我想用PHP创建一个JSON对象,像下面。它将返回一个字符串作为结果sql查询的响应。
{"Orders":[
{"DeliveryId":"DeliveryId","CustomerName":"CustomerName","PhoneNumber":"PhoneNumber","Address":"Address"},
{"DeliveryId":"DeliveryId","CustomerName":"CustomerName","PhoneNumber":"PhoneNumber","Address":"Address"}
]
}
我的代码
<?php
mysql_connect("mysql12.000webhost.com","a4602996_longvan","longvan2012");
mysql_select_db("a4602996_lv");
$id=$_POST[user];
$sql=mysql_query("select * from testlongvan where Status = 'PACKED'");
$json = array();
if(mysql_num_rows($sql)){
while($row=mysql_fetch_row($sql)){
$json['Orders'][]=$row;
}
}
//while($row=mysql_fetch_assoc($sql))
//$output[]=$row;
print(json_encode($json));
mysql_close();
?>
但是,当使用我的代码的结果不是我所期待的:
{ “订单”: [ “longvan”,“10/12/2012“,”Be34433jh“,”Long Van“,”115 Pham Viet Chanh,quan Binh Thanh“,”http://longvansolution.tk/image/sample.jpg“,”PACKED“,”0909056788“], [“takeshi”,“24/12/2012”,“BF6464633”,“Vn-zoom”,“16 nguyen cuu van,quan binh thanh”,“http:// long vansolution.tk/image/hoadon3.jpg","PACKED","098897657" ] ]}
你能帮助我吗?
有一个在关于字符串类型的问题匹配的问题+答案:HTTP ://stackoverflow.com/questions/28261613/convert-mysql-result-to-json-with-correct-types –
删除您的托管密码 –