這是我的代碼MySQL的取回陣列布爾錯誤
<?php
$result = mysql_query("SELECT * FROM post WHERE username = ".$username." ORDER BY ID DESC ");
while($row = mysql_fetch_array($result)){
?>
<div class="post">
<a href="/p/<?php echo $row['ID']; ?>" class="post-title"><?php echo $row['title']; ?> - (Rating: <?php echo $row['rank']; ?>)</a>
<p class="post-content"><?php echo $row['description']; ?><br /><br />On <?php echo $row['date']; ?></p>
</div>
<?php }; ?>
但我得到這個錯誤:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /*/programs/user.php on line 77
檢查查詢是否正確。 – Dev
它是,當我運行搜索PHPMyAdmin並顯示PHP代碼,這是我得到:$ sql =「SELECT * FROM'vb_posts' WHERE'username' = \'joshblease \'ORDER BY'ID' DESC」 ; –
在$ username上使用strip斜槓,在PHP之外運行查詢,並檢查它是否返回一些東西。 – Dev