所以,我试图创建一个可以执行活动CRUD的实时交互式表格,但由于某种原因,我尝试为添加功能插入的值未正确读取到$ _POST 。
这是表格:
<?php
include "../db.php";
$connection = DB::CreateConnection();
$output = '';
$sql = "SELECT * FROM users ORDER BY userid ASC";
$result = mysqli_query($connection, $sql);
$output .= '
<div class="table-responsive">
<table class="table table-bordered">
<tr>
<th width="10%">Id</th>
<th width="40%">First Name</th>
<th width="40%">Last Name</th>
<th width="10%">Delete</th>
</tr>';
if(mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_array($result))
{
$output .= '
<tr>
<td>'.
$row["userid"].'</td>
<td class="userfirstname" data-id1="'.$row["userid"].'" contenteditable>'.$row["userfirstname"].'</td>
<td class="userlastname" data-id2="'.$row["userid"].'" contenteditable>'.$row["userlastname"].'</td>
<td><button type="button" name="delete_btn" data-id3="'.$row["userid"].'" class="btn btn-xs btn-danger btn_delete">x</button></td>
</tr>';
}
$output .= '
<tr>
<td></td>
<td id="userfirstname" contenteditable></td>
<td id="userlastname" contenteditable></td>
<td><button type="button" name="btn_add" id="btn_add" class="btn btn-xs btn-success">+</button></td>
</tr>';
}
else
{
$output .= '<tr>
<td colspan="4">Data not Found</td>
</tr>';
}
$output .= '</table>
</div>';
echo $output;
?>
这里是实时数据编辑文件:
<!-- This should be where matt's code goes to edit via popup or live -->
<head>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
</head>
<body>
<div class="container">
<br />
<br />
<br />
<div class="table-responsive">
<h3 align-"center">Live Table</h3>
<br />
<div id="live_data"></div>
</div>
</div>
</body>
<script type="text/javascript">
function fetch_data()
{
$.ajax({
url:"UViewTABLE.php",
method:"POST",
success:function(data){
$('#live_data').html(data);
}
});
}
fetch_data();
$(document).on('click', '#btn_add', function(){
var userfirstname = $('#userfirstname').text();
var userlastname = $('#userlastname').text();
if(userfirstname == '')
{
alert("Enter First Name");
return false;
}
if(userlastname == '')
{
alert("Enter Last Name");
return false;
}
$.ajax({
url:"UViewTABLEinsert.php",
method:"POST",
data:{userfirstname:userfirstname, userlastname:userlastname},
dataType:"text",
success:function(data)
{
alert(data);
fetch_data();
}
})
});
</script>
这是插入文件:
<?php
include "../db.php";
$connection = DB::CreateConnection();
$sql = "INSERT INTO users (userfirstname, userlastname) VALUES('".$_POST["userfirstname"]."', '".$_POST["userlastname"]."')";
if(mysqli_query($connection, $sql))
{
echo 'Data Inserted';
}
?>
我得到的错误是:
注意:未定义索引:userfirstname in 第4行的/var/www/html/UOM/UViewTABLEinsert.php
注意:未定义的索引:userlastname in 第4行的/var/www/html/UOM/UViewTABLEinsert.php
并且,我知道这意味着在我的编辑文件中,我没有正确定义这些索引,但我看不出它有什么问题。我已经研究了这个问题3天了,如果有人能够提供帮助,请这样做,显然我不知道自己做错了什么。
答案 0 :(得分:0)
只需从ajax选项中删除dataType
,因为您明确将数据类型设置为文本,数据未在键值对中提交,插入后,以JSON发送响应可能有效。有关$.ajax
的{{1}}选项的更多信息,请参阅this。如果要显式设置dataType
,请将响应保持为与ajax请求中设置的相同类型。