paynod Posted March 1, 2011 Share Posted March 1, 2011 Hi,I'm hoping someone can help me:). When I add foreign keys to the database in MySQL I CAN change the primary key field and the corresponding indexed field will automatically change giving me the same value which is exactly what i want. The issue is when I try to insert the values using my php code i get the 'cannot update child error'. I have not selected a value for the unique_id field as I want it to inherit the value from the primary key. Here's my code as simplified as I could do it. <?php //this calls the class productupload and instantiates the function insert_form to insert values into the db. $product = new productUpload(); $product->insert_form();?> //This calls the insert_form function and inserts info into tables <?php class productUpload extends DatabaseObject{ protected static $table_name="itm_details"; protected static $db_fields =array('id','unique_id','itm_cat','itm_make','itm_model','itm_desc','itm_cond','itm_date_from','itm_date_to','itm_add_date'); public $id; public $unique_id; public $itm_cat; public $itm_make; public $itm_model; public $itm_desc; public $itm_cond; public $itm_date_from; public $itm_date_to; public $itm_add_date; Heres the function insert_form which grabs the form submitted data. Note I have not defined a value for $unique_id public function insert_form(){ global $database; //set the object attributes to the form the parameters if(isset($_POST['submit'])) { $result = array( $this->itm_cat =(!empty($_POST['itm_cat'])) ? trim($_POST['itm_cat']) : NULL, $this->itm_make =(!empty($_POST['itm_make'])) ? trim($_POST['itm_make']) : NULL, $this->itm_model = (!empty($_POST['itm_model'])) ? trim($_POST['itm_model']) : NULL, $this->itm_desc =(!empty($_POST['itm_desc'])) ? trim($_POST['itm_desc']) : NULL, $this->itm_cond =(!empty($_POST['itm_cond'])) ? trim($_POST['itm_cond']) : NULL, $this->itm_date_from =(!empty($_POST['itm_date_from'])) ? trim($_POST['itm_date_from']) : NULL, $this->itm_date_to =(!empty($_POST['itm_date_to'])) ? trim($_POST['itm_date_to']) : NULL, $this->itm_add_date = date("Y-m-d H:m:s")); //check date is numeric //if(is_numeric($_POST['itm_date_from'])) { //$this->itm_date_from = $_POST['itm_date_from']; //} //check date is numeric //if(is_numeric($_POST['itm_date_to'])) { //$this->itm_date_to = $_POST['itm_date_to']; //} if($result){ $result = $this->create(); } } } //here's the create function referred to above which sanitises the posted data and inserts it into the db. public function create() { global $database; $attributes = $this->sanitised_attributes(); $sql = "INSERT INTO ".self::$table_name."("; $sql .= join(", ", array_keys($attributes)); $sql .= ") VALUES ('"; $sql .= join("', '", array_values($attributes)); $sql .= "')"; if($database->query($sql)) { $this->id = $database->insert_id(); return true; } else { return false; } } Quote Link to comment Share on other sites More sharing options...
redarrow Posted March 1, 2011 Share Posted March 1, 2011 You got the database setup incorrectly then. your need to post the id as a varchar in the database feild. mysql wont do it for you. it will make a primary key as one field as you no, but afther that you controll all inserts. Quote Link to comment Share on other sites More sharing options...
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