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我有两个表用户和图像表我只想选择图像表和用户表中的个人档案图片,我可以选择3个字段。我尝试使用内部连接,但我看不到任何图像显示,并没有错误。波纹管是我的代码如何从两个表格中进行选择,并只显示一个表格中的一个项目和第二个表格中的多个项目?
<?Php
$target = "image_uploads/";
$image_name = (isset($_POST['image_name']));
$query ="select * from
tish_user inner join tish_images
on tish_user.user_id = tish_images.user_id";
$result= $con->prepare($query);
$result->execute();
$table = <<<ENDHTML
<div style ="text-align:center;">
<h2>Client Review Software</h2>
<table id ="heredoc" border ="0" cellpaddinig="2" cellspacing="2" style = "width:100%" ;
margin-left:auto; margin-right: auto;>
<tr>
<th>Name</th>
<th>Last Name</th>
<th>Ref No</th>
<th>Cell</th>
<th>Picture</th>
</tr>
ENDHTML;
while($row = $result->fetch(PDO::FETCH_ASSOC)){
$date_created = $row['date_created'];
$user_id = $row['user_id'];
$username = $row['username'];
$image_id = $row['image_id'];
#this is the Tannery operator to replace a pic when an id do not have one
$photo = ($row['image_name']== null)? "me.png":$row['image_name'];
#display image
$table .= <<<ENDINFO
<tr>
<td><a href ="client_details.php?user_id=$user_id">$username </a></td>
<td>$image_id</td>
<td></td>
<td>c</td>
<td><img src="'.$target.$photo.'" width="100" height="100">
</td>
</tr>
ENDINFO;
}
?>
请提供表结构,一些样本的数据和预期的结果 – 2013-02-21 14:26:16
@FathahRehmanP确定可以使简单我想从tish_images tish_user和IMAGE_NAME用户名,这是要显示 – humphrey 2013-02-21 14:31:29
如果您提供创建的表脚本形象这两个表,那将是非常有用的 – 2013-02-21 14:36:32