Issue in using arrays as parameters of method of PHP custom class - php

Question

To explain my issue, here is a simple code at first:
public function sql($data) {
if (is_array($data)) {
$cells = $data['cells'];
$from = $data['from'];
$where = $data['where'];
$joins = $data['joins'];
$order_by = $data['order_by'];
$o_type = $data['order_by_type'];
$limit = $data['limit'];
/*****************************/
if ($cells == '') { $cells = "*"; }
if ($where != '') { $where = "where ".$where; }
if ($oredr_by != '') { $order_by = "order by ".$order_by." ".$o_type; }
if ($limit != '') { $limit = "limit ".$limit; }
//
$sql = "select ".$cells." from ".$from." ".$joins." ".$where." ".$order_by." ".$limit;
$run = mysqli_query($_SESSION['con'], $sql);
}else{
$run = mysqli_query($_SESSION['con'], $data);
}
}
When I start using this method, I pass a multidimensional array as a parameter, like this:
$sql = $class->sql([ "from" => "table", "order_by" => "id", "order_by_type" => "asc" ]);
/* This will generate and run this query: select * from table order by id asc */
// Notice that I've only used 3 keys, not the all above.
In Apache server, it works OK perfectly when I just use some of the keys of array, but in XAMPP it doesn't because it says that I have to pass all the parameters (cells, from, where, joins, ...) even if they are empty.
Please help me to resolve this, and thanks.

Answer 1

You can use isset to check if an array key is present, then get it's value like.
public function sql($data) {
if (is_array($data)) {
$cells = '';
if(isset($data['cells']) {
$cells = $data['cells'];
}
....
/*****************************/
if ($cells == '') { $cells = "*"; }
if ($where != '') { $where = "where ".$where; }
if ($oredrby != '') { $orderby = "order by ".$orderby." ".$od_type; }
if ($limit != '') { $limit = "limit ".$limit; }
$sql = "select ".$cells." from ".$table." ".$joins." ".$where." ".$orderby." ".$limit;
$run = mysqli_query($_SESSION['con'], $sql);
}else{
$run = mysqli_query($_SESSION['con'], $data);
}
}
Or simply just do error_reporting(1) before calling this function or in your index.php.

Answer 2

The problem is this.
$arr = ["a"];
echo $arr["b"];
You will get an error notice.
Notice: Undefined index: b
If you want to avoid this use it in this way.
$arr = ["a"];
$arr = ["b"] = "";
echo $arr["b"];

Answer 3

Change $from to $table, you have not $table variable
$from = $data['from'];
to
$table = $data['from'];
Also you have spelling mistake biggest spelling mistake which is very difficult to find.orderby and oredrby