无法使用sql删除表中的数据

1 php mysql sql

我试图从表中删除行.表名是username用户(用户登录)

删除页面在这里

require('db.php');
if(isset($_POST["deletebtn"])){
   $x = $_SESSION['username'];
   $id=$_POST["statusid"];
   $row=$conn->query("select id from $x where id='$id'");
   if ($conn->query($row) === TRUE) {
      $sql = "delete from $x where id ='$id'";
      //  echo 'success';
      header( "Refresh:3; url=admin2.php", true, 303);
   }else{
      echo "not";
   }
 }
Run Code Online (Sandbox Code Playgroud)

这是我的内容代码

$sql = "SELECT id,date,status FROM $x order by id DESC;";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
   while($row = $result->fetch_assoc()) {
      $id = $row['id'];
      echo "<div class='border'> ";
      echo "";
      echo "<form action='deletepage.php?action=$id'>";
      echo "<input type='hidden' name='statusid' value='$id' >";
      echo  "<p> " . $row["status"] ."</p><br>";
      echo "<input type='submit' value='delete' class='btn1 pull-right' name='deletebtn'> ";
      echo "<a class=' btn1 pull-right' href='#'>edit &nbsp;&nbsp;</a>";
      echo  "<small> " . $row["date"]. "</small><br>";
      echo "</form>";
      echo "</div>";            
    } 
} else {............}
Run Code Online (Sandbox Code Playgroud)

use*_*204 5

您没有在sql上执行查询

$sql = "delete from $x where id ='$id'";
Run Code Online (Sandbox Code Playgroud)

像执行它一样

$del= $conn->query($sql);
Run Code Online (Sandbox Code Playgroud)

查询后