Php 如何从两个表中进行选择,并从一个表中仅显示一个项目,从第二个表中显示多个项目?
我有两个表user和Image table,我只想在Image表中选择profile图片,在user表中我可以选择3个字段。我尝试使用内部连接,但我看不到任何图像显示和没有错误。下面是我的密码Php 如何从两个表中进行选择,并从一个表中仅显示一个项目,从第二个表中显示多个项目?,php,mysql,Php,Mysql,我有两个表user和Image table,我只想在Image表中选择profile图片,在user表中我可以选择3个字段。我尝试使用内部连接,但我看不到任何图像显示和没有错误。下面是我的密码 <?Php $target = "image_uploads/"; $image_name = (isset($_POST['image_name'])); $query ="select * from tish_user inner join tish_images on tish_user.
<?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;
}
?>
我看不出上面的代码有什么问题,所以请检查一些东西
请提供带有一些示例数据的表结构以及您的期望值result@FathahRehmanP好的,让我们简单一点,我希望用户名来自tish_用户,图像名称来自tish_图像这是要显示的图像如果您提供两个表的创建表脚本,那么它将非常有用useful@FathahRehmanP