Question

我正在尝试获取此td的ID ....当我检查我的网络控制台时，它说id = undefined。我不确定我做错了什么......

<td data-id="' . $fetch['userID'] . '" >
    <button onClick = "deleterecord();">delete</button>
</td>

的javascript

function deleterecord() {

     //var del_id = $(this).attr("id");
     var del_id = $(this).closest('td').data('id');
     var dataString = 'id=' + del_id;

     if (confirm("Are you sure you want to delete this record?")) {
            $.ajax({
                type: "POST",
                url: "deletemember.php",
                data: dataString,
                success: function () {
                }
            });
            $(this).parents(".record").animate("fast").animate({
                opacity: "hide"
            }, "slow");
        }
        return false;
}

deletemember.php

    $id = $_GET['id'];

    $stmt = $conn->prepare("DELETE li.*
                           FROM `loginInfo` li
                           WHERE li.userID = ?");
    $stmt->bind_param("s", $id);
    $stmt->execute();
    if ($stmt){
        header('Location: memberinfo.php');       
    } else echo 'Delete failed. Try again';

Answer 1

<强> HTML

<td>
    <button onClick = "deleterecord('123');">delete</button>
</td>

<强>的javascript

function deleterecord(userID) {

     if (confirm("Are you sure you want to delete this record?")) {
            $.ajax({
                type: "POST",
                url: "deletemember.php",
                data: {userID:userID},
                success: function () {
                }
            });
            $(this).parents(".record").animate("fast").animate({
                opacity: "hide"
            }, "slow");
        }
        return false;

<强> deletemember.php

$id = $_POST['userID'];

    $stmt = $conn->prepare("DELETE li.*
                           FROM `loginInfo` li
                           WHERE li.userID = ?");
    $stmt->bind_param("s", $id);
    $stmt->execute();
    if ($stmt){
        header('Location: memberinfo.php');       
    } else echo 'Delete failed. Try again';

Answer 2

此代码将根据需要返回data-id。请根据您的方便修改以下代码。

<!DOCTYPE html>
<html>
<head>
<script src="http://ajax.aspnetcdn.com/ajax/jQuery/jquery-1.12.0.min.js">
</script>
<script>
$(document).ready(function(){
    $('#delete_button').on('click', function(){
       alert( $(this).closest('td').attr('data-id') );
});
});
</script>
</head>
<body>
<table>
<tr>
<td data-id="456" >
    <button id="delete_button" >delete</button>
</td>
<tr>
</table>
</body>
</html>

Answer 3

尝试修改HTML：

<td data-id="<?php echo $fetch['userID']?>" > <button onClick = "deleterecord(<?php echo $fetch['userID']?>);">delete</button> </td> 现在在JAvascript中：

function deleterecord(deleteid) {
   if (confirm("Are you sure you want to delete this record?")) {
        $.ajax({
            type: "GET",
            url: "deletemember.php",
            data: {id:deleteid},
            success: function () {
            }
        });
        $(this).parents(".record").animate("fast").animate({
            opacity: "hide"
        }, "slow");
    }
    return false;

得到td的id

3 个答案: