douci1677 2014-09-10 06:13
浏览 54
已采纳

php的json输出中的未定义变量

I just don't get what's wrong in my code!

In browser, it shows that undefined variable!

But I declared it in a while loop before!

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

The PHP Code that I wrote is something like below:

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']);
    }
}

Code:

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

P.S. $retrieve1 variable here is a variable that I used to assign a mysql query that generally retrieves information from my database!

  • 写回答

2条回答 默认 最新

  • douyimiao1993 2014-09-10 06:19
    关注

    Add

       $dhkBlood = array();
    

    Before

     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']);
       }
      }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(1条)

报告相同问题?

悬赏问题

  • ¥100 set_link_state
  • ¥15 虚幻5 UE美术毛发渲染
  • ¥15 CVRP 图论 物流运输优化
  • ¥15 Tableau online 嵌入ppt失败
  • ¥100 支付宝网页转账系统不识别账号
  • ¥15 基于单片机的靶位控制系统
  • ¥15 真我手机蓝牙传输进度消息被关闭了,怎么打开?(关键词-消息通知)
  • ¥15 装 pytorch 的时候出了好多问题,遇到这种情况怎么处理?
  • ¥20 IOS游览器某宝手机网页版自动立即购买JavaScript脚本
  • ¥15 手机接入宽带网线,如何释放宽带全部速度