SELECT没有';在INSERT之后,无法从数据库中检索更新的数据


SELECT doesn't retrieve updated data from DB after INSERT

在一个页面上,我向自己发送AJAX请求。在PHP的成功函数中收到这个请求后,我将在一些MySQL数据库的表"some_table"中添加新记录。之后,在相同的成功函数中,我从该表中选择数据,并通过JQuery将其显示在页面上。问题是所有的旧记录都显示了,但没有显示我刚刚添加的记录:

$.ajax({
    url: 'http://somesite.com', 
    type: "POST",
    data: ({param1: value1, param2: value2, act: "I"}),
    success: function(data){
        <?php $mysqli = new mysqli("some_server", "some_database", "some_login", "some_password"); 
              $query = "INSERT INTO some_table VALUES ("'.$_POST["param1"].'","'.$_POST["param2"].'")";
              $mysqli->query($query);
              $mysqli->close();
        ?>
        alert("New record was added");
        <?php echo '$("#left_region").html('''');';
              $mysqli = new mysqli("some_server", "some_database", "some_login", "some_password"); 
              $result = $mysqli->query("SELECT some_column1, some_column2 FROM some_table"); 
              while($obj = $result->fetch_object()) { 
                  echo '$("#left_region").append("<p class='"some_class'"><a href='"#'">'.$obj->some_column1.'</a></p>");';
              } 
              $result->close();
              $mysqli->close();
        ?>
    }
});

有人能告诉我怎么解决吗?

如果页面已经加载,则不能期望页面再次执行代码

举个例子:

客户端:

$.post("some_url.php",{param1: value1, param2: value2, act: "I"},
      function(response){
         if(isNaN(response)==="true")
             $("#left_region").html(response);
         else
             alert("An error ocurred");
      }
)

服务器端:

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli("some_server", "some_database", "some_login", "some_password");
if ($_POST['act']) {
    $stmt = $mysqli->prepare("INSERT INTO some_table VALUES (?, ?)");
    $stmt->bind_param('ss', $_POST["param1"], $_POST["param2"]);
    $stmt->execute();
    $last_inserted_id = $mysqli->insert_id;
    $stmt = $mysqli->prepare("SELECT some_column1, some_column2 FROM some_table where some_column=?");
    $stmt->bind_param('s', $last_inserted_id);
    $stmt->execute();
    $result = $stmt->get_result();
    echo $result->fetch_assoc();
}

试试这样的方法(如果有人发现错误,请更正,谢谢)。