通过单击Enter并从Bootstrap模态窗口中单击Submit按钮来提交表单数据

2021/01/22 21:11 · jquery ·  · 0评论

这是我的模式窗口中的表格:

我没有关闭按钮,并且在按esc时已禁用了关闭窗口的功能。我希望能够在按提交或按Enter键时提交表单数据。

<div class="modal hide fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true" data-backdrop="static" data-keyboard="false">
    <div class="modal-header">
        <h3 id="myModalLabel">Enter your Credentials</h3>
    </div>
    <div class="modal-body">
        <form id="login-form" class="form-horizontal" accept-charset="UTF-8"  data-remote="true"">
            <div class="control-group">
                <label class="control-label" for="inputEmail">Email</label>
                <div class="controls">
                    <input type="email" id="email" name="email" value=""  placeholder="Email">
                </div>
            </div>
            <div class="control-group">
                <label class="control-label" for="inputPassword">Password</label>
                <div class="controls">
                    <input type="password" id="passwd" name="passwd" value="" placeholder="Password">
                </div>
            </div>
            <div class="control-group">
                <div class="controls">
                    <label class="checkbox">
                        <input type="checkbox"> Remember me
                    </label>
                    <input type="submit"  id="login" value="Login"class="btn btn-primary" />

                </div>
            </div>
        </form>          
    </div>
</div>

这是我正在使用的脚本:

$(document).ready(function(){
    $('#myModal').modal('show');
});

function submitLogin() {
    $.ajax({
        url:"login_mysql.php",
        type:'POST',
        dataType:"json",
        data: $("#login-form").serialize()
    }).done(function(data){
        //do something
    });
}

$('#passwd').keypress(function(e) {
    if (e.which == '13') {
        submitLogin();
    }
});

$('#login').click(function(){
    submitLogin();
});

在输入我的密码后按Enter键或单击“提交”按钮,将重新加载同一页面,并且ajax脚本不会运行。有人可以告诉我我的脚本是否与模态窗口的默认设置冲突?

监听表单submit事件-它由enter和click事件触发。

标记

<form id="yourForm">
  <input type="text" name="username">
  <input type="password" name="password">
  <input type="submit" value="Submit">
</form>

JS

$('#yourForm').submit(function(event){

  // prevent default browser behaviour
  event.preventDefault();

  //do stuff with your form here
  ...

});
  1. 从登录按钮将type =“ submit”更改为type =“ button”。

编辑:2.尝试删除data-keyboard =“ false”。

我已经完成了提交表单将在模式对话框中打开的操作,然后在字段中提交了整个条目。如果单击提交按钮,则将在数据库上完成输入,并且该页面立即被重定向到包含新数据的同一页面。无需刷新即可查看最近输入的数据。希望这可以帮助。

<button class="btn btn-primary btn-lg" data-toggle="modal" data-target="#myModal">ADD CONTENT</button>
<!-- Modal -->
  <div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
    <div class="modal-dialog">
      <div class="modal-content">
        <div class="modal-header">
          <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button>
          <h4 class="modal-title" id="myModalLabel">Please Add Content</h4>
        </div>
      <div class="modal-body">
        <form role="form" action="" method="POST">
          <!-- YOUR HTML FORM GOES HERE--->
        <button type="submit" name="submit" class="btn btn-primary btn-lg"id="sub" onclick="SUBMISSION()" >Submit </button>
        </fieldset>
        </form>
<?php
  if(isset($_POST['submit']))
  {
    SUBMISSION();
  }
  function SUBMISSION()
  {
    // UR CONNECTION ESTABLISHMENT CODE GOES HERE
    // SQL QUERY FOR INSERTION IN FIELDS

    echo"<script type=\"text/javascript\">
      document.location.href='http://localhost/dict/pages/YOURPAGE.php';
    </script>";

    $conn->CLOSE();
  }
?>

        </div>                      
       </div>
     <!-- /.modal-content -->
     </div>
   <!-- /.modal-dialog -->
   <!-- /.panel-body -->
   </div>
   <!-- /.panel -->
 </div>
 <!-- /.col-lg-12 -->

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