
    }else{
        echo "<center><br/>Here is a list of all your friends:<br/>";
        echo "<table>";
        while($row = mysqli_fetch_array($result)){
            echo "<tr>";
            echo "<td>Pro Pic: <img style='width:200px; height:200px' alt='No Profile Picture' src='uploads/" .$row['Picture']. "' /></td>";
            echo "<td>Name :" .$row['Name']. "</td>";
            echo "<td>Surname :" .$row['Surname']. "</td>";
            echo "<td><form method='post' action='viewFriend.php'>";
            echo     "<input type='hidden' name='friendId' value='".$row['FriendID']."'/>";
            echo     "<input type='submit' name='View' value='View Profile'/>";
            echo "</form></td>";
            echo "</tr>";
        }
        echo "</table></center>";
    }