php - php json 输出中 undefined variable

Question

我只是不明白我的代码出了什么问题!

在浏览器中，显示 undefined variable !

但是我之前在 while 循环中声明了它!

In browser it shows: Notice: Undefined variable: dhkBlood in C:\xampp\htdocs\JSONdata.php on line 371*

我编写的 PHP 代码如下所示:

if($retrieve1){
    while($row = mysql_fetch_assoc($retrieve1))
    {   
        $dhkBlood[] = array("ID" => $row['PID'], "PlaceName" => $row['PName'], "Address" => $row['Address'], "DeploymentName"  => $row['DName'], "Latitude" => $row['Latitude'], "Longitude" => $row['Longitude']);
    }
}

代码:

370. $result = array();
371. $result["dhakaBlood"]=$dhkBlood;
372. $finalResult = array();
373. $finalResult['data']=$result;
374. echo json_encode($finalResult);

附注这里的$retrieve1变量是我用来分配一个mysql查询的变量，该查询通常从我的数据库中检索信息!

Answer 1

添加

   $dhkBlood = array();

之前

 if($retrieve1){
  while($row = mysql_fetch_assoc($retrieve1))
   {   
    $dhkBlood[] = array("ID" => $row['PID'], "PlaceName" => $row['PName'], "Address" => $row['Address'], "DeploymentName"  => $row['DName'], "Latitude" => $row['Latitude'], "Longitude" => $row['Longitude']);
   }
  }