$db_item = $wpdb->get_results($wpdb->prepare( "SELECT * FROM wp_wowhead_items WHERE name LIKE %s", "但是下面的那个不起作用!"Hello";
$db_item = $wpdb->get_results($wpdb->prepare( "SELECT * FROM wp_wowhead_items WHERE name LIKE
我使用LIKE进行搜索,我在phpMyAdmin中尝试并返回结果,但是当我在php中使用它时,它返回空的结果。$search = "ip";$query = "SELECT * FROM product WHERE product_name LIKE '%$search%' LIMIT$start,30";if(empty($result))else
我已经数百次将mysql数组结果赋值给变量,但是由于某些原因,在这个特定的脚本中变量赋值对我不起作用,我不知道为什么。
查询在数据库上工作,并从循环代码中生成结果。$pcode = mysql_query("SELECT * FROM table WHERE suburb like('%$suburb%') && state like('%$state%') &&category like
我有一个mysql查询不起作用。它给出了以下错误:
$select = mysqli_query($sql, "SELECT title FROMcategory WHERE id LIKE (SELECT categorie_id FROM categories_sub WHERE file LIKE '".
("$host", "$username", "$password") or die ("cannot connect server"); if(isset($_REQUEST['submit'])){
$sql=" SELECT * FROM $tbl_name WHERE name like$q=mysql<