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How to process JSON data returned by ajax in PHP

一个新手Original: 2017-10-09 09:22:061631browse

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<script src="jquery-1.11.2.min.js"></script>
<title>无标题文档</title>
</head>
<body>
<select id="nation"></select>
</body>

<script type="text/javascript">
$.ajax({
    url:"jsonchuli.php",
    dataType:"JSON",
    success: function(data){
        var str ="";
            /*for(var i=0;i<data.length;i++)//关联数组里面的大小写和数据库里面的大小写是要保持一致的。
            {
                str = str+"<option value=&#39;"+data[i].Code+"&#39;>"+data[i].Name+"</option>";
            }
            $("#nation").html(str);*/
            
            for(var k in data)//这种方法相当于foreach方法遍历，上面注释的方法是for循环。
            {
                str = str+"<option value=&#39;"+data[k].Code+"&#39;>"+data[k].Name+"</option>";
            }
            $("#nation").html(str);
        }
    })

</script>

</html>

<?php
include("./DBDA.class.php");
$db = new DBDA();
$sql = "select * from nation";
//需要使用关联数组
//数组内容的编码格式必须是utf8的，其它的是不可以的。
echo json_encode($db->Query($sql,0));//返回的是关联数组。json_encode返回的是json数据。

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