在PHP中使用引導模式和AJAX更新mysql數據庫中的數據

Question

我正在嘗試使用引導模式形式在沒有頁面刷新的情況下更新mysql數據庫中的用戶詳細信息，但是當我按下編輯按鈕時，模式沒有打開並且什么也沒有發生。

這是我的編輯按鈕。 我的編碼或語法有問題嗎？

   <button type="button" class="glyphicon glyphicon-edit btn btn-sm btn-default edit_user " id="'.$row["uID"].'" name="edit""></button>

這是我的添加用戶模式。

<!-- Add user modal -->
<div id="add_user_Modal" class="modal fade">
<div class="modal-dialog">
    <div class="modal-content">
        <div class="modal-header">
            <button type="button" class="close" data-dismiss="modal">&times;</button>
            <h4 class="modal-title">Add New User/Admin</h4>
        </div>
        <div class="modal-body">
            <form method="post" id="add_user_form">
                <label>Enter Employee ID</label>
                <input type="text" name="user_id" id="user_id" class="form-control" required/>
                <br />
                <label>Enter Employee Name</label>
                <input type="text" name="username" id="username" class="form-control" required/>
                <br />
                <label>Enter Employee Email</label>
                <input type="email" name="email" id="email" class="form-control" required/>
                <br />
                <label>Password</label>
                <input type="password" name="password" id="password" class="form-control" required/>
                <br />
                <label>Confirm Password</label>
                <input type="password" name="confirm_password" id="confirm_password" class="form-control" required/>
                <br />
                <label>Account Type</label>
                <select name="account" id="account" class="form-control">
                    <option value="Guest">Guest</option>
                    <option value="Admin">Admin</option>

                </select>
                <br />
<input type="hidden" name="uid" id="uid" />
                <input type="submit" name="insert" id="insert" value="Insert" class="btn btn-success" />

            </form>
        </div>
        <div class="modal-footer">
            <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
        </div>
    </div>
</div>

這是ajax的Java腳本，AJAX錯誤為“請求的JSON解析失敗”。

   $(document).ready(function(){
    //add
    $('#add_user_form').on("submit", function(event){
        event.preventDefault();
        $.ajax({

                url:"add_user.php",
                method:"POST",
                data:$('#add_user_form').serialize(),
                beforeSend:function(){
                    $('#insert').val("Inserting");
                },
                success:function(data){
                    alert("succees")
                    $('#add_user_form')[0].reset();
                    $('#add_user_Modal').modal('hide');
                    $('#datatable-buttons').html(data);
                },

        });
    });


$(document).on('click', '.edit_user', function(){
        var uid = $(this).attr("id");
        $.ajax({
            url:"fetch_user_details.php",
            method:"POST",
            data: {uid:uid},
            dataType:"json",
            success:function(data){
                console.log(data),

                $('#username').val(data.username);
                $('#email').val(data.email);
                $('#password').val(data.password);
                $('#account').val(data.account);
                $('#user_id').val(data.user_id);
                $('#insert').val("Update");
                $('#add_user_Modal').modal('show');
            },
            error: function(jqXHR, exception) {

                var msg = '';
                if (jqXHR.status === 0) {
                    msg = 'Not connect.\n Verify Network.';
                } else if (jqXHR.status == 404) {
                    msg = 'Requested page not found. [404]';
                } else if (jqXHR.status == 500) {
                    msg = 'Internal Server Error [500].';
                } else if (exception === 'parsererror') {
                    msg = 'Requested JSON parse failed.';
                } else if (exception === 'timeout') {
                    msg = 'Time out error.';
                } else if (exception === 'abort') {
                    msg = 'Ajax request aborted.';
                } else {
                    msg = 'Uncaught Error.\n' + jqXHR.responseText;
                }
                $('#post').html(msg);
            }
        });
    });

Answer 1

我認為您引用的ID錯誤。 這可能是重復的13183630

Answer 2

你的按鈕應該是這樣的

<button type="button" class="glyphicon glyphicon-edit btn btn-sm btn-default 
edit_user " id="'.$row["uID"].'" name="edit" data-toggle="modal" data- target="#myModal">Open Modal</button>