我在MySql中创建了两个表,一个表保存了博客内容,另一个表保存了博客ID和不同的图片路径。我在一个postID中保存了多张图片。现在,我试图同时获取图片和博客文章,但无法正常工作。下面的代码...

blogposts.php

<?php
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($conn,"SELECT  * , upload_data.FILE_NAME from blog_posts LEFT JOIN upload_data ON blog_posts.postID=upload_data.postID  ; ");
while($row = mysqli_fetch_array($result))
{?>

  <div class="b-slider j-smallslider" data-height="382">
                    <ul><?php
                    if (!empty($FILE_NAME)) {
                   $result = mysqli_query($conn,"SELECT  * from upload_data where postID=postID ; ");
                while($row = mysqli_fetch_array($result))

                    {?>
                            <li data-transition="3dcurtain-vertical" data-slotamount="12">
                            <img data-retina src="articles/user_data/<?php echo( htmlspecialchars( $row['FILE_NAME'] ) ); ?>">
                        </li> <?php }} ?>
                     </ul>
                </div>
 <div class="b-blog-one-column__info">
        Title : <a href="#" class="f-more"><?php echo( htmlspecialchars( $row['postTitle'] ) ); ?></a>, <a href="#" class="f-more"><?php echo( htmlspecialchars( $row['postCat'] ) ); ?></a>
        <span class="b-blog-one-column__info_delimiter"></span>
        Tag : <a href="#" class="f-more">Nllam</a>
        <span class="b-blog-one-column__info_delimiter"></span>
        <a href="#" class="f-more f-primary"><i class="fa fa-comment"></i>12 Comments</a>
    </div>
<?php } ?>


问题是我无法获取照片。 :(

最佳答案

使用此查询

$result = mysqli_query($conn,"SELECT  blog_posts.* , upload_data.FILE_NAME from blog_posts,upload_data WHERE blog_posts.postID=upload_data.postID");


然后使用print_r()函数打印结果,并检查其获取结果与否。

10-05 21:36