mysql_fetch_array（）＆mysql_free_result（）錯誤

Question

我正在設計一個網站，它允許用戶註冊和登錄 - 登錄後，用戶會看到其他鏈接（僅對註冊用戶可見）。其中一個環節包括， 「我的書籤」 - 「我的書籤」（的index.php）的代碼如下：

我的書籤（的index.php）代碼：

<?php 
    session_start(); 
    //check session first 
    if (!isset($_SESSION['email'])){ 
    echo "You are not logged in!"; 
     exit(); 
    } 
    else{ 
    //include the header 
    include ("../includes/header.php"); 
    require_once ('../../mysql_connect.php'); 
    echo ("<center>"); 
    echo ("<h3>Bookmark</h3><p>"); 
    echo ("<a href=add.php>Add a record</a><p>"); 
    echo ("<a href=searchform.php>Search records</a><p>"); 
    //Set the number of records to display per page 
    $display = 5; 
    //Check if the number of required pages has been determined 
    if(isset($_GET['p'])&&is_numeric($_GET['p'])){//Already been determined 
    $pages = $_GET['p']; 
    } 
    else{//Need to determine 
    //Count the number of records; 
    $query = "SELECT COUNT(id) FROM bookmark"; 
    $result = @mysql_query($query); 
    $row = @mysql_fetch_array($result, MYSQL_NUM); 
    $records = $row[0]; //get the number of records 
    //Calculate the number of pages ... 
    if($records > $display){//More than 1 page is needed 
     $pages = ceil($records/$display); 
    }else{ 
     $pages = 1; 
    } 
}// End of p IF. 

//Determine where in the database to start returning results ... 
if(isset($_GET['s'])&&is_numeric($_GET['s'])){ 
    $start = $_GET['s']; 
}else{ 
    $start = 0; 
} 

//Make the paginated query; 
$query = "SELECT * FROM bookmark LIMIT $start, $display"; 
$result = @mysql_query ($query); 

//Table header: 
echo "<table cellpadding=5 cellspacing=5 border=1><tr> 
<th>Title</th><th>Comment</th><th>URL</th><th>*</th><th>*</th></tr>"; 

//Fetch and print all the records... 
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { 
    echo "<tr><td>".$row['title']."</td>"; 
    echo "<td>".$row['comment']."</td>"; 
    echo "<td><a href=".$row['url']." target=_blank>".$row['url']."</a></td>"; 
    echo "<td><a href=deleteconfirm.php?id=".$row['id'].">Delete</a></td>"; 
    echo "<td><a href=updateform.php?id=".$row['id'].">Update</a></td></tr>"; 
} // End of While statement 
echo "</table>"; 
mysql_free_result ($result); // Free up the resources.   
mysql_close(); // Close the database connection. 

//Make the links to other pages if necessary. 
if($pages>1){ 
    echo '<br/><table><tr>'; 
    //Determine what page the script is on: 
    $current_page = ($start/$display) + 1; 
    //If it is not the first page, make a Previous button: 
    if($current_page != 1){ 
     echo '<td><a href="index.php?s='. ($start - $display) . '&p=' . $pages. '"> Previous </a></td>'; 
    } 
    //Make all the numbered pages: 
    for($i = 1; $i <= $pages; $i++){ 
     if($i != $current_page){ // if not the current pages, generates links to that page 
      echo '<td><a href="index.php?s='. (($display*($i-1))). '&p=' . $pages .'"> ' . $i . ' </a></td>'; 
     }else{ // if current page, print the page number 
      echo '<td>'. $i. '</td>'; 
     } 
    } //End of FOR loop 
    //If it is not the last page, make a Next button: 
    if($current_page != $pages){ 
     echo '<td><a href="index.php?s=' .($start + $display). '&p='. $pages. '"> Next </a></td>'; 
    } 

    echo '</tr></table>'; //Close the table. 
}//End of pages links 
//include the footer 
include ("../includes/footer.php"); 
} 
?> 

對不起格式問題，我還沒有完全想出如何有效地複製和粘貼代碼。

誤區一：

警告：

不管怎麼說，我的問題是由兩個錯誤，我接受，這是派生mysql_fetch_array（）預計參數1是資源，布爾在給定的上線50 /home/kethcart/public_html/440/finalproject/htdocs/bookmark/index.php

誤區之二：

警告：了mysql_free_result（）預計參數1是資源，在布爾給出/home/kethcart/public_html/440/finalproject/htdocs/bookmark/index.php在線路58

我明白關於這個話題已經有了足夠多的問題，但我似乎無法將它們應用於我的情況。如果有人可以提供關於如何解決這個問題的建議，甚至可以提供關於另一個問題或解決方案的鏈接，那就太棒了！

我很感謝大家的幫助和耐心 - 我不知道沒有這個社區我會做什麼。

感謝，

Rockmandew

Answer 1

快速的答案。您收到錯誤是因爲您的查詢出了問題。在使用mysql_query時，請務必注意，如果查詢失敗，您將得到BOOLEAN返回FALSE。因此，爲什麼你被告知... expects parameter 1 to be resource, boolean given...

要避免此類問題，請嘗試使用die語句來獲得可能錯誤的「反饋」。喜歡的東西：

$query = "SELECT COUNT(id) FROM bookmark"; 
$result = mysql_query($query) or die ("Error in query: $query. <br />".mysql_error()); 

參見：

• info about your @ symbol
• http://us3.php.net/
• http://us3.php.net/die || http://us3.php.net/manual/en/function.exit.php
• http://us3.php.net/mysql_query