与PHP中使用mysqli的系统相似和不同

我试图在php中创建一个喜欢/不喜欢的按钮。我的代码工作得很好。我有一个喜欢的按钮，当我点击它，它传输数据到我的"喜欢"表在我的数据库。我唯一的问题是，当我点击一个喜欢的按钮，我不能自动切换到不喜欢的按钮。知道怎么做吗?

";$ _SESSION("pid")= $行("id");userid美元= $ _SESSION("标识");echo $标识;?>">

<?php
require '../connection/config.php';
$id=$_GET['id'];
$userid=$_SESSION['userid'];
$postid=$id;
{
$sql="select * from post inner join likenew on post.id=likenew.id where userid='$userid' and postid='$id'";
echo  $sql;
$result=mysqli_query($conn,$sql);
$row=mysqli_fetch_array($result);
if($row['status']=='')
{
$sql="insert into likenew(postid,userid) values('$id','$userid')";
$result=mysqli_query($conn,$sql);
echo $sql;
}
if($result)
{
$sql="update likenew set status=1 where postid='$id' and userid='$userid'";
$result=mysqli_query($conn,$sql);
echo $sql;
}
elseif($row['status']=='1')
{
$sql="update likenew set status=0,status2='unlike' where postid='$id' and userid='$userid'";
$result=mysqli_query($conn,$sql);
echo $sql;
}
elseif($row['status']=='0')
{
$sql="update likenew set status=1,status2='like' where postid='$id' and userid='$userid'";
$result=mysqli_query($conn,$sql);
echo $sql;
}
}



/*if($result)
{
    header('location:likedisplay.php');
}
    else
    {
        echo "error";
    }
}

    if($num==1){
        $sql="insert into likenew(postid,userid) values('$id','$userid')";
        $result1=mysqli_query($conn,$sql1);
    }
    else
{
    $sql1="update likenew set status=1 where postid='$id' and userid='$userid'";
$result=mysqli_query($conn,$sql);
if($result)
{
    header('location:likedisplay.php');
}
    else
    {
        echo "error";
    }
}*/
?>