2016-03-06 97 views
0

我有两个表博客和tbl_users。博客包含专栏'blogid','消息','标题','名称','创建'&'image_ctgy'。 tbl_users包含列'userPic','image_ctgy','userId'。下面的代码基于相同的image_ctgy显示博客消息和与该博客消息相关联的图像。然而,我对PHP的知识是有限的,我很难显示标题,名称&创建列以及博客消息。任何帮助,将不胜感激。PHP sql左连接表从表1回声列

$sql = "SELECT blog.message as blogmessage, tbl_users.userPic as tblPics 
FROM blog 
LEFT JOIN tbl_users ON(blog.image_ctgy = tbl_users.image_ctgy) 
ORDER by blog.blogid DESC"; 
$stmt = $db->prepare($sql); 
$stmt->execute(); 

$blogs = array(); 
while (($row = $stmt->fetch(PDO::FETCH_ASSOC))) 
{ 
$blogmessage = $row['blogmessage']; 
$blogs[$blogmessage] [] = $row['tblPics']; 
} 
?> 

<html> 
<body> 
<?php 
    foreach ($blogs as $blogmessage => $tblPics) 
    { 
?> 
    <h2><?php echo $blogmessage; ?></h2> 

    <ul> 
<?php 
    foreach ($tblPics as $tblPic) 
    { 
?> 
     <li> <img src="cms3/user_images/<?php echo $tblPic; ?>"width="50px" height="50px"/></li> 

<?php 
    } 
?> 
    </ul> 
<?php 
} 
?> 
</body> 
</html> 

回答

1

您可以使用此代码来获取领域爵位,姓氏,具有消息和userPic一起创建:

<?php 
$sql = "SELECT blog.message as blogmessage, blog.title, blog.name, blog.created, tbl_users.userPic as tblPics 
FROM blog 
LEFT JOIN tbl_users ON(blog.image_ctgy = tbl_users.image_ctgy) 
ORDER by blog.blogid DESC"; 
$stmt = $db->prepare($sql); 
$stmt->execute(); 

$blogs = array(); 
while (($row = $stmt->fetch(PDO::FETCH_ASSOC))) 
{ 
    // here you can get along with blogmessage another desired columns: 
    // title, name, created 
$blogmessage = $row['blogmessage']; 
$blogs[$blogmessage] [] = $row['tblPics']; 
} 
?> 

<html> 
<body> 
<?php 
foreach ($blogs as $blogmessage => $tblPics) 
{ 
    ?> 
    <h2><?php echo $blogmessage; ?></h2> 

    <ul> 
     <?php 
     foreach ($tblPics as $tblPic) 
     { 
      ?> 
      <li> <img src="cms3/user_images/<?php echo $tblPic; ?>"width="50px" height="50px"/></li> 

      <?php 
     } 
     ?> 
    </ul> 
    <?php 
} 
?> 
</body> 
</html> 
+0

感谢您的建议 - 但没有能够把标题,姓名和创建(日期)成功。你能提供建议的编码吗?我已经在stackoverflow上进行了大量搜索,但还没有找到解决方案。 – Lee