设置代码文件：

                              if($User_Name != $row['USERNAME'])
                              {
                                    $insert = "UPDATE USER_LOGIN SET PASSWORD = '".$New_Password."', USERNAME = '".$User_Name."' WHERE USERNAME = '".$_COOKIE['username']."'";
                                    $result = oci_parse($con, $insert);

                                    // Executes a statement.
                                    $check = oci_execute($result);

                                    if($check)
                                    {
                                        echo "Saved";

                                        // Commit the changes to the table.
                                        oci_commit($con);
                                    }

                                    else
                                    {
                                        // Rollback changes to table.
                                        oci_rollback($con);
                                    }
                              }

管理员面板：

            <div class="user">
                <img src="../../img/Users/Users/Admin.jpg" align="left"/>
                <a class="name">
                    <?php 
                        session_start();

                        if(isset($_COOKIE['username']))
                        {
                            echo $_COOKIE['username'];
                        }

                        else
                        {
                            echo $_SESSION['username'];
                        }
                    ?>
                    <span class="sm">Administrator</span>
                </a>
            </div>

我的ajax功能是这样的：

<!-- Ajax function is used to post the data to databse and show result in the table without refresh -->
<script>
var frm = $('.wrefresh');

frm.submit(function (ev)
{

    ev.preventDefault();

    var postdata = $(this).serialize();
    var path = $(this).attr("action");
    var mehtodtype = $(this).attr("method").toUpperCase();

    // Clear fields data. (password).
    $('form :input[type=password]').attr('value','');

    // Remove selected pictures field.
    $('#Picture').attr('src', '../../img/User No-Frame.png');

    $.ajax
    ({
        type: mehtodtype,
        url: path,
        data: postdata,

        success: function(data) 
        {

            if(data === "Please Fill Entries.")
            {
                $('#fullname').css("border-color", "rgb(219, 46, 46)");
                $('#old').css("border-color", "rgb(219, 46, 46)");
                $('#new').css("border-color", "rgb(219, 46, 46)");
                $('#confirm').css("border-color", "rgb(219, 46, 46)");

                $('#showmediv').show();
            }
        }
    });
});
</script>

HTML：

                            <form class="wrefresh" action="../Code Files/User.php" method="post">

                                    <div id="photo_settings2" style="margin-left:74px;">
                                            <img id="Picture" src="../../img/User No-Frame.png"/>       
                                    </div>

                                    <br><br><br><br>

                                    <div id='Upload_Panel' style="margin-left: 32px;">
                                            <input name='file' type='file' id='file_browse' onchange="readURL(this,'Picture')" style="cursor: pointer;"/>
                                    </div>

                                    <div id="delete" style="margin-top: -40px; margin-left: 198px; cursor: pointer">
                                            <img src="../../img/Delete_Panel.png">
                                    </div>

                                    <div style="margin-left:144px; margin-top:-15px"></div>

                                    <br><br> 

                                    <div class="row-form" style="margin-top:10px">
                                            <div class="span4">Username:</div>
                                            <div class="span8"><input id="fullname" style="border: 1px solid #918C8C" type="text" class="cNav" name="uname" value=""/></div>
                                    </div>

                                    <div class="row-form">
                                            <div class="span4">Old Password:</div>
                                            <div class="span8"><input id="old" style="border: 1px solid #918C8C" type="password" class="cNav" name="opass" value=""/></div>
                                    </div>

                                    <div class="row-form">
                                            <div class="span4">New Password:</div>
                                            <div class="span8"><input id="new" style="border: 1px solid #918C8C" type="password" class="cNav" name="npass" value=""/></div>
                                    </div>

                                    <div class="row-form">
                                            <div class="span4">Confirm:</div>
                                            <div class="span8"><input id="confirm" style="border: 1px solid #918C8C" type="password" class="cNav" name="cpass" value=""/></div>
                                    </div> 

                                    <br>


                                    <div class="row-fluid" style="margin-left:112px; margin-top:-05px">
                                            <div class="span9">
                                                    <button type="submit" class="btn btn-success" style="height:28px">Save</button>
                                                    <span id="showmediv" class="label label-important" style="display: none; margin-left: 14px; margin-top:-35px">Please Enter Fields</span>

                                            </div>

                                    </div>

                            </form>

Answer 1

您需要更改您在

中显示的用户名

<a class="name">

在ajax成功回调中这样：

我不确定它是否会起作用。

$.ajax
    ({
        type: mehtodtype,
        url: path,
        data: postdata,

        success: function(data) 
        {

            if(data === "Please Fill Entries.")
            {
                $('#fullname').css("border-color", "rgb(219, 46, 46)");
                $('#old').css("border-color", "rgb(219, 46, 46)");
                $('#new').css("border-color", "rgb(219, 46, 46)");
                $('#confirm').css("border-color", "rgb(219, 46, 46)");

                $('#showmediv').show();
               //get new name
                var userName = $("#fullname").val();
               // show new username
                $(".name").html(userName);
            }
        }
    });

再一件事：

当您发送ajax呼叫时，您需要在SESSION中设置新用户名。

Answer 2

我希望你试着在没有刷新的情况下发布表单，因为我没有理解你想要实现的目标

您需要做的是使用ajax提交表单。试试这个插件并添加以下代码

http://jquery.malsup.com/form/#ajaxSubmit

下载插件

并在表单中添加此

 <form onsubmit="javascript:login();return false;">

function action(){
  $('#form').ajaxSubmit({

target:'#output_login',
url:'./php/page.php'


});
 return false;

}

动态更改用户名设置php

设置代码文件：

管理员面板：

我的ajax功能是这样的：

HTML：

2 个答案: