我想用PHP創建一個JSON像下面。它將返回一個字符串json作爲結果sql查詢的響應。我該怎麼做?
{"Orders":[
{"DeliveryId":"DeliveryId","CustomerName":"CustomerName","PhoneNumber":"PhoneNumber","Address":"Address"},
{"DeliveryId":"DeliveryId","CustomerName":"CustomerName","PhoneNumber":"PhoneNumber","Address":"Address"}
]
}
我的代碼
<?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,當使用我的代碼我收到的結果,不會導致我想:
{ 「訂單」: [ 「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:// longvansolution.tk/image/hoadon3.jpg","PACKED","098897657" ] ]} 你能幫幫我!
你能告訴我們你的SQL查詢? – jamis0n
爲什麼downvote?原因? – Elbek
['json_encode()'](http://php.net/manual/en/function.json-encode.php) –