php - 通过 php 从 mysql 结果创建 json 对象

Question

可能重复：
JSON 编码 MySQL 结果

我想使用 php 创建一个 json 对象，如下所示。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时，结果不是我所期望的：

{"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 ://longvansolution.tk/image/hoadon3.jpg","PACKED","098897657"] ]}

你能帮助我吗？

Answer 1

您必须为每一行创建一个数组以指定字段名称和值。

$json['Orders'][] = array('DeliveryId' => $row[0], 'CustomerName' => $row[1], ...);

或者，如果表列名称正是您想要在 JSON 中使用的名称，则使用mysqli_fetch_assoc()函数：

$rows = array();
while($r = mysqli_fetch_assoc($sql)) {
    $rows[] = $r;
}
$data = array('Orders' => $rows);
print json_encode($data);