PHP mysqli Fatal error: Call to a member function execute() on boolean [closed]

问题:

Quick question. Why does this error occur? I can't find a type or anything. (my code is a bit messy though)

//Where i fill the names for the dropdown boxes <select name="user_item_1_'.$user_item_box_id.'"> <select name="user_item_2_'.$user_item_box_id2.'">  //Fill the variables with the names $trade_item_id_1_1 = $_POST['user_item_1_1']; $trade_item_id_1_2 = $_POST['user_item_1_2']; $trade_item_id_1_3 = $_POST['user_item_1_3']; $trade_item_id_1_4 = $_POST['user_item_1_4'];  $trade_item_id_2_1 = $_POST['user_item_2_1']; $trade_item_id_2_2 = $_POST['user_item_2_2']; $trade_item_id_2_3 = $_POST['user_item_2_3']; $trade_item_id_2_4 = $_POST['user_item_2_4'];  //Query $query_insert_items = "INSERT INTO trade_item (item_id_1_1, item_id_1_2, item_id_1_3, item_id_1_4, item_id_2_1, item_id_2_2, item_id_2_3, item_id_2_4) VALUES (?,?,?,?,?,?,?,?)"; $stmt = $db->prepare($query_insert_trade); $stmt->bind_param("iiiiiiii", $trade_item_id_1_1, $trade_item_id_1_2, $trade_item_id_1_3, $trade_item_id_1_4, $trade_item_id_2_1, $trade_item_id_2_2, $trade_item_id_2_3, $trade_item_id_2_4); $stmt->execute();   

Thanks!

回答1:

mysqli_prepare() returns a statement object or FALSE if an error occurred. Seems that prepare failed and assigned a boolean to $stmt. I speculate that prepare failed because $query_insert_trade isn't defined. You probably mean

$stmt = $db->prepare($query_insert_items); 

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