Question

我相信你们之前有很多人已经看过这个问题，我已经将适用于他人的代码应用到我自己的代码中，但仍然没有，我发现很难确定哪些是错误的。这是我的用户端，其中显示了mysql信息和删除按钮的表：

<?php 
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "gpdb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("connection failed: " . $conn->connect_error);
}

$id = $_SESSION['id'];
$sql = "SELECT AppID, AppDate FROM appointment WHERE PatientID ='$id'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
echo ("<form  action='delete.php' method='post'><table class='jobs-list'>
<tr>
<th class='job-position'><h4>Appointment ID</h4></th>
<th class='job-position'><h4>Appointment Date and Time</h4></th>
<th class='job-position'><h4>Edit Appointments</h4></th>
</tr>");

while($row = mysqli_fetch_array($result))
{
echo ('<tr>');
echo ('<td>' . $row['AppID'] . '</td>');
echo ('<td>' . $row['AppDate'] . '</td>');
echo ("<td><input type='submit' id='deleteItem' value='".$row['AppID']."'/>   </td>");
echo ('</tr>');
}
echo ('</table></form>');

}
else{
echo('<h4><b>You currently have no appointments</b></h4>');
}
?>
<script>

$('#deleteItem').on('click', function(){
var pass = $(this).val(); 
console.log(pass);
$.post( "delete.php", { 
    'pass':pass
}).done(function( data ) {
    alert( "You cancelled : " + data );
});
});
</script>

这是delete.php文件：

<?php
session_start();
//$servername = "localhost";
//$username = "root";
//$password = "";
//$dbname = "gpdb";
//$conn = new mysqli($servername, $username, $password, $dbname);
//if ($conn->connect_error) {
//die("connection failed: " . $conn->connect_error);
 //}
 print_r($_POST['pass']);
//if(isset($_POST['pass']) && $_POST['pass'] != "")
//{ 
//$deleteItem = $_POST['pass'];
//$sql = "DELETE FROM appointment WHERE AppID='{$deleteItem}'";
//$result = $conn->query($sql);
//header("location: http://localhost/gp/account.php");
//}else{
//echo('Something wrong');
//}

?>

我目前正在读取delete.php上的“错误”，因此delete.php没有返回。

提前感谢您的任何帮助：）

Answer 1

问题在于您的代码。

你应该像这样回复按钮：

echo ("<td><input type='button' class='deleteItem' name='delete_".$row['AppID']."' value='".$row['AppID']."'/>")

你不应该在你的javascript中使用php变量，你的javascript应该是这样的

<script>

$('.deleteItem').on('click', function(){
    var pass = $(this).val(); 
    $.post( "delete.php", { 
        'pass':pass
    }).done(function( data ) {
        alert( "You cancelled : " + data );
    });
});
</script>

然后在您的delete.php中，您应该检查$_POST['pass'] NOT $_POST['deleteItem']

所以代码将是

<?php
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "gpdb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("connection failed: " . $conn->connect_error);
 }
if(isset($_POST['pass']) && is_numeric($_POST['pass']))
 {
$deleteItem = $_POST['pass'];
$sql = "DELETE FROM appointment WHERE AppID='{$deleteItem}'";
$result = $conn->query($sql);
header("location: http://localhost/gp/account.php");
}else{
echo('Something wrong');
}

?>

希望这会有所帮助。

Answer 2

这种情况是你的敌人

if(isset($_POST['deleteItem']) and is_numeric($_POST['deleteItem']))

因为$_POST['deleteItem']不是数字，因为值类似于Delete 1，这就是为什么你总是看到＆＃34;错误的东西＆＃34;

Answer 3

您使用按钮类型提交它将重新加载页面。使用event.preventDefault();检查下面的代码，或者将submit类型更改为button

<script>
$('#deleteItem').on('click', function(event){
 event.preventDefault();
var pass = $row['AppID']; 
$.post( "delete.php", {  'pass':pass }).done(function( data ) {
alert( "You cancelled : " + data );
});
});
</script>

Answer 4

终于解决了这个问题，谢谢你的帮助，特别是@Gopakumar Gopalan。我真的需要那个发布调用，但似乎表单方法正在保留函数，所以我从表中删除了表单标签：

<?php 

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "gpdb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("connection failed: " . $conn->connect_error);
}

$id = $_SESSION['id'];
$sql = "SELECT AppID, AppDate FROM appointment WHERE PatientID ='$id'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
echo ("<table class='jobs-list'>
<tr>
<th class='job-position'><h4>Appointment ID</h4></th>
<th class='job-position'><h4>Appointment Date and Time</h4></th>
<th class='job-position'><h4>Edit Appointments</h4></th>
</tr>");

while($row = mysqli_fetch_array($result))
{
echo ('<tr>');
echo ('<td>' . $row['AppID'] . '</td>');
echo ('<td>' . $row['AppDate'] . '</td>');
echo ("<td><input type='submit' class='deleteItem'    name='delete_".$row['AppID']."' value='".$row['AppID']."'/></td>");
echo ('</tr>');
}
echo ('</table>');

}
else{
echo('<h4><b>You currently have no appointments</b></h4>');
}
?>
<script>

$('.deleteItem').on('click', function(){
var pass = $(this).val();
$.post( "delete.php", { 
   'pass':pass
}).done(function( data ) {
   location.reload();
});
});
</script>

删除表单标记后，delete.php文件可识别$ _POST [＆＃39; pass＆＃39;]; ：

<?php
session_start();

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "gpdb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("connection failed: " . $conn->connect_error);
}
if(isset($_POST['pass']) && $_POST['pass'] != "")
{ 
$deleteItem = $_POST['pass'];
$sql = "DELETE FROM appointment WHERE AppID='{$deleteItem}'";
$result = $conn->query($sql);
header("location: http://localhost/gp/account.php");
}else{
echo('Something wrong');
}

?>

希望这个答案也可以帮助别人。干杯！

Answer 5

#deleteItem首先意味着您正在访问和触发id中的事件，并且页面中没有具有此类名称的ID

第二，如果你将id =“deleteItem”放在循环中会有多个同名的id错误

你可以做的是在一个函数中发送id，它将发送单击删除按钮的行的确切id

<input type='button' name='deleteItem' value='Delete" onclick="deleteSelected('" .$row['AppID']."')'/>

现在在函数内部调用ajax，如下所示

<script>
    function deleteSelected(delete_id)
    {
        $.post( "delete.php", { 
       'deleteItem':delete_id
        }).done(function( data ) {
    alert( "You cancelled : " + data );
        });
    }
</script>

现在将操作页面上的if条件更改为

if(isset($_POST['deleteItem']) && is_numeric($_POST['deleteItem']))
{
$deleteItem = $_POST['deleteItem'];
$sql = "DELETE FROM appointment WHERE AppID='{$deleteItem}'";
$result = $conn->query($sql);
header("location: http://localhost/gp/account.php");
}else{
echo('Something wrong');
}

使用删除按钮删除mysql记录

5 个答案: