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Asked:2020-11-19 22:10:27 +0000 UTC2020-11-19 22:10:27 +0000 UTC

AJAX 不起作用，网站滞后

  • 772

当我开始在文本字段中书写时，它没有被写入，它被更新并且网站开始严重滞后。该怎么办？

[聊天.php]

    <?php> session_start(); 
       header('Content-Type: text/html; charset=utf-8');
?>
<html>
    <head>
        <meta charset="utf-8">
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>

    </head>
    <script type="text/javascript">
   // setTimeout("window.location.reload()",3000);
    </script>
    <body>

       <script>
          $(document).ready (function (
            ){  show();setInterval('show()',5000);

                $("#button").bind("click", function (){
                var message = $("#text").val();
                  $.ajax ({
                      url: "chat.php",
                      type: "POST",
                      data: {message},
                      dataType: "json"

                  });
                });
            });

                 function show()
                {
                    $.ajax({
                        url: "chat.php",
                        cache: false,
                        success: function(html){
                            $("#content").html(html);
                        }
                    });
                }
        </script>
        <div id="content">
            <?php
                include("config.php");
                if(isset($_POST['message'])){
                  $sql = "insert into `messages` (`message`) values ('".$_SESSION['login'].": ".$_POST['message']."')";
                  mysql_query($sql);

                }
                $sql = "select message from `messages` ORDER BY id desc LIMIT 10";
                $res = mysql_query($sql);


                if($res)
                {
                    while($row = mysql_fetch_object($res))
                    {
                     printf("<ul type='none'><li>%s</li></ul>",$row->message);
                    }
                }
                else
                {
                  echo "<p> <b>Error: ".mysql_error()."</b> </p>";
                  exit();
                }
             ?>

            <input type='text' id="text" name='message'>
            <button id="button">Saada</button>
        </div>
    </body>
</html>
ajax

1 个回答

  1. Best Answer

    像这样尝试

    <?php
    session_start();
    header('Content-Type: text/html; charset=utf-8');
?>
<html>
<head>
    <meta charset="utf-8">
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
    <script type="text/javascript">
        // setTimeout("window.location.reload()",3000);
        $(document).ready(function () {
            show();
            // в setInterval надо передать название функции без скобок, со скобками вы передаете результат выполненной функции
            setInterval(show, 5000);

            $("#button").bind("click", function () {
                var message = $("#text").val();
                $.ajax({
                    url: "chat.php",
                    type: "POST",
                    data: {message},
                    dataType: "json"
                });
            });

            function show() {
                $.ajax({
                    url: "chat.php",
                    cache: false,
                    success: function (html) {
                        $("#content").html(html);
                    }
                });
            }
        }); // перенес сюда закрытие $(document).ready
    </script>
</head>
<body>
<div id="content">
    <?php
        include("config.php");
        if (isset($_POST['message'])) {
            $sql = "insert into `messages` (`message`) values ('" . $_SESSION['login'] . ": " . $_POST['message'] . "')";
            mysql_query($sql);

        }
        $sql = "select message from `messages` ORDER BY id desc LIMIT 10";
        $res = mysql_query($sql);

        if ($res) {
            while ($row = mysql_fetch_object($res)) {
                printf("<ul type='none'><li>%s</li></ul>", $row->message);
            }
        } else {
            echo "<p><b>Error: " . mysql_error() . "</b></p>";
            exit();
        }
    ?>
    </div>
    <input type='text' id="text" name='message'>
    <button id="button">Saada</button>

</body>
</html>
    • 1

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