实际上我在行号-34中遇到错误(指向函数check_id,if(codition))并且我不明白为什么会显示错误,这是什么意思和帮助SomeOne解释我以及我如何解决它(我是PHP新手): 这是我的php文件:
<?php
include_once("Order.php");
include("connect.php");
$query="SELECT * FROM `orders`";
$filter_Result=mysqli_query($con,$query);
$newOrders=Array();
while($row=mysqli_fetch_array($filter_Result))
{
$order;
$orderId= $row['id']; //fetch row id
echo "hello".$orderId;
if(check_id($newOrders,$orderId)){
echo "kim";
//$order= Add the order object
}
else{
echo " jim<br>";
$order=new Order($row['id'], $row['tableId'], $row['createdDate']);
array_push($newOrders,$order);
}
$item=new Item($row['ProductId'], $row['ProductName'], $row['Quantity']);
//include("Order.php");
$order->AddItem($item,null,null,null);
}
function check_id($newOrders,$orderId){
$length=count($newOrders);
echo $length;
for($i=0; $i<$length; $i++){
if($newOrders[$i].$orderId == $orderId)
return true;
}
return false;
}
?>