我正在bootstrap上创建一个表，从同一数据库中的两个表获取该表中的数据，这是我的数据库的图片
这是我用来从数据库获取值的代码
<table class="table table-hover table-striped">
<thead>
<tr>
<th>Id</th>
<th>File-name</th>
<th>Purpose</th>
<th>Recieved-By </th>
<th>Processed-By</th>
<th>Address</th>
<th>Contact-No</th>
<th>Recieved-Date</th>
<th>Update-date</th>
<th>status</th>
</tr>
</thead>
<tbody>
<?php
$p_query= "SELECT id, file_name, recieved_by, processed_by, purpose, address, contact_no, date,update_date,reason FROM files INNER JOIN update_table ON files.id=update_table.id";
$con=mysqli_connect("localhost","root","","fileprogramsysteeem");
$p_run=mysqli_query($con,$p_query);
if(mysqli_num_rows($p_run)){
while($row=mysqli_fetch_array($p_run))
{
$id=$row['id'];
$file=$row['file_name'];
$purpose=$row['purpose'];
$recieve=$row['recieved_by'];
$processed=$row['processed_by'];
$address=$row['address'];
$contact=$row['contact_no'];
$date=$row['date'];
$up_date=$row['update_date'];
$status=$row['reason'];
?>
<tr>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $id;?></a></td>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $file;?></a></td>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $purpose;?></a></td>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $recieve;?></a></td>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $processed;?></a></td>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $address;?></a></td>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $contact;?></a></td>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $date;?></a></td>
<td><a href="post.php?post_id=<?php echo $id?>"><?php echo $up_date;?></a></td>
<td style="color:#337AB7;"><?php echo $id?>"><?php echo $status;?></td>
</tr>
<?php
}
}
?>
</tbody>
</table>
现在应用此代码后我收到此错误:
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result
我必须从表文件中选择file_name、receive_by、processed_by、目的、地址、contact_no、日期以及update_date、原因表update_table
最佳答案
$p_query= "SELECT files.* , update_table.* FROM files INNER JOIN update_table ON files.id=update_table.id";
$con=mysqli_connect("localhost","root","","fileprogramsysteeem");
试试这个可能会有帮助
关于php - 使用基于id的join子句从两个表中获取，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/46969814/