Question

我在Delivary（）类下创建了一个公共函数View（）。但在第40行，我返回$ arr。现在它显示Undefined变量：arr当我的数据表为空时。 注意：未定义的变量：第40行的C：\ xampp \ htdocs \ online shop \ dataAccessLayer \ dalDelivary.php中的arr

我的代码是：

class Delivary {

public $place;
public $address;
public $amount;
public $date;
public $payment;
public $walletnumber;
public $user_id;

public function DB()
{
    $connection = mysqli_connect("localhost","root","","project");
    return $connection;
}

public function Insert()
{
    $sql = "INSERT INTO delivery (place, address, amount, `date`,payment, walletnumber, user_id) VALUES 
    ('$this->place', '$this->address', '$this->amount', '$this->date', '$this->payment', '$this->walletnumber', '$this->user_id');";
    if(mysqli_query($this->DB(), $sql))
    {
        return true;
    }
        return false;
}

public function View()
{
    $sql = "SELECT delivery.delivery_id, service.area, delivery.address, delivery.amount, delivery.date, delivery.payment, delivery.walletnumber, delivery.user_id FROM delivery, service WHERE service.service_id = delivery.place";
    $result = mysqli_query($this->DB(),$sql);
    while($d = mysqli_fetch_row($result))
        {
            $arr[] = $d; 
        }
        return $arr; //line 40
}
public function Delete()
    {
        $sql = "DELETE FROM delivery WHERE delivery_id = '".$this->delivery_id."'";
        if(mysqli_query($this->DB(), $sql))
        {
            return true;
        }
        return false;
    }}

Answer 1

在将数组定义为数组之前，您将像数组一样访问$arr。要消除此通知（这不是警告而不是错误），只需在使用之前定义数组：

$arr = [];
while($d = mysqli_fetch_row($result))
{
     $arr[] = $d; 
}
return $arr; //line 40

确保您正在检查mysqli_query函数返回时的错误：

$conn = $this->DB();
$sql = "SELECT delivery.delivery_id, service.area, delivery.address, delivery.amount, delivery.date, delivery.payment, delivery.walletnumber, delivery.user_id FROM delivery, service WHERE service.service_id = delivery.place";
$result = mysqli_query($conn, $sql);
if ( !$result ) {
    echo "There was an error executing this query: ", mysqli_error($conn);
    exit;
}
$arr = [];
while($d = mysqli_fetch_row($result))
{
    $arr[] = $d; 
}
return $arr;

Answer 2

更改

public function View()
{
    $sql = "SELECT delivery.delivery_id, service.area, delivery.address, delivery.amount, delivery.date, delivery.payment, delivery.walletnumber, delivery.user_id FROM delivery, service WHERE service.service_id = delivery.place";
    $result = mysqli_query($this->DB(),$sql);
    while($d = mysqli_fetch_row($result))
        {
            $arr[] = $d; 
        }
        return $arr; //line 40
}

要

    public function View()
    {
        $arr = [];
        //Please correct the following SQL statement as per your requirement.
        $sql = "SELECT delivery.delivery_id, service.area, delivery.address, delivery.amount, delivery.date, delivery.payment, delivery.walletnumber, delivery.user_id FROM delivery, service WHERE service.service_id = delivery.place";
        $result = mysqli_query($this->DB(),$sql);
        while($d = mysqli_fetch_row($result))
            {
                $arr[] = $d; 
            }
        return $arr; //line 40
    }

错误应该去了。由于您在while循环中直接推送到$ arr，因此如果查询返回0条记录，PHP无法找到$ arr。由于您的查询返回0条记录，因此它不会执行while loop中的代码，因此PHP不知道$ arr。

显示通知：＆＃34;未定义的变量：第40行和第34行的arr;当数据表为空时

2 个答案: