$(document).ready(function(){
$('.button').click(function(){
var clickBtnValue = $(this).val();
var ajaxurl = 'functions/delivered.php',
data = {'action': clickBtnValue};
$.post(ajaxurl, data, function (response) {
// Response div goes here.
alert("action performed successfully");
});
});
});
这是ajax代码,我正在尝试更新数据库
functions / delivered.php:
<?php
$db=new mysqli("localhost","root","","restaurant")
if(isset($_POST['onum'])){
$onum=$_POST['onum'];
$query="UPDATE `revenue` SET `payment`='y' WHERE onum=$onum";
$db->query($query);
}
?>
我认为您的错误在这里
"UPDATE `revenue` SET `payment`='y' WHERE onum=$onum";
将此行更改为
'UPDATE revenue SET payment = 'y' WHERE onum='.$onum.'';