Question

我的页面加载时，document.ready函数没有响应。而且我也面临着.submit功能的问题。请帮忙。请帮我改写这段代码。 我只是想在php中设计一个表单来检查给定的电子邮件ID是否已经存在于数据库中或者是否使用jquery ajax 请帮忙 .... ！我的学校项目迟到了。

 $(document).ready(function(){
                alert("Working..");
                $("#userSubmit").submit(function(){
                    alert("Working ..");
                    var email=document.getElementById("userEamil").value;
                    $.ajax({
                        type:'POST',
                        data:"SELECT * FROM user WHERE userEmail='"+email+"'",
                        url:'common1.php',
                        success:function(data)
                        {
                            $result=$.parseJSON($data);
                            alert($result);
                            return false;
                        },
                        error:function(data)
                        {
                            alert(data);
                            return false;
                        }
                    });
                });
            });

Answer 1

您的代码中有一些错误，我希望以下代码可以完成这项工作。如有进一步的问题，请查看$.ajax部分。

$(document).ready(function() {
        // alert("Working..");
        $("#userSubmit").submit(function(event) {
            event.preventDefault();
            alert("Working ..");
            var email = document.getElementById("#userEamil").value;
            $.ajax({
                type: 'POST',
                data: "SELECT * FROM user WHERE userEmail='" + email + "'",
                url: 'common1.php',
                success: function(data) {
                    $result = $.parseJSON($data);
                    alert($result);
                    return false;
                },
                error: function(data) {
                    alert(data);
                    return false;
                }
            });
        });
    });

<!-- begin snippet: js hide: false console: true babel: false -->

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form id="userSubmit" action="">
            <input type="email">
            <input type="submit" value="submit">
        </form>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form action="">
    <input type="email" >
    <input id="userSubmit" type="submit" value="submit">
</form>

Answer 2

您可以使用解决方案https://jsfiddle.net/1uor0gxf/1/

＆＃13;

$(document).ready(function(){
    $("#userSubmit").click(function(){
        var email = $("userEamil").val();
        $.ajax({
            type:'POST',
            data:`SELECT * FROM user WHERE userEmail='${email}'`,
            url:'common1.php',
            success:function(data)
            {
                $result=$.parseJSON($data);
                alert($result);
                return false;
            },
            error:function(data)
            {
                alert(data);
                return false;
            }
        });
    });
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form action="">
  <input type="text" id="userEamil" />
  <input type="button" id="userSubmit" value="Submit" />
</form>

＆＃13;

Answer 3

<script>
   $(document).ready(function() {
      //alert("Working..");
      $("#userSubmit").submit(function() {
           //alert("Working ..");
                  var email = $(this).closest("form").find("input[name='emailbox']").val();
                  console.log(email);
                  $.ajax({
                    type: 'POST',
                    data: "SELECT * FROM user WHERE userEmail='" + email + "'",
                    url: 'common1.php',
                    success: function(data) {
                      $result = $.parseJSON($data);
                      alert($result);
                      return false;
                    },
                    error: function(data) {
                      alert(data);
                      return false;
                    }
                  });
                });
              });
            </script>

<html>
    <body>
        <form id="userSubmit">
            <input type="email" id="userEmail" name="emailbox">
            <input type="submit" value="Submit">    
        </form>
    </body>
</html>

Jquery document.ready没有响应

3 个答案: