php - ajax $.get的data参数无法传递
使用$.get方法，点击一次button获取两条数据，点击一次之后变量+1，之后把这个值传递到getajax.php页面的数据库查询语句中作为查询条件，如果把"page="+pageno改成page=1。程序能正常运行，但现在是"page="+pageno，运行getajax.php,首先提示“false”,就是var_dump($result)的结果，然后提示“Fatal error: Call to a member function fetch_assoc() on a non-object in ”，这个应该是$result出错了？后推应该是查询语句的错误吗，那么这个"page="+pageno应该怎么写才正确？求指教
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<script src="jquery.js" type="text/javascript"></script>
<title></title>
</head>
<body>
<p id="showdata"></p>
<!-- 点击一次，value值变化一次 -->
<input type="hidden" id="currentresult" value="1" />
<button id="show">Load</button>
<script type="text/javascript">
$(document).ready(function(){
$(document).on('click','#show',function(e){
var pageno = $('#currentresult').val();
$.get("getajax.php","page="+pageno ,function(data){
pageno++;
$('#currentresult').val(pageno);
$('#showdata').append(data); $('#showdata').html(data);
});
});
});
</script>
</body>
</html>getajax.php
<!-- This is ajax data. -->
<?php
$con = mysqli_connect("localhost", "root", "", "maroon5");
$page=isset($_GET["page"])?$_GET["page"]:0;
$pageNo = $page;
$startLimit = ($pageNo-1)*2;
$query = "SELECT * from tour LIMIT $startLimit,2 ";
$result = mysqli_query($con,$query);//如果查询失败，则返回false
var_dump($result);
while($row = $result -> fetch_assoc()){
?>
<p>
<h4>month:<?php echo $row["month"]; ?>day:<?php echo $row["day"]; ?></h4>
</p>
<?php
}
?>
