dsfjnxjlbqv9812 2019-02-18 07:29
浏览 143
已采纳

如何防止页面刷新? jQuery Ajax PHP

I have made a form where I submit new employee data to the database using jquery ajax and php. The data gets submitted and appears in the database, but how do I get it so the data will display on my page without the page refreshing?

I want the data to appear on the page without the page refreshing.

My problem: I can submit the data, but the page refreshes. If I make it so the page doesn't refresh, it won't submit the data.

    function createNewEmployee(){
require ('connection.php');
    $first = $_POST['first'];
    $last = $_POST['last'];
    $sin = $_POST['sin'];
    $pwd = $_POST['pwd'];
    $sql = "INSERT INTO employee (firstname, lastname, sin_, pass_) VALUES ('$first', '$last', '$sin', '$pwd');";
    mysqli_query($conn, $sql);
}




  <form id="addform" action="addemployee.php" method="POST">
<p>Add New Employee:<p>
    <input type="text" name="first" placeholder="First Name">
    <br>
    <input type="text" name="last" placeholder="Last Name">
    <br>
    <input type="text" name="sin" placeholder="SIN Number">
    <br>
    <input type="password" name="pwd" placeholder="Password">
    <br>
    <button type="button" id="submitbtn">Add</button>
    $(document).ready(function() {

  $("#submitbtn").click(function() {
    var first = $("#name").val();
    var last = $("#last").val();
    var sin = $("#sin").val();
    var pwd = $("pwd").val();

    $.ajax({
      type: "POST",
      data: {first:first,last:last,sin:sin,pwd:pwd},
      url: "addemployee.php",
      success: function(result) {
        $("#resultadd").html(response);
      }

    });
  });
});
  • 写回答

3条回答 默认 最新

  • duanou1904 2019-02-18 08:03
    关注

    Add return false; after your ajax request to prevent the page from refreshing.

    Code Snippet

    
    $(document).ready(function() {
      $("#submitbtn").click(function() {
        var first = $("#name").val();
        var last = $("#last").val();
        var sin = $("#sin").val();
        var pwd = $("pwd").val();
        $.ajax({
          type: "POST",
          data: {first:first,last:last,sin:sin,pwd:pwd},
          url: "addemployee.php",
          success: function(result) {
            $("#resultadd").html(response);
          }
    
    });
    
    return false; }); });
    

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(2条)

报告相同问题?

悬赏问题

  • ¥15 metadata提取的PDF元数据,如何转换为一个Excel
  • ¥15 关于arduino编程toCharArray()函数的使用
  • ¥100 vc++混合CEF采用CLR方式编译报错
  • ¥15 coze 的插件输入飞书多维表格 app_token 后一直显示错误,如何解决?
  • ¥15 vite+vue3+plyr播放本地public文件夹下视频无法加载
  • ¥15 c#逐行读取txt文本,但是每一行里面数据之间空格数量不同
  • ¥50 如何openEuler 22.03上安装配置drbd
  • ¥20 ING91680C BLE5.3 芯片怎么实现串口收发数据
  • ¥15 无线连接树莓派,无法执行update,如何解决?(相关搜索:软件下载)
  • ¥15 Windows11, backspace, enter, space键失灵