使用$ .get()从PHP导入数组的问题

时间:2012-06-13 23:06:42

标签: php jquery json get

打印时,数组值返回为'undefined',即使使用print_r($items)我也可以看到数组不为空。这可能是什么问题?

MySQL的:

表格称为'之夜',列名为'名称','价格','日','排队跳','关闭'和'门'。他们都是人口稠密的。

eventinfo.php:

<?php

header('Cache-Control: no-cache, must-revalidate');
header('Expires: Mon, 26 Jul 1997 05:00:00 GMT');
header('Content-type: application/json');

include('functions.php');
connect();

$night = $_POST['club'];
// this echos fine
$night = mysql_real_escape_string($night);

$query = "SELECT * FROM nights WHERE name = '$night'";

    $result = mysql_query($query);
    $items = array();

    if($result && mysql_num_rows($result) > 0) { 
        while ($row = mysql_fetch_array($result)) { 
        $items[] = array("price"=>$row['price'], "day"=>$row['day'], "queue jump"=>$row['queue jump'], "closing"=>$row['closing'], "doors"=>$row['doors']);
        }
    } 

    mysql_close(); 
    // convert into JSON format and print

    echo json_encode($items);
?>

JS:

<script type="text/javascript">

    $(document).ready(function() {
// for simplicity's sake, but ultimately I'd like to load all the information
            $('#right_inside').html('<h2>' + $('#club').val() + '</h2>');
    });

    $('#club').change(function(event) {
        $.ajax({
            type: "post",
            url: "eventinfo.php",
            data:  $(this).serialize(),
            success: function(data) {
                $('#right_inside').html('<h2>' + $('#club').val() + '</h2><p>Day: ' + data.day + '</p>');
                },
            dataType: "json"
        });
    });

</script>

1 个答案:

答案 0 :(得分:5)

您的JSON结构错误。数据实际上是一个数组:

尝试将data.day更改为data[0].day

注意:您应该在ajax成功中使用JSON.stringify(data)来查看JSON的结构。

$.ajax({
  type: "post",
  url: "eventinfo.php",
  data:  $(this).serialize(),
  success: function(data) {
     alert(JSON.stringify(data));
     //or
     console.log(JSON.stringify(data));
  },
   dataType: "json"
 });