我刚刚开始学习代码并正在编写一些代码来回显一个数组,它给了我这两个错误:
警告:array_keys() 期望参数 1 为数组,在第 70 行的 C:\xampp\htdocs\ogmt\rest_server_api.php 中给出 null
警告:在第 71 行的 C:\xampp\htdocs\ogmt\rest_server_api.php 中为 foreach() 提供的参数无效。
我该如何解决?这是我的代码:
if($Result1){
// script to get business no, amount & merchant id,output to merchant page
$query="SELECT * FROM customer_order WHERE order_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");
}
$response=array('details'=>$data);
}
}
else{
$response=array('details'=>"query failed!");
}
}
//print output to merchant page to complete payment
$keys = array_keys($data);
foreach($response['details'] as $keys=>$value){
echo "$keys: $value</br>";
}
?>