dongsha2792 2013-08-16 07:10
浏览 73

如何在页面中回显变量数组?

I am just starting learning codes and am writing some codes to echo an array and it give me this error "Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\ogmtest_server_api.php on line 62" HOW CAN I SOLVE THIS? This is my code:

   if($Result1){
    // script to get business no, amount & merchant id,output to merchant page
      $query="SELECT * FROM customer_order WHERE insert_time=(SELECT max(order_time)from customer_order)";
      $result=mysql_query($query);
      while($row=mysql_fetch_assoc($result)){
          $amount=$row['amount'];
          $id=$row['merchant_id'];
          $payment_mode=$row['mobile_service'];
          switch($payment_mode){               
           case 'TIGO-PESA':
            $result1=mysql_query("SELECT * FROM mobile_client WHERE mobile_service='TIGO-PESA'");
            while($row1=mysql_fetch_assoc($result1)){ 
                 $data=array(
                   'Business no'=>$row1['business_no'],
                   'Payment Mode'=>$payment_mode,
                   'Total Amount Tsh'=>$amount,
                   'Merchant ID'=>$id
                   );
                   }break;

            case 'M-PESA':
            $result1=mysql_query("SELECT * FROM mobile_client WHERE mobile_service='M-PESA'");
            while($row1=mysql_fetch_assoc($result1)){ 
                 $data=array(
                   'Business no'=>$row1['business_no'],
                   'Payment Mode'=>$payment_mode,
                   'Total Amount Tsh'=>$amount,
                   'Merchant ID'=>$id
                   );
                   }break;
            case 'AIRTEL-MONEY':
            $result1=mysql_query("SELECT * FROM mobile_client WHERE mobile_service='AIRTEL-MONEY'");
            while($row1=mysql_fetch_assoc($result1)){ 
                 $data=array(
                   'Business no'=>$row1['business_no'],
                   'Payment Mode'=>$payment_mode,
                   'Total Amount Tsh'=>$amount,
                   'Merchant ID'=>$id
                    );
                   }break;

           default:
           $data=array('error'=>"no payment mode selected");        
        }
     }
           foreach( $data as $value ){
              echo $value;
                 }      
             }
            else{
      echo "wrong";
    }
}
  • 写回答

2条回答 默认 最新

  • douji5397 2013-08-16 07:15
    关注

    you can use print_r() to print the array. like this:

    echo '<pre>';
    print_r($data);
    echo '</pre>';
    

    if $data is an array. So you can debug it by yourself.

    Note that first declare $data as array at the top of script. like: $data = array();

    评论

报告相同问题?

悬赏问题

  • ¥100 任意维数的K均值聚类
  • ¥15 stamps做sbas-insar,时序沉降图怎么画
  • ¥15 unity第一人称射击小游戏,有demo,在原脚本的基础上进行修改以达到要求
  • ¥15 买了个传感器,根据商家发的代码和步骤使用但是代码报错了不会改,有没有人可以看看
  • ¥15 关于#Java#的问题,如何解决?
  • ¥15 加热介质是液体,换热器壳侧导热系数和总的导热系数怎么算
  • ¥100 嵌入式系统基于PIC16F882和热敏电阻的数字温度计
  • ¥15 cmd cl 0x000007b
  • ¥20 BAPI_PR_CHANGE how to add account assignment information for service line
  • ¥500 火焰左右视图、视差(基于双目相机)