警告：mysqli_fetch_array（）期望参数1为mysqli_result，布尔在E：\ wamp \ www \ Demo 22 \ gallery.php中给出404行？

Question

我给页面中的图像显示PHP代码，而我没有在数据库中，我有发生所谓的这个错误

警告插入图片：mysqli_fetch_array（）预计参数1被mysqli_result，布尔在E中给出： \ wamp \ www \ Demo 22 \ gallery.php在404行上？

这是代码：

<?php 
include("includes/connect.php"); 

$select_posts = "select * from album1 where post_id=1"; 

$run_posts = mysqli_query($con,$select_posts); 


while($row=mysqli_fetch_array($run_posts)){ 

    $post_title = $row['post_title']; 
    $post_name = $row['post_name']; 
    $post_image = $row['post_image'];`enter code here` 
?> 

<p><?php echo $post_title; ?></p> 
<a href="gallery1.php"><img class="gallery1" src="image/album1/<?php echo $post_image; ?>"width="160" height="120" ></a> 
<p><?php echo $post_name; ?></p> 

<?php } ?> 

Answer 1

这意味着你以前到mysqli_query调用返回false，而不是参照结果集。这发生在SQL中有语法错误时。尝试手动运行SQL或通过调用echo mysqli_error($con)来报告错误。