$dupnamecheck = "SELECT product_name FROM products WHERE product_name = $name";
$dupnameresult = mysqli_query($link,$dupnamecheck);
$data = mysqli_fetch_array($dupnameresult, MYSQLI_NUM);
if($data[0] > 1) {
echo "Product name is duplicated. <br/>";
}
else {
$query = ("INSERT INTO products (id, product_name, product_description, category_id, brand_id, views, price)
VALUES (null, '".$name."','".$desc."','".$cat."', '".$brand."', '0','".$price."')");
$status = mysqli_query($link,$query) or die(mysqli_error($link));
if($status){
$message = "Successfully Added! Please check on products page or on inventory page in admin panel." ;
}
else {
$message = "FAILED!";
}
}
我試圖用product_name
作爲條件來檢查我的產品表中重複的名字。我有這個錯誤: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
我有以下的代碼,但它似乎並沒有爲我工作
'NULL'是有效的,他也在fetch_array上發生錯誤,而不是在插入 –
啊好吧,我明白了。對不起,如果我的問題很愚蠢,我正試圖解決這個問題近一個小時。 – user3087717
謝謝btw傢伙 – user3087717