将变量与链接连接


Concatenating variable with link

我很难在引号中连接引号这个问题可能对任何人都没有用,但我真的需要这个来解决。

这是代码,我希望它能正常工作,但似乎我在某个地方犯了错误。

$query = "SELECT message,time,user_id FROM post ORDER BY time DESC";
                $query_run = mysqli_query($con,$query);
                while($row = mysqli_fetch_assoc($query_run))
                {
                    $message = $row['message'];
                    $user_id = $row['user_id'];
                    echo "<img src= 'https://graph.facebook.com/'.'$user_id'.'/picture?width=50&height=40' class = 'post_dp'>";
                }

我不知道如何将user_id与链接连接起来。

$query = "SELECT message,time,user_id FROM post ORDER BY time DESC";
            $query_run = mysqli_query($con,$query);
            while($row = mysqli_fetch_assoc($query_run))
            {
                $message = $row['message'];
                $user_id = $row['user_id'];
                echo "<img src= 'https://graph.facebook.com/". $user_id ."/picture?width=50&height=40' class = 'post_dp'>";
            }

如果字符串以双引号开始,则必须以双引号结束。

echo "<img src= 'https://graph.facebook.com/".'$user_id'."/picture?width=50&height=40' class = 'post_dp'>";

但是,由于php双引号可以转换变量,您根本不需要关闭它们并重新打开:

echo "<img src= 'https://graph.facebook.com/$user_id/picture?width=50&height=40' class = 'post_dp'>";