通过mysql从php创建json对象 [英] Create json Object by php from mysql result

问题描述

可能重复:
JSON编码MySQL结果

Possible Duplicate:
JSON encode MySQL results

我想使用php创建如下的json对象.它将返回一个字符串作为来自结果sql查询的响应.

I want to use php to create a json object like below. it will return a string as a response from result sql query.

{"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，结果却不是我期望的:

But when use my code the result is not what I expected:

{订单":[ ["longvan"，"2012年10月12日"，"Be34433jh"，"Long Van"，"115 Pham Viet Chanh和quan Binh Thanh"，"http://longvansolution.tk/image/sample.jpg"，包装"，"0909056788"]， ["takeshi"，"2012/12/24"，"BF6464633"，"Vn-zoom"，"16 nguyen cuu van，quan binh thanh"，"http://longvansolution.tk/image/hoadon3.jpg"， 已打包"，"098897657"] ]}

{"Orders":[ ["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"] ]}

你能帮我吗?

推荐答案

您必须为每行创建一个数组以指定字段名称和值.

You have to create an array for each row to specify the field name and value.

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

或者如果表列名称正是您所使用的名称，则使用 mysqli_fetch_assoc()函数想要在您的JSON中使用:

Or use mysqli_fetch_assoc() function if the table column name is exactly what you want to use in your JSON:

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

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