2014-04-10 20 views
0

將不勝感激,如果有人能幫我找出這個錯誤。在我的第一頁上,我們稱它爲「index.php」,我的身體開始處有代碼連接到我的數據庫,沒有問題。進行連接並在頁面上顯示數據庫中的變量。我有一個使用相當一致的代碼,用於連接另一個頁面,但我在我的網頁上收到此錯誤:PHP連接問題在一個頁面上,但不是以前(相同的連接代碼)

警告:請求mysql_query()[function.mysql查詢]:無法通過連接到本地MySQL服務器在/var/sites/s/ssangar.com/public_html/PMWebsite/projects.php在線36上的套接字'/var/lib/mysql/mysql.sock'(2)

警告:mysql_query()[函數。 mysql-query]:無法在/var/sites/s/ssangar.com/public_html/PMWebsite/projects.php上建立到服務器的鏈接36行

警告:mysql_numrows()期望參數1是資源,布爾在/ var/sites/s/ssangar.com/public_html/PMWebsite/projects.php上線37

我的PHP代碼:

<?php session_start(); ?> 
<!DOCTYPE html> 
<html lang="en"> 
<head> 
... 
</head> 

<body> 

    <?php 
    $connect = mysqli_connect("server", "username", "password", "database"); 
    $user = $_SESSION['username']; 
    $query="SELECT * FROM Projects"; 
    $result=mysql_query($query); 
    $num=mysql_numrows($result); 
    ?> 



... 

      <?php 
      while($row = mysqli_fetch_array($result)){ 
      echo "<div class='col-xs-6 col-sm-3 placeholder'>"; 
      echo "<h4>". $row['title'] . "</h4><br/>"; 
      echo "". $row['description'] ."<br/>"; 
      echo "Due: ". $row['deadline'] .""; 
      echo "<span class='text-muted'></span>"; 
      echo " </div>"; 
      } 
      mysqli_close($connect); 
      ?> 

     </div> 

</body> 
</html> 

我還收到錯誤爲在底部附近,顯示數據庫中的內容的代碼:

警告:mysqli_fetch_array()預計參數1被mysqli_result,布爾線在給定/var/sites/s/ssangar.com/public_html/PMWebsite/projects.php 83

回答

0

通過擺脫$ num = ....行,解決了它的問題!

新代碼:$connect = mysqli_connect("server", "username", "pw", ""); $user = $_SESSION['username']; $query = "SELECT * FROM Projects"; $result = mysqli_query($connect, $query);