asked on

How to Avoid Undefined Index / Offset Errors In PHP

hi,
Plz find the code

public function getLeaveBalance($emp_code, $leave_type)
{
$query = mysqli_query($this->_dbconnect, "SELECT balance FROM leave_balances WHERE code = '$leave_type' AND emp_code = '$emp_code'");
$result = array();
$i = 0;
while($res = $query->fetch_assoc())
   {
  $result[$i] = $res;
  $i++;
}
[b]  return $result[0]['balance'];                   [/b]
	}

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error occurs line ( return $result[0]['balance']; ) and I don't want ignore by using
error_reporting = E_ALL & ~E_NOTICE

ASKER CERTIFIED SOLUTION

Mukesh Yadav

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prashanth ag

ASKER

Hi mukesh yadav
thanks for your feed back ,It solved error
but is shows null vales (balance column) in front end back end balance column had values

plz Find my DB
emp_code code balance
1003 CL 1.0
1003 EL 35.0
1003 SL 4.0

Marco Gasi

Since that query returns one field of a unique row, I'd rewrite your code this way:

public function getLeaveBalance($emp_code, $leave_type)
{
$query = mysqli_query($this->_dbconnect, "SELECT balance FROM leave_balances WHERE code = '$leave_type' AND emp_code = '$emp_code'");
[b]  return $query->fetch_object()->balance;[/b]
}

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prashanth ag

ASKER

Hi all,
some values are not in the column (balance)
thanks to all

prashanth ag

ASKER

I try this code return $query->fetch_object()->balance;
but it shows Notice: Trying to get property of non-object in

Marco Gasi

Try this:

public function getLeaveBalance($emp_code, $leave_type)
{
return mysqli_query($this->_dbconnect, "SELECT balance FROM leave_balances WHERE code = '$leave_type' AND emp_code = '$emp_code'")->fetch_object()->balance;
}

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prashanth ag

ASKER

Marco Gasi
Thanks for your feedback

Ray Paseur

Notice: Trying to get property of non-object

When your script gives this message in the given context, this usually means the query failed. Information about how to run a query and test for success or failure is available in this article.
https://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/PHP_Databases/A_11177-PHP-MySQL-Deprecated-as-of-PHP-5-5-0.html

prashanth ag

ASKER

my appln throws one(new) more error

Fatal error: Call to a member function fetch_assoc() on a non-object in
line 4

$query = mysqli_query($this->_dbconnect, $query);
		$result = array();
		$i = 0;
		[b]while($res = $query->fetch_assoc()){[/b]
			$result[$i] = $res;
			$i++;
		}
		return $result;

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Marco Gasi

As I wrote in my last message, that error tells that the query is malformed.
You are in an object, right? So you could write something like this:

public function getLeaveBalance($emp_code, $leave_type)
{
$mysqli = $this->_dbconnect;
return $mysqli->query("SELECT balance FROM leave_balances WHERE code = '$leave_type' AND emp_code = '$emp_code'")->fetch_object()->balance;
}