調用一個非對象的成員函數query（）

Question

我對PHP很陌生，對MySQLi和都很陌生我試圖向包含5個變量的數據庫插入一行。

if($_POST['submit']) { 
      $title = mysqli_escape_string($_POST['title']); 
      $content = $_POST['content']; 
      $author = $_SESSION['fullname']; 
      $publishedtime = time(); 
      $pageID = $_POST['pageid']; 
      $connection = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME); 

      if ($connection->errno) { 
       printf("Connection failed: %s\n", $connection->error); 
       exit(); 
      } 

      $author = $_SESSION['fullname']; 
      $publishedtime = time(); 
      $q = "INSERT INTO posts (title, content, author, publishedtime, pageID) VALUES ('".$title."', '".$content."', '".$author."', '".$publishedtime."', '".$pageID."')"; 

      if (!$dbc->query($q)) { 
       echo "INSERT failed: (" . $dbc->errno . ") " . $dbc->error; 
      }  
      echo "Newest user id = ",$dbc->insert_id; 

      $connection->close(); 

     } 
     else { 
      addPostForm(); 
     } 

，我得到這些錯誤：

Warning: mysqli_escape_string() expects exactly 2 parameters, 1 given in /admin/manage.php on line 11

Fatal error: Call to a member function query() on a non-object in /admin/manage.php on line 27

我可以讀取數據庫的罰款，但不能插入進去。

感謝

Answer 1

如果使用real_escape_string創建mysqli對象之後，你可以只使用

$connection->real_escape_string($_POST['title']) 

http://www.php.net/manual/en/mysqli.real-escape-string.php

---

至於查詢錯誤......哪裏$dbc來從？！你會想要使用：

$connection->query(...);