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我有問題得到我的表使用ajax行的值,到目前爲止,每當我得到它的值時,它不給我任何回報,這裏是我如何得到它。得到ajax中的一行的值
select.php 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>
';
}
edit.php
<?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);
}
?>
AJAX用於更新
$(document).on('click', '.btn_delete', function(){
var id=$(this).data("id4");
var payment=$(this).data("id3");
var firstname=$(this).data("id2");
var lastname=$(this).data("id1");
if(confirm("Is this really remitted?"))
{
$.ajax({
url:"edit.php",
type:"POST",
data:{id:id,payment:payment},
dataType: "text",
success:function(data){
alert(data);
fetch_data();
}
});
}
});
嗨在你的頁面select.php你的$輸出的回聲和edit.php相同? – 2017-09-15 01:06:11
@headmax - 你的意思是輸出表先生? – jared
我的意思是echo json_encode($ oupout)和頭部(「Content-type:application/json」); – 2017-09-15 01:11:58