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PHP Fatal error: Call to a member function prepare() on null

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Trying to implement with the prepare statement, and it gives me this error.

"PHP Fatal error:  Call to a member function prepare() on null."

Anybody knows what is wrong? 

	function listMerchant() {
		$merchant = array();
		$merchantList = $this->mysqli->prepare("SELECT * FROM core_merchant");
		while($merchantRows = $merchantList->fetch_array()) {
			$merchant[] = $merchantRows;
		return $merchant;
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ummm.... I have other functions in the class using  $this->mysqli-query(), and they work fine. Somehow the prepare() is not working. I am new to prepare(), and not sure my syntax is correct or not. 

Edited by bbmak
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  • Solution

A prepared statement doesn't have a fetch_array() method. Read the manual on how prepared statements work.


Actually, you don't need this at all, because you have no external input. The entire method can be boiled down to

function listMerchant() {
    return $this->mysqli->query('SELECT col1, col2, ... FROM core_merchant')->fetch_all(MYSQLI_BOTH);
Edited by Jacques1
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