Show Notice: "Undefined variable: arr on line 40" when data table is empty

Question

I made a public function View() under a class Delivary(). But in line 40 I return $arr. Now it show Undefined variable: arr when my data table is empty. Notice: Undefined variable: arr in C:\xampp\htdocs\online shop\dataAccessLayer\dalDelivary.php on line 40

My code is:

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;
    }}

What if `mysqli_fetch_row` don't get any data. What `$arr` will return; — lkdhruw, Jul 07 '17 at 04:49

Hannan · Accepted Answer · 2017-07-10T13:41:39.800

Change

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
}

TO

    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
    }

And the error should go. Since you are directly pushing to $arr in the while loop, PHP is not able to find $arr in case where your query returns 0 records. Since your query returns 0 records, it doesnt execute the code in while loop and hence PHP is not aware of $arr.

After Use your code my database don't show anything – Syed Mohammad Ibrahim Khalil Jul 07 '17 at 04:52 — Syed Mohammad Ibrahim Khalil, Jul 07 '17 at 04:52

Blue · Answer 2 · 2017-07-07T05:01:53.563

0

You're accessing $arr like an array before it's been defined as one. To eliminate this notice (Which is more a warning than an error), simply define the array before you use it:

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

Make sure you're checking for errors from the return of the mysqli_query function:

$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;

edited Jul 07 '17 at 05:01

answered Jul 07 '17 at 04:37

Blue

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After Use your code my database don't show anything – Syed Mohammad Ibrahim Khalil Jul 07 '17 at 04:51
Pet peeve: Delivary is spelled wrong. It should be Delivery. Check my updated answer for how to check for mysql errors. – Blue Jul 07 '17 at 05:01
Thanks for your help. Now my problem is solved. just change sql query "select * from delivery"; and add $arr = []; – Syed Mohammad Ibrahim Khalil Jul 07 '17 at 05:13
What do you mean just change sql query? – Blue Jul 07 '17 at 05:20
Change $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"; to $sql = "SELECT * from delivery"; – Syed Mohammad Ibrahim Khalil Jul 07 '17 at 05:26
1

It's not my job to change your SQL query. You posted an issue about a NOTICE error from your $arr not being defined. That issue has been solved. Just because you're getting a SQL error, which appears to be another issue entirely, doesn't negate that this answer has solved your question. Changing the sql statement offers absolutely nothing that would help future readers. – Blue Jul 07 '17 at 05:37
Here is also a array problem. your solve was also right. need to add $arr = []; which i was also mention please check. – Syed Mohammad Ibrahim Khalil Jul 07 '17 at 08:17

Show Notice: "Undefined variable: arr on line 40" when data table is empty

2 Answers2