我在使用ajax获取表格行的值时遇到问题,到目前为止每次获取其值时,我都没有给予任何回报,这就是我得到它的方式。
while($row = mysqli_fetch_array($result))
{
$output .= '
<tr>
<td>'.$row["wmID"].'</td>
<td class="first_name" data-id1="'.$row["userFirstname"].'" >'.$row["userFirstname"].'</td>
<td class="last_name" data-id2="'.$row["userLastname"].'" >'.$row["userLastname"].'</td>
<td class="payment" data-id3="'.$row["payment"].'" >'.$row["payment"].'</td>
<td><button type="button" name="delete_btn" data-id4="'.$row["wmID"].'" class="btn btn-xs btn-danger btn_delete">Remit</button></td>
</tr>
';
}
<?php
include_once('../connection.php');
$sql = "UPDATE tbl_request SET isRemitted = 'Yes' WHERE wmID = '".$_POST['id']."'";
$query = "INSERT INTO tbl_remit ('rAmount', 'rDate', 'rTime', 'wmID') VALUES
('".$_POST['payment']."',$date,$time,'".$_POST['id']."')";
if(mysqli_query($conn, $sql) && mysqli_query($conn, $query))
{
echo 'Payment Remitted' .mysqli_error($conn);
} else{
echo 'Rermittance failed' .mysqli_error($conn);
}
?>
$(document).on('click', '.btn_delete', function(){
var id=$(this).data("id4");
var payment=$(this).data("id3");
var firstname=$(this).data("id2");
var lastname=$(this).data("id1");
if(confirm("Is this really remitted?"))
{
$.ajax({
url:"edit.php",
type:"POST",
data:{id:id,payment:payment},
dataType: "text",
success:function(data){
alert(data);
fetch_data();
}
});
}
});
答案 0 :(得分:0)
将所有数据放入按钮以触发值
<td><button type="button" name="delete_btn" data-id1="'.$row["userFirstname"].'" data-id2="'.$row["userLastname"].'" data-id3="'.$row["payment"].'" data-id4="'.$row["wmID"].'" class="btn btn-xs btn-danger btn_delete">Remit</button></td>