Question

我的php代码如下：

$query = "SELECT *
        FROM `address`
        WHERE customer_id =$customer_id
        ORDER BY `default` DESC";

$result=mysql_query($query);

$value = mysql_num_rows($result);

if($value>=1)
{
    while($row = mysql_fetch_array($result)) 
    {

        $details =  array(
        'status'=>'sucess', 
        'message'=>'address available',
        'id' => $row['id'], 
        'customer_id' =>$row['customer_id'],
        'at' => $row['at'],
        'name'=>$row['name'],
        'mobile'=>$row['mobile'],
        'city'=>$row['city'],
        'address'=>$row['address'],
        'latlog'=>$row['latlog'],
        'default'=>$row['default']
        );          
    }
    echo  json_encode($details);
}

其输出如下：

{"status":"sucess","message":"address available","id":"52","customer_id":"14","at":"Home","name":"Shhsh","mobile":"99989998","city":"Calicut","address":"Gsggsgs","latlog":"76.3007429,76.3007429","default":"Yes"}

它是json对象。我要求json是：

{"status":"sucess","message":"address available","details":[{"id":"52","customer_id":"14","at":"Home","name":"Shhsh","mobile":"99989998","city":"Calicut","address":"Gsggsgs","latlog":"76.3007429,76.3007429","default":"Yes"}]
}

我应该对所需的json格式进行哪些更改？那就是json对象包含json数组。我应该做什么改变才能获得所需的json格式。我是新来的。提前感谢。

Answer 1

您需要在外部创建一个数组以附加结果数据，然后在将其解码为JSON之前将其放入主数组中。

像这样，

if($value>=1)
{
    $details=array();
    while($row = mysql_fetch_array($result)) 
    {

        $details[]=  array(
        'id' => $row['id'], 
        'customer_id' =>$row['customer_id'],
        'at' => $row['at'],
        'name'=>$row['name'],
        'mobile'=>$row['mobile'],
        'city'=>$row['city'],
        'address'=>$row['address'],
        'latlog'=>$row['latlog'],
        'default'=>$row['default']
        );          
    } 
    $main_array=array();
    $main_array['status']='sucess';
    $main_array['message']='address available';
    $main_array['details']=$details;

    echo  json_encode($main_array);
}

这将生成这样的JSON，http://json-parser.com/deec3edd

Answer 2

请在我认为不需要时删除

$query ="SELECT id,customer_id,at,name,mobile,
        city,address,latlog,default
        FROM `address`
        WHERE customer_id =$customer_id
        ORDER BY `default` DESC";

$result=mysql_query($query);

$value = mysql_num_rows($result); 

$json=array();
        $json['status']='sucess';
        $json['message']='address available';
        $json['details']=$value;

        echo jsone_encode($json);

Answer 3

将此代码替换为您的代码...它为您工作...谢谢

while($row = mysql_fetch_array($result)) 
{

    $details =  array(

    'id' => $row['id'], 
    'customer_id' =>$row['customer_id'],
    'at' => $row['at'],
    'name'=>$row['name'],
    'mobile'=>$row['mobile'],
    'city'=>$row['city'],
    'address'=>$row['address'],
    'latlog'=>$row['latlog'],
    'default'=>$row['default']
    );          
}
$json= array('status'=>'sucess','message'=>'address available','details'=> $details);
echo  json_encode($json);

Answer 4

您要做的是生成详细信息数组并将其作为数组添加到响应中，这样您就可以控制循环外的状态（例如，如果它不返回任何数据）：

if($value>=1){
    $details=array();
    while($row = mysql_fetch_array($result)) 
    {

        $details[]=  array(
        'id' => $row['id'], 
        'customer_id' =>$row['customer_id'],
        'at' => $row['at'],
        'name'=>$row['name'],
        'mobile'=>$row['mobile'],
        'city'=>$row['city'],
        'address'=>$row['address'],
        'latlog'=>$row['latlog'],
        'default'=>$row['default']
        );          
    } 
    $main_array=array();
    $message = 'address available';
}else{
    $message = 'no address';
    $details = null;
}

echo json_encode(array(
  'success' => $value>=1,
  'message' => $message, 
  'details' => $details
));

Answer 5

您可以编辑代码

while($row = mysql_fetch_array($result)) 
{

    $details =  array(
    'status'=>'sucess', 
    'message'=>'address available',
    'id' => $row['id'], 
    'customer_id' =>$row['customer_id'],
    'at' => $row['at'],
    'name'=>$row['name'],
    'mobile'=>$row['mobile'],
    'city'=>$row['city'],
    'address'=>$row['address'],
    'latlog'=>$row['latlog'],
    'default'=>$row['default']
    );          
}
echo  json_encode($details);

到

$datails = array();
while($row = mysql_fetch_array($result)) 
{
  $detail =  array(
  'status'=>'sucess', 
  'message'=>'address available',
  'details'=>[
    'id' => $row['id'], 
    'customer_id' =>$row['customer_id'],
    'at' => $row['at'],
    'name'=>$row['name'],
    'mobile'=>$row['mobile'],
    'city'=>$row['city'],
    'address'=>$row['address'],
    'latlog'=>$row['latlog'],
    'default'=>$row['default']
 ]
 );
 array_push($datails,$datail);         
}
echo  json_encode($details);

如何显示Mysql php结果json格式？

5 个答案: